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20.计算:
(1)$\sqrt{18} + $
$ -8\sqrt{\frac{1}{2}}$;
(2)$(\sqrt{3}-1)^2 + \sqrt{12} ÷ \sqrt{3}$;
(3)解分式方程:$\frac{1}{x-1} + \frac{x}{1-x} =1$;
(4)已知:$A=(\frac{1}{x+1} - \frac{1}{x^2-1}) ÷ \frac{x-2}{x^2+2x+1}$;
①当$x=\sqrt{3}+1$时,先化简,再求值;
②代数式$A$的值能不能等于3,并说明理由.
(1)$\sqrt{18} + $
$ -8\sqrt{\frac{1}{2}}$;(2)$(\sqrt{3}-1)^2 + \sqrt{12} ÷ \sqrt{3}$;
(3)解分式方程:$\frac{1}{x-1} + \frac{x}{1-x} =1$;
(4)已知:$A=(\frac{1}{x+1} - \frac{1}{x^2-1}) ÷ \frac{x-2}{x^2+2x+1}$;
①当$x=\sqrt{3}+1$时,先化简,再求值;
②代数式$A$的值能不能等于3,并说明理由.
答案
20.(1)原式$=3\sqrt{2}+\sqrt{2}-1-4\sqrt{2}=-1$;
(2)原式$=4-2\sqrt{3}+2=6-2\sqrt{3}$;
(3)两边都乘$x-1$,得:$1-x=x-1$,
解得:$x=1$,
检验:当$x=1$时,$x-1=0$,
$\therefore x=1$是原分式方程的增根,
则原分式方程无解;
(4)①原式$=[\dfrac{x-1}{(x+1)(x-1)}-\dfrac{1}{(x+1)(x-1)}]·\dfrac{(x+1)^2}{x-2}$
$=\dfrac{x-2}{(x+1)(x-1)}·\dfrac{(x+1)^2}{x-2}$
$=\dfrac{x+1}{x-1}$,
当$x=\sqrt{3}+1$时,原式$=\dfrac{\sqrt{3}+1+1}{\sqrt{3}+1-1}=\dfrac{\sqrt{3}+2}{\sqrt{3}}=\dfrac{3+2\sqrt{3}}{3}$;
②若代数式A的值为3,则$\dfrac{x+1}{x-1}=3$,
解得$x=2$,
当$x=2$时,原式没有意义,
$\therefore$代数式A的值不可能为3.
(2)原式$=4-2\sqrt{3}+2=6-2\sqrt{3}$;
(3)两边都乘$x-1$,得:$1-x=x-1$,
解得:$x=1$,
检验:当$x=1$时,$x-1=0$,
$\therefore x=1$是原分式方程的增根,
则原分式方程无解;
(4)①原式$=[\dfrac{x-1}{(x+1)(x-1)}-\dfrac{1}{(x+1)(x-1)}]·\dfrac{(x+1)^2}{x-2}$
$=\dfrac{x-2}{(x+1)(x-1)}·\dfrac{(x+1)^2}{x-2}$
$=\dfrac{x+1}{x-1}$,
当$x=\sqrt{3}+1$时,原式$=\dfrac{\sqrt{3}+1+1}{\sqrt{3}+1-1}=\dfrac{\sqrt{3}+2}{\sqrt{3}}=\dfrac{3+2\sqrt{3}}{3}$;
②若代数式A的值为3,则$\dfrac{x+1}{x-1}=3$,
解得$x=2$,
当$x=2$时,原式没有意义,
$\therefore$代数式A的值不可能为3.
