【解析】设$$ x = \sqrt { x - 3 } $$ $$ y = \sqrt { y + 1 } $$ 则原方程变为$$ x + y = \frac { 1 } { 2 } ( x ^ { 2 } + y ^ { 2 } + 2 ) $$ $$ 2 x + 2 y = x ^ { 2 } + y ^ { 2 } + 2 $$ $$ x ^ { 2 } + y ^ {...
【题目】给出下列说法:①实数集可以表示为{R};②方程$$ \sqrt { 2 x - 1 } + | 2 y + 1 | = 0 $$的解集是$$ \left\{ - \frac { 1 } { 2 } , \frac { 1 } { 2 } \right\} ; $$③方程组$$\left\{ \begin{matrix} x + y = 3 , \\ x - y = - 1 ...
11.已知$\left\{\begin{array}{l}{x=2}\\{y=1}\end{array}\right.$是二元一次方程组$\left\{\begin{array}{l}{ax+by=8}\\{bx-ay=1}\end{array}\right.$的解.则4a-5b的平方根为( )A.$\sqrt{2}$B.2C.±$\sqrt{2}$D.±2
Draw the canonical unit circle, then: the integration domain is the left half unit disk, and from here the integral is: ∫01∫π/23π/2r2cosθdθdr You can check this ... 更多结果 共享 复制 已复制到剪贴板 示例 二次方程式 ...
Evaluate (complex solution) y=z∣x∣(−∣x∣+2)andz∣x∣(−∣x∣+2)=4 Solve for z ⎩⎪⎨⎪⎧z=x(2−x)4,z=−x(x+2)4,y=4andx>0andx<2y=4andx<0andx>−2 Share Copy ...
答案见上【详解】设$$ \mu = \sqrt { 2 x - 1 } ( x \geq \frac { 1 } { 2 } ) $$ 则$$ x = \frac { 1 + \mu ^ { 2 } } { 2 } ( \mu > 0 ) $$,∴$$ y = \frac { 1 + \mu ^ { 2 } } { 2 } + \mu = \frac { ( \mu + 1 ) ^ { 2 }...
二、选择题(每小题5分,共20分)9. 若集合$$ A = \left\{ x | y = \sqrt { 2 x - 1 } \right\}
2.已知点p(x.y)满足$\left\{\begin{array}{l}x+y-2\sqrt{2}≥0\\ x≤2\sqrt{2}\\ y≤2\sqrt{2}\end{array}\right.$过点p(x.y)向圆x2+y2=1做两条切线.切点分别是点A和点B.则当∠APB最大时.$\overrightarrow{PA}•\overrightarrow{PB}$的值是( )A.2B.3C.$\frac{5}{2}$D
2.定义函数max$\left\{{f}\right\}=\left\{{\begin{array}{l}{f}\\{g}\end{array}}$.则max{sinx.cosx}的最小值为( )A.$-\sqrt{2}$B.$\sqrt{2}$C.$-\frac{{\sqrt{2}}}{2}$D.$\frac{{\sqrt{2}}}{2}$
试题来源: 解析 B \because 双曲线方程可得双曲线的焦点在x轴上,且{{a}^{2}}=3,{{b}^{2}}=1, 由此可得c=\sqrt{{{a}^{2}}+{{b}^{2}}}=2, \therefore 该双曲线的焦点坐标为\left( \pm 2,0 \right) 故选:\text B反馈 收藏 ...