1981 AMC 12 真题（P版）

1 答案 E

由 \(\sqrt{x+2}=2\) 可得 \(x+2=4\)。因此 \((x+2)^2=4^2=16\)。

所以正确答案是 E。

English

From \(\sqrt{x+2}=2,\) we get \(x+2=4.\) Therefore \((x+2)^2=4^2=16.\)

Therefore, the correct answer is E.

2 答案 C

设正方形的边长为 \(s\)。在直角三角形 \(EBC\) 中，\[ s^2+1^2=2^2\text{。}\]因此 \(s^2=3\)，这恰好就是正方形的面积。

所以正确答案是 C。

English

Let the square’s side length be \(s.\) In right triangle \(EBC,\) \[ s^2+1^2=2^2. \] Hence \(s^2=3,\) which is exactly the area of the square.

Therefore, the correct answer is C.

3 答案 D

通分到公分母 \(6x\)，得到 \[ \begin{aligned} \frac1x+\frac1{2x}+\frac1{3x} &=\frac6{6x}+\frac3{6x}\\ &\quad+\frac2{6x}\\ &=\frac{11}{6x} \end{aligned}\text{。}\]

所以正确答案是 D。

English

Using the common denominator \(6x\) gives \[ \begin{aligned} \frac1x+\frac1{2x}+\frac1{3x} &=\frac6{6x}+\frac3{6x}\\ &\quad+\frac2{6x}\\ &=\frac{11}{6x}. \end{aligned} \]

Therefore, the correct answer is D.

4 答案 C

关系式 \(3L=4S\) 给出 \(S=\frac{3L}{4}\)。因此 \[ L-S=L-\frac{3L}{4}=\frac L4=8\text{，}\]所以 \(L=32\)。

所以正确答案是 C。

English

The relation \(3L=4S\) gives \(S=\frac{3L}{4}.\) Thus \[ L-S=L-\frac{3L}{4}=\frac L4=8, \] so \(L=32.\)

Therefore, the correct answer is C.

5 答案 C

由于 \(AB\parallel CD\)，同旁内角互补，所以 \(\angle ABC=180^\circ-110^\circ=70^\circ\)。因此 \(\angle ABD=70^\circ-30^\circ=40^\circ\)。又因 \(AD=BD\)，三角形 \(ABD\) 中 \(\angle DAB=\angle ABD=40^\circ\)。所以 \[ \begin{aligned} \angle ADB &=180^\circ-40^\circ-40^\circ\\ &=100^\circ \end{aligned}\text{。}\]

所以正确答案是 C。

English

Since \(AB\parallel CD,\) consecutive interior angles give \(\angle ABC=180^\circ-110^\circ=70^\circ.\) Hence \(\angle ABD=70^\circ-30^\circ=40^\circ.\) Because \(AD=BD,\) triangle \(ABD\) has \(\angle DAB=\angle ABD=40^\circ.\) Therefore \[ \begin{aligned} \angle ADB &=180^\circ-40^\circ-40^\circ\\ &=100^\circ. \end{aligned} \]

Therefore, the correct answer is C.

6 答案 A

令 \(A=y^2+2y-1\)。原方程变为 \(\frac{x}{x-1}=\frac{A}{A-1}\)。交叉相乘得 \(xA-x=Ax-A\)，所以 \(x=A=y^2+2y-1\)。

所以正确答案是 A。

English

Let \(A=y^2+2y-1.\) The equation becomes \(\frac{x}{x-1}=\frac{A}{A-1}.\) Cross-multiplication gives \(xA-x=Ax-A,\) so \(x=A=y^2+2y-1.\)

Therefore, the correct answer is A.

7 答案 B

最小公倍数是 \(\operatorname{lcm}(2,3,4,5)=60\)。在不超过 \(100\) 的正整数中，只有 \(60\) 是 \(60\) 的倍数。因此个数为 \(1\)。

所以正确答案是 B。

English

The least common multiple is \(\operatorname{lcm}(2,3,4,5)=60.\) Among the positive integers through \(100,\) only \(60\) is a multiple of \(60.\) Thus the count is \(1.\)

Therefore, the correct answer is B.

8 答案 A

有 \[ x^{-1}+y^{-1}+z^{-1}=\frac{xy+yz+zx}{xyz} \] 以及 \[ \begin{aligned} &(xy)^{-1}+(yz)^{-1}\\ &\quad+(zx)^{-1} =\frac{x+y+z}{xyz} \end{aligned}\text{。}\]所有对称和因子约去后，剩下 \(\frac{1}{x^2y^2z^2}=x^{-2}y^{-2}z^{-2}\)。

所以正确答案是 A。

English

We have \[ x^{-1}+y^{-1}+z^{-1}=\frac{xy+yz+zx}{xyz} \] and \[ \begin{aligned} &(xy)^{-1}+(yz)^{-1}\\ &\quad+(zx)^{-1} =\frac{x+y+z}{xyz}. \end{aligned} \] All symmetric sum factors cancel, leaving \(\frac{1}{x^2y^2z^2}=x^{-2}y^{-2}z^{-2}.\)

Therefore, the correct answer is A.

9 答案 A

若边长为 \(s\)，则空间对角线满足 \(s\sqrt3=a\)，所以 \(s^2=\frac{a^2}{3}\)。表面积为 \(6s^2=6(\frac{a^2}{3})=2a^2\)。

所以正确答案是 A。

English

If the side length is \(s,\) then the space diagonal is \(s\sqrt3=a,\) so \(s^2=\frac{a^2}{3}.\) The surface area is \(6s^2=6(\frac{a^2}{3})=2a^2.\)

Therefore, the correct answer is A.

10 答案 E

将 \(x\) 与 \(y\) 互换，原式 \(y=ax+b\) 变为 \(x=ay+b\)。解得 \[ y=\frac{x-b}{a}=\frac1a x-\frac ba\text{。}\]

所以正确答案是 E。

English

Interchanging \(x\) and \(y\) in \(y=ax+b\) gives \(x=ay+b.\) Solving, \[ y=\frac{x-b}{a}=\frac1a x-\frac ba. \]

Therefore, the correct answer is E.

11 答案 C

若三边为 \(s-d\)、\(s\)、\(s+d\)，则勾股定理给出 \((s-d)^2+s^2=(s+d)^2\)，从而有 \(s=4k\) 且 \(d=k\)，其中 \(k\) 为正整数。因此三边为 \(3k\)、\(4k\)、\(5k\)。在选项中，\(81=3\cdot27\) 可以作为一条边。

所以正确答案是 C。

English

If the sides are \(s-d,\) \(s,\) \(s+d,\) the Pythagorean equation gives \((s-d)^2+s^2=(s+d)^2,\) hence \(s=4k\) and \(d=k\) for some positive integer \(k.\) Thus the sides are \(3k,\) \(4k,\) \(5k.\) Of the choices, \(81=3\cdot27\) can occur.

Therefore, the correct answer is C.

12 答案 E

最终数额为 \[ M\left(1+\frac p{100}\right) \left(1-\frac q{100}\right)\text{。}\]它大于 \(M\) 当且仅当 \((100+p)(100-q)\gt10000\)。展开后得到 \(p(100-q)\gt100q\)。由于 \(100-q\gt0\)，这等价于 \(p\gt\frac{100q}{100-q}\)。

所以正确答案是 E。

English

The final amount is \[ M\left(1+\frac p{100}\right) \left(1-\frac q{100}\right). \] It exceeds \(M\) exactly when \((100+p)(100-q)\gt10000.\) Expanding gives \(p(100-q)\gt100q.\) Since \(100-q\gt0,\) this is \(p\gt\frac{100q}{100-q}.\)

Therefore, the correct answer is E.

13 答案 E

至少损失 \(90\%\) 意味着 \(0.9^n\le0.1\)。现在 \[ \begin{aligned} \log_{10}0.9 &=\log_{10}9-1\\ &=2(0.477)-1\\ &=-0.046 \end{aligned}\text{。}\]因此 \(n(-0.046)\le-1\)，所以 \(n\ge21.739\ldots\)。最小的整数是 \(22\)。

所以正确答案是 E。

English

At least \(90\%\) lost means \(0.9^n\le0.1.\) Now \[ \begin{aligned} \log_{10}0.9 &=\log_{10}9-1\\ &=2(0.477)-1\\ &=-0.046. \end{aligned} \] Thus \(n(-0.046)\le-1,\) so \(n\ge21.739\ldots.\) The least integer is \(22.\)

Therefore, the correct answer is E.

14 答案 A

若首项为 \(a\)，公比为 \(r\)，则 \[ 91=a(1+r)(1+r^2+r^4)\text{。}\]由于 \(a(1+r)=7\)，可得 \(1+r^2+r^4=13\)。令 \(t=r^2\ge0\)，则 \(t^2+t-12=0\)，所以 \(t=3\)。前四项之和为 \(a(1+r)(1+r^2)=7(4)=28\)。

所以正确答案是 A。

English

If the first term is \(a\) and ratio is \(r,\) then \[ 91=a(1+r)(1+r^2+r^4). \] Since \(a(1+r)=7,\) we have \(1+r^2+r^4=13.\) With \(t=r^2\ge0,\) \(t^2+t-12=0,\) so \(t=3.\) The first four terms sum to \(a(1+r)(1+r^2)=7(4)=28.\)

Therefore, the correct answer is A.

15 答案 B

令 \(A=\log_b2\)、\(B=\log_b3\)、\(u=\log_bx\)。对等式两边取以 \(b\) 为底的对数，得到 \(A(A+u)=B(B+u)\)。由于 \(A\ne B\)，\[ A+B+u=0\text{。}\]因此 \(u=-\log_b6=\log_b(\frac{1}{6})\)，所以 \(x=\frac{1}{6}\)。

所以正确答案是 B。

English

Let \(A=\log_b2,\) \(B=\log_b3,\) and \(u=\log_bx.\) Taking base-\(b\) logarithms gives \(A(A+u)=B(B+u).\) Since \(A\ne B,\) \[ A+B+u=0. \] Therefore \(u=-\log_b6=\log_b(\frac{1}{6}),\) so \(x=\frac{1}{6}.\)

Therefore, the correct answer is B.

16 答案 E

这个三进制表示有偶数位，因此九进制表示中最左边的一位来自 \(12_3\)。这一组的值为 \(1\cdot3+2=5\)，所以九进制表示的首位是 \(5\)。

所以正确答案是 E。

English

The representation has an even number of base-three digits, so its leftmost base-nine digit comes from \(12_3.\) That pair has value \(1\cdot3+2=5,\) so the first base-nine digit is \(5.\)

Therefore, the correct answer is E.

17 答案 B

把 \(x\) 替换为 \(\frac{1}{x}\)，得到 \(f(\frac{1}{x})+2f(x)=\frac{3}{x}\)。解这两个线性方程，得到 \[ f(x)=\frac2x-x\text{。}\]因此 \(f(-x)=-f(x)\)，要使两者相等，必须有 \(f(x)=0\)。于是 \(\frac{2}{x}-x=0\)，即 \(x^2=2\)。\(x=\sqrt2\) 与 \(x=-\sqrt2\) 都满足条件，所以恰有两个实数解。

所以正确答案是 B。

English

Replacing \(x\) by \(\frac{1}{x}\) gives \(f(\frac{1}{x})+2f(x)=\frac{3}{x}.\) Solving the two linear equations yields \[ f(x)=\frac2x-x. \] Hence \(f(-x)=-f(x),\) so equality requires \(f(x)=0.\) Thus \(\frac{2}{x}-x=0,\) or \(x^2=2.\) Both \(x=\sqrt2\) and \(x=-\sqrt2\) work, giving exactly two real numbers.

Therefore, the correct answer is B.

18 答案 C

该方程关于原点对称，并有解 \(x=0\)。在 \((0,\pi)\) 内还有一个正数解。对每个 \(k=1,2,\ldots,15\)，区间 \((2k\pi,(2k+1)\pi)\) 上的正半波先升到直线 \(\frac{x}{100}\) 上方，再降到其下方，因而产生两个解。在 \(31\pi\) 之后没有正数解，因为下一个正半波开始时已经超过 \(100\)，而 \(|\sin x|\le1\)。所以正数解有 \(1+2(15)=31\) 个，负数解同样多，再加上零本身，共有 \(31+31+1=63\) 个。

所以正确答案是 C。

English

The equation is odd-symmetric and has the solution \(x=0.\) On \((0,\pi)\) there is one further positive solution. For each \(k=1,2,\ldots,15,\) the positive sine hump on \((2k\pi,(2k+1)\pi)\) rises above \(\frac{x}{100}\) and then falls below it, giving two solutions. There are no positive solutions beyond \(31\pi\), because the next positive hump begins above \(100\) while \(|\sin x|\le1.\) Thus there are \(1+2(15)=31\) positive solutions, the same number negative, and zero itself: \(31+31+1=63.\)

Therefore, the correct answer is C.

19 答案 B

延长 \(BN\)，与 \(AC\) 相交于 \(E\)。直角三角形 \(ABN\) 与 \(AEN\) 全等，因为它们共有 \(AN\)，且该线平分顶点 \(A\) 处的角。因此 \(BE\) 被 \(N\) 平分，且 \(AE=AB=14\)。由于 \(AC=19\)，所以 \(EC=5\)。在三角形 \(BEC\) 中，点 \(N\) 和 \(M\) 分别是 \(BE\) 和 \(BC\) 的中点，因此 \(MN=\frac{EC}{2}=\frac{5}{2}\)。

所以正确答案是 B。

English

Extend \(BN\) to meet \(AC\) at \(E.\) The right triangles \(ABN\) and \(AEN\) are congruent because \(AN\) is common and bisects the angle at \(A.\) Hence \(BE\) is bisected by \(N\) and \(AE=AB=14.\) Since \(AC=19,\) \(EC=5.\) In triangle \(BEC,\) points \(N\) and \(M\) are the midpoints of \(BE\) and \(BC,\) so \(MN=\frac{EC}{2}=\frac{5}{2}.\)

Therefore, the correct answer is B.

20 答案 B

设 \(\theta\gt0\) 为光线起初与 \(AD\) 之间的锐角。连续的外角关系表明，每次反射都会使相应的锐角增加 \(8^\circ\)。最后垂直射到边界时，\[ 90^\circ=\theta+(8n+8)^\circ\text{。}\]因此 \(\theta=82^\circ-8n^\circ\gt0\)，所以 \(n\lt\frac{82}{8}=10.25\)。最大可能整数为 \(10\)，此时 \(\theta=2^\circ\)。

所以正确答案是 B。

English

Let \(\theta\gt0\) be the initial acute angle between the ray and \(AD.\) Successive exterior-angle relations increase the corresponding acute angle by \(8^\circ\) at each reflection. At the final perpendicular strike, \[ 90^\circ=\theta+(8n+8)^\circ. \] Thus \(\theta=82^\circ-8n^\circ\gt0,\) so \(n\lt\frac{82}{8}=10.25.\) The greatest integer possible is \(10,\) attained when \(\theta=2^\circ.\)

Therefore, the correct answer is B.

21 答案 D

已知条件给出 \((a+b)^2-c^2=3ab\)，所以 \[ c^2=a^2+b^2-ab\text{。}\]若 \(\theta\) 是边 \(c\) 所对的角，则余弦定理给出 \(c^2=a^2+b^2-2ab\cos\theta\)。因此 \(2ab\cos\theta=ab\)，所以 \(\cos\theta=\frac{1}{2}\)，且 \(\theta=60^\circ\)。

所以正确答案是 D。

English

The condition gives \((a+b)^2-c^2=3ab,\) so \[ c^2=a^2+b^2-ab. \] If \(\theta\) is opposite \(c,\) the law of cosines says \(c^2=a^2+b^2-2ab\cos\theta.\) Therefore \(2ab\cos\theta=ab,\) so \(\cos\theta=\frac{1}{2}\) and \(\theta=60^\circ.\)

Therefore, the correct answer is D.

22 答案 D

平行于坐标轴的直线有 \(3\cdot4^2=48\) 条。对于沿面对角线的方向，先用 \(3\) 种方法选择两个变化的坐标，再选择 \(2\) 种对角线斜率之一，并用 \(4\) 种方法固定剩余坐标，共有 \(3\cdot2\cdot4=24\) 条。最后，立方体有 \(4\) 条空间对角线。因此总数为 \(48+24+4=76\)。

所以正确答案是 D。

English

There are \(3\cdot4^2=48\) axis-parallel lines. For face-diagonal directions, choose the pair of varying coordinates in \(3\) ways, choose one of \(2\) diagonal slopes, and fix the remaining coordinate in \(4\) ways, giving \(3\cdot2\cdot4=24.\) Finally, the cube has \(4\) space diagonals. Thus the total is \(48+24+4=76.\)

Therefore, the correct answer is D.

23 答案 C

等边三角形的外接圆半径为 \(R=\frac{12}{\sqrt3}=4\sqrt3\)。若小圆半径为 \(r\)，其圆心在高上，离顶点的距离为 \(2r\)，这个顶点是 \(A\)，因为圆心到任一边的距离为 \(r\)，而半角为 \(30^\circ\)。由下端的内切关系可得 \(2r=2R-r\)，所以 \(r=\frac{2R}{3}=\frac{8\sqrt3}{3}\)。两边上的切点之间的距离为 \(PQ=r\sqrt3=8\)。

所以正确答案是 C。

English

The circumradius of the equilateral triangle is \(R=\frac{12}{\sqrt3}=4\sqrt3.\) If the smaller radius is \(r,\) its center lies on the altitude and is \(2r\) from \(A,\) because its distance to either side is \(r\) and the half-angle is \(30^\circ.\) Internal tangency at the bottom gives \(2r=2R-r,\) so \(r=\frac{2R}{3}=\frac{8\sqrt3}{3}.\) The tangency points on the two sides are separated by \(PQ=r\sqrt3=8.\)

Therefore, the correct answer is C.

24 答案 D

方程为 \(x^2-2x\cos\theta+1=0\)，其根为 \(e^{i\theta}\) 和 \(e^{-i\theta}\)。因此 \(x=e^{\pm i\theta}\)，且 \(x^{-1}=e^{\mp i\theta}\)。由棣莫弗定理，\[ \begin{aligned} x^n+x^{-n} &=e^{in\theta}+e^{-in\theta}\\ &=2\cos n\theta \end{aligned}\text{。}\]

所以正确答案是 D。

English

The equation is \(x^2-2x\cos\theta+1=0,\) whose roots are \(e^{i\theta}\) and \(e^{-i\theta}.\) Thus \(x=e^{\pm i\theta}\) and \(x^{-1}=e^{\mp i\theta}.\) By De Moivre’s theorem, \[ \begin{aligned} x^n+x^{-n} &=e^{in\theta}+e^{-in\theta}\\ &=2\cos n\theta. \end{aligned} \]

Therefore, the correct answer is D.

25 答案 A

令 \(AB=c\)、\(AD=y\)、\(AC=b\)、\(AE=z\)。由于 \(AD\) 平分 \(\angle BAE\)，且 \(AE\) 平分 \(\angle DAC\)，角平分线定理给出 \[ \frac{c}{z}=\frac{2}{3},\qquad \frac{y}{b}=\frac{3}{6}\text{，}\]所以 \(z=\frac{3c}{2}\)，且 \(b=2y\)。分别对三角形 \(ADB\)、\(ADE\) 和 \(AEC\) 应用余弦定理，并令它们共有的 \(\cos\alpha\) 值相等，得到 \[ \begin{gathered} 3c^2-2y^2=12,\\ 9c^2-8y^2=-72 \end{gathered}\text{。}\]因此 \(y^2=54\)，且 \(c^2=40\)。三边分别为 \(AB=2\sqrt{10}\)、\(BC=11\)、\(AC=2\sqrt{54}=6\sqrt6\)，所以最短边为 \(2\sqrt{10}\)。

所以正确答案是 A。

English

Let \(AB=c,\) \(AD=y,\) \(AC=b,\) and \(AE=z.\) Since \(AD\) bisects \(\angle BAE,\) and \(AE\) bisects \(\angle DAC,\) the angle-bisector theorem gives \[ \frac{c}{z}=\frac{2}{3},\qquad \frac{y}{b}=\frac{3}{6}, \] so \(z=\frac{3c}{2}\) and \(b=2y.\) Applying the law of cosines to triangles \(ADB,\) \(ADE,\) and \(AEC\) and equating their common \(\cos\alpha\) values gives \[ \begin{gathered} 3c^2-2y^2=12,\\ 9c^2-8y^2=-72. \end{gathered} \] Hence \(y^2=54\) and \(c^2=40.\) The sides are \(AB=2\sqrt{10},\) \(BC=11,\) and \(AC=2\sqrt{54}=6\sqrt6,\) so the shortest is \(2\sqrt{10}.\)

Therefore, the correct answer is A.

26 答案 D

卡萝尔在第一轮获胜的概率为 \((\frac{5}{6})^2(\frac{1}{6})=\frac{25}{216}\)。完整一轮都没有出现六点的概率为 \((\frac{5}{6})^3=\frac{125}{216}\)。因此所求的等比级数为 \[ \frac{\frac{25}{216}}{1-\frac{125}{216}} =\frac{25}{91}\text{。}\]

所以正确答案是 D。

English

Carol wins in the first round with probability \((\frac{5}{6})^2(\frac{1}{6})=\frac{25}{216}.\) A complete round with no six has probability \((\frac{5}{6})^3=\frac{125}{216}.\) Therefore the desired geometric series is \[ \frac{\frac{25}{216}}{1-\frac{125}{216}} =\frac{25}{91}. \]

Therefore, the correct answer is D.

27 答案 C

缩放使 \(AB=AC=DG=1\)。等弦对应的弧相等，由此可知三角形 \(DEC\) 是 \(30^\circ\)-\(60^\circ\)-\(90^\circ\) 三角形，并且 \(AE=DE\)。令 \(AE=DE=x\)。则 \(CE=1-x=\frac{2x}{\sqrt3}\)，所以 \(x=2\sqrt3-3\)。等弦还给出 \(AF=FG=EF=\frac{1-x}{2}\)。因此 \[ \begin{aligned} \frac{[AFE]}{[ABC]} &=\frac{x(1-x)}2\\ &=7\sqrt3-12 \end{aligned}\text{。}\]

所以正确答案是 C。

English

Scale so \(AB=AC=DG=1.\) The equal-chord arc relations show that triangle \(DEC\) is \(30^\circ\)-\(60^\circ\)-\(90^\circ\) and \(AE=DE.\) Put \(AE=DE=x.\) Then \(CE=1-x=\frac{2x}{\sqrt3},\) so \(x=2\sqrt3-3.\) Equal chords also give \(AF=FG=EF=\frac{1-x}{2}.\) Hence \[ \begin{aligned} \frac{[AFE]}{[ABC]} &=\frac{x(1-x)}2\\ &=7\sqrt3-12. \end{aligned} \]

Therefore, the correct answer is C.

28 答案 D

将一个允许的多项式记为 \(g\)。当 \(x\ge0\) 时，\[ \begin{aligned} g(x)&=x^3+a_2x^2\\ &\quad+a_1x+a_0\\ &\ge f(x) \end{aligned}\text{，}\]其中 \(f(x)=x^3-2x^2-2x-2\)。因此，任何正根都不会超过最大根 \(\rho\)（即 \(f\) 的最大根）。取 \(a_2=a_1=a_0=-2\) 时 \(g=f\)，所以 \(\rho\) 本身可以取到。由于 \[ \begin{gathered} f\left(\frac52\right)=-\frac{31}{8}\lt0,\\ f(3)=1\gt0 \end{gathered}\text{，}\]可得 \(\frac{5}{2}\lt\rho\lt3\)。

所以正确答案是 D。

English

Write an allowed polynomial as \(g.\) For \(x\ge0,\) \[ \begin{aligned} g(x)&=x^3+a_2x^2\\ &\quad+a_1x+a_0\\ &\ge f(x), \end{aligned} \] where \(f(x)=x^3-2x^2-2x-2.\) Thus no positive root can exceed the largest root \(\rho\) of \(f.\) Taking \(a_2=a_1=a_0=-2\) gives \(g=f,\) so \(\rho\) itself is attained. Since \[ \begin{gathered} f\left(\frac52\right)=-\frac{31}{8}\lt0,\\ f(3)=1\gt0, \end{gathered} \] we have \(\frac{5}{2}\lt\rho\lt3.\)

Therefore, the correct answer is D.

29 答案 E

由于 \(x\ge0\)，平方一次得到 \(a-\sqrt{a+x}=x^2\)。令 \(y=\sqrt{a+x}\)。则 \(y=a-x^2\)，且 \(a=y^2-x\)，所以 \[ 0=(y+x)(y-x-1)\text{。}\]由于 \(y+x\gt0\)，可得 \(y=x+1\)，即 \(\sqrt{a+x}=x+1\)。因此 \(a+x=x^2+2x+1\)，且 \[ x^2+x+1-a=0\text{。}\]唯一的非负根为 \(x=\frac{\sqrt{4a-3}-1}{2}\)，并且它满足原方程。因此它也就是所有实数解之和。

所以正确答案是 E。

English

Since \(x\ge0,\) squaring once gives \(a-\sqrt{a+x}=x^2.\) Let \(y=\sqrt{a+x}.\) Then \(y=a-x^2\) and \(a=y^2-x,\) so \[ 0=(y+x)(y-x-1). \] Since \(y+x\gt0,\) we have \(y=x+1,\) or \(\sqrt{a+x}=x+1.\) Therefore \(a+x=x^2+2x+1\) and \[ x^2+x+1-a=0. \] The only nonnegative root is \(x=\frac{\sqrt{4a-3}-1}{2},\) and it satisfies the original equation. It is therefore also the sum of all real solutions.

Therefore, the correct answer is E.

30 答案 D

韦达定理给出 \(a+b+c+d=0\)。例如，\[ \frac{a+b+c}{d^2}=-\frac1d\text{，}\]类似地，新方程的根为 \(-\frac{1}{a}\)、\(-\frac{1}{b}\)、\(-\frac{1}{c}\)、\(-\frac{1}{d}\)。令 \(x=-\frac{1}{r}\)，并代入 \(r^4-br-3=0\)。两边乘以 \(x^4\)，得到 \(1+bx^3-3x^4=0\)，即 \[ 3x^4-bx^3-1=0\text{。}\]

所以正确答案是 D。

English

Vieta’s formulas give \(a+b+c+d=0.\) Thus, for example, \[ \frac{a+b+c}{d^2}=-\frac1d, \] and similarly the new roots are \(-\frac{1}{a},\) \(-\frac{1}{b},\) \(-\frac{1}{c},\) \(-\frac{1}{d}.\) Put \(x=-\frac{1}{r}\) in \(r^4-br-3=0.\) Multiplying by \(x^4\) gives \(1+bx^3-3x^4=0,\) or \[ 3x^4-bx^3-1=0. \]

Therefore, the correct answer is D.