《2025年复习计划风向标暑八年级数学北师大版》

1. 已知$x= \sqrt {5}+2$,求$x^{2}-x-2$的值.

答案： 解:$x^{2}-x-2=(x-2)(x+1)$.当$x=\sqrt{5}+2$时,原式$=(\sqrt{5}+2-2)× (\sqrt{5}+2+1)=5+3\sqrt{5}$.

2. 若$x,y$为实数,且$y= \sqrt {1-4x}+\sqrt {4x-1}+\frac {1}{2}$,求$\sqrt {\frac {x}{y}+2+\frac {y}{x}}+\sqrt {\frac {x}{y}+2-\frac {y}{x}}$的值.

答案： 解:依题意,得$1-4x\geqslant 0$且$4x-1\geqslant 0$.$\therefore 1-4x=0$.$\therefore x=\dfrac{1}{4}$,$y=\dfrac{1}{2}$.$\therefore \dfrac{x}{y}=\dfrac{1}{2}$,$\dfrac{y}{x}=2$.$\therefore \sqrt{\dfrac{x}{y}+2+\dfrac{y}{x}}+\sqrt{\dfrac{x}{y}+2-\dfrac{y}{x}}=\sqrt{\dfrac{1}{2}+2+2}+\sqrt{\dfrac{1}{2}+2-2}=\sqrt{\dfrac{9}{2}}+\sqrt{\dfrac{1}{2}}=2\sqrt{2}$.

3. 已知$x= \sqrt {3}+2,y= \sqrt {3}-2$,求$x^{3}-xy^{2}$的值.

答案： 解:$\because x=\sqrt{3}+2$,$y=\sqrt{3}-2$,$\therefore x+y=2\sqrt{3}$,$x-y=4$.$\therefore x^{3}-xy^{2}=x(x^{2}-y^{2})=x(x+y)(x-y)=(\sqrt{3}+2)× 2\sqrt{3}× 4=24+16\sqrt{3}$.

4. 已知$x= \frac {\sqrt {3}-1}{2},y= \frac {\sqrt {3}+1}{2}$,求$\frac {x}{y}+\frac {y}{x}$的值.

答案： 解:$\because x=\dfrac{\sqrt{3}-1}{2}$,$y=\dfrac{\sqrt{3}+1}{2}$,$\therefore x+y=\sqrt{3}$,$xy=\dfrac{1}{2}$.$\therefore \dfrac{x}{y}+\dfrac{y}{x}=\dfrac{x^{2}+y^{2}}{xy}=\dfrac{(x+y)^{2}-2xy}{xy}=\dfrac{(\sqrt{3})^{2}-2× \dfrac{1}{2}}{\dfrac{1}{2}}=4$.

5. 已知$x+y= -5,xy= 4$,求$\sqrt {\frac {x}{y}}+\sqrt {\frac {y}{x}}$的值.

答案： 解:$\because x+y=-5$,$xy=4$,$\therefore x＜0$,$y＜0$,$\therefore \sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{y}{x}}=\dfrac{\sqrt{xy}}{-y}+\dfrac{\sqrt{xy}}{-x}=\dfrac{x\sqrt{xy}}{-xy}+\dfrac{y\sqrt{xy}}{-xy}=\dfrac{(x+y)\sqrt{xy}}{-xy}$.当$x+y=-5$,$xy=4$时,原式$=\dfrac{-5× \sqrt{4}}{-4}=\dfrac{5}{2}$.