第80页 - 经纶学典学霸八年级数学上下册【江苏国标】

B

3

$解:分式的分子、分母都除以xy$

$则\frac{2x-3xy+2y}{x+2xy+y}=\frac{(2x-3xy+2y)÷xy}{(x+2xy+y)÷xy}=\frac {\frac {2}{y}-3+\frac {2}{x}}{\frac {1}{y}+2+\frac {1}{x}}\ $

$∵\frac {1}{x}+\frac {1}{y}=3$

$∴原式=\frac{2×3-3}{3+2}=\frac{3}{5}$

$解:\frac{x^{2}-1}{x}=x-\frac {1}{x}$

$∵ x^{2}-3x-1=0$

$等式两边同除以x,即x-3-\frac {1}{x}=0$

$∴原式=3$

$解:\frac{x^{4}+1}{x^{2}}=(x-\frac{1}{x})^{2}+2$

$=11$

$解:分式的分子、分母都除以x^{2}$

$则原式=\frac {2}{x^{2}-3+\frac {1}{x^{2}}}=\frac {1}{4}$

$ 解:∵\frac{xy}{x+y}=-3,\frac{yz}{y+z}=\frac{4}{3},\frac{zx}{z+x}=-\frac{4}{3} $

$ ∴\frac{x+y}{xy}=-\frac{1}{3},\frac {y+z}{yz}=\frac{3}{4},\frac{z+x}{zx}=-\frac{3}{4} $

$ 即\frac{1}{x}+\frac {1}{y}=-\frac{1}{3} $

$ \frac {1}{y}+\frac{1}{z}=\frac{3}{4} $

$ \frac{1}{z}+\frac{1}{x}=-\frac{3}{4} $

$ 解得\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=-\frac{1}{6} $

$ ∴\frac{xyz}{xy+yz+zx}=\frac{1}{\frac {1}{z}+\frac {1}{x}+\frac {1}{y}}=-6 $

$解:(1)设\frac{x}{2}=\frac{y}{3}=\frac{z}{6}=k$

$则x=2k、y=3k、z=6k$

$∴\frac{x+2y-z}{x-2y+3z}=\frac{2k+6k-6k}{2k-6k+18k}=\frac{1}{7}$

$∴分式\frac{x+2y-z}{x-2y+3z}的值为\frac{1}{7}$

$(2)设\frac {y+z}{x}=\frac {z+x}{y}=\frac {x+y}{z}=k$

$则\begin{cases}{y+z=kx① }\ \\ { x+z=ky② } \\{ x+y=kz③} \end{cases}\ $

$①+②+③得2x+2y+2z= k(x+y+z)$

$∵x+y+z≠0,∴k=2$

$原式=\frac{2z-z}{2z+z}=\frac{1}{3}$