题目内容
20.解方程组:$\left\{\begin{array}{l}{{x}^{2}+{y}^{2}=10}\\{{x}^{2}-3xy+2{y}^{2}=0}\end{array}\right.$.分析 由x2-3xy+2y2=0可得x=y或x=2y.分别代入x2+y2=10解出即可.
解答 解:$\left\{\begin{array}{l}{{x}^{2}+{y}^{2}=10}&{①}\\{{x}^{2}-3xy+{2y}^{2}=0}&{②}\end{array}\right.$,
由②可得x=y或x=2y.
把x=y代入①可得y2=5,解得y=$±\sqrt{5}$.
∴$\left\{\begin{array}{l}{x=\sqrt{5}}\\{y=\sqrt{5}}\end{array}\right.$,$\left\{\begin{array}{l}{x=-\sqrt{5}}\\{y=-\sqrt{5}}\end{array}\right.$.
把x=2y代入①可得y2=2,解得y=±$\sqrt{2}$.
∴$\left\{\begin{array}{l}{x=2\sqrt{2}}\\{y=\sqrt{2}}\end{array}\right.$,$\left\{\begin{array}{l}{x=-2\sqrt{2}}\\{y=-\sqrt{2}}\end{array}\right.$.
综上可得原方程组的解为:$\left\{\begin{array}{l}{x=\sqrt{5}}\\{y=\sqrt{5}}\end{array}\right.$,$\left\{\begin{array}{l}{x=-\sqrt{5}}\\{y=-\sqrt{5}}\end{array}\right.$,$\left\{\begin{array}{l}{x=2\sqrt{2}}\\{y=\sqrt{2}}\end{array}\right.$,$\left\{\begin{array}{l}{x=-2\sqrt{2}}\\{y=-\sqrt{2}}\end{array}\right.$.
点评 本题考查了“代入消元法”解方程组,考查了计算能力,属于中档题.
A. | 0∈{(0,1)} | B. | 1∈{(0,1)} | C. | (0,1)∈{(0,1)} | D. | (0,1)∈{0,1} |
A. | $\frac{a}{p}$[(1+p)10-(1+p)] | B. | $\frac{a}{p}$[(1+p)9-1] | C. | $\frac{a}{p}$[(1+p)9-(1+p)] | D. | $\frac{a}{p}$[(1+p)8-1] |
A. | y=$\sqrt{{x}^{2}+1}$ | B. | y=2x+1 | C. | y=x2+x+1 | D. | y=$\frac{1}{\sqrt{{x}^{2}-1}}$ |