第90页 - 通城学典课时作业本八年级数学苏科版江苏专版

1

$解：原式=x-1.$

$由题意，得x-1≠0，x-2≠0，x^2-2x+1≠0，$

$ ∴x≠1，x≠2.$

$∴x的值只能取3.$

$此时原式=2$

$-\frac{65}{36} $

$解：∵x^2+y^2=4xy，$

$∴x^2+2xy+y^2=6xy，x^2-2xy+y^2=2xy.$

$∴(x+y)^2=6xy，(x-y)^2=2xy.$

$∴\frac{(x+y)^2}{(x-y)^2}=\frac{6xy}{2xy}，即(\frac{x+y}{x-y})^2=3.$

$∵y＞x＞0.$

$∴\frac{x+y}{x-y}＜0.$

$∴\frac{x+y}{x-y}=-\sqrt{3}$

3

$解：原式=\frac{3}{a-b}$

$∵a-b-1=0，$

$∴a-b=1.$

$∴原式=3$

$解：∵\frac{1}{a}+\frac{1}{b}=\frac{1}{2}，\frac{1}{b}+\frac{1}{c}=\frac{1}{3}，\frac{1}{a}+\frac{1}{c}=\frac{1}{4}，$

$∴ \frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a}+\frac{1}{c}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}，$

$即\frac{2}{a}+\frac{2}{b}+\frac{2}{c}=\frac{13}{12}$

$∴\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{13}{24}.\ $

$根据题意，得 abc ≠0，\ $

$∴\frac{abc}{ab+bc+ac} =\frac{abc÷abc}{(ab+bc+ac)÷abc}=\frac{1}{\frac{1}{c}+\frac{1}{a}+\frac{1}{b}}=\frac{24}{13}$