17.已知 $x+\dfrac{1}{x}=5$，试求下列各式的值：
(1)① $x^2+\dfrac{1}{x^2}=$ ； ② $x^4+\dfrac{1}{x^4}=$ .
(2)① $x^2+\dfrac{1}{x^2}=(x\pm\dfrac{1}{x})^2$ ； ② $x^4+\dfrac{1}{x^4}=(x^2\pm\dfrac{1}{x^2})^2$ .
(3)已知 $x^2-4x-1=0$，求 $x^4+\dfrac{1}{x^4}$ 的值.

17.解:(1)①23 ②527
(2)①∓2 ②∓2
(3)由题意,得$x≠0$,$\because x^2-4x-1=0$.
$\therefore x^2-1=4x$,$\therefore x-\frac{1}{x}=4$,
$\therefore x^2+\frac{1}{x^2}=(x-\frac{1}{x})^2+2=4^2+2=18$,
$\therefore x^4+\frac{1}{x^4}=(x^2+\frac{1}{x^2})^2-2=18^2-2=322$.

1.(辽宁省)将方程$\frac{1}{x-1}+3=\frac{3x}{1-x}$去分母，两边同乘$(x-1)$后的式子为()

A.$1+3=3x(1-x)$
B.$1+3(x-1)=-3x$
C.$x-1+3=-3x$
D.$1+3(x-1)=3x$

1.B

2.(黑龙江省)如果关于x的分式方程$\frac{2x - m}{x + 1}=1$的解是负数，那么实数m的取值范围是()

A.$m ＜ -1$
B.$m ＞ -1$且$m≠0$
C.$m ＞ -1$
D.$m ＜ -1$且$m≠-2$

2.D

3.(河北省)已知A为整式,若计算$\frac{A}{xy+y^2}-\frac{y}{x^2+xy}$的结果为$\frac{x-y}{xy}$,则A=()

A.$x$
B.$y$
C.$x+y$
D.$x-y$

3.A 点拨:$\because \frac{A}{xy+y^2}-\frac{y}{x^2+xy}$的结果为$\frac{x-y}{xy}$,$\therefore \frac{y}{x^2+xy}+\frac{x-y}{xy}=\frac{A}{xy+y^2}$,
$\therefore \frac{y^2}{xy(x+y)}+\frac{(x-y)(x+y)}{xy(x+y)}=\frac{x^2}{xy(x+y)}=\frac{x}{xy+y^2}=\frac{A}{xy+y^2}$,$\therefore A=x$,故选A.

4.(长沙市)要使分式$\frac{6}{x-19}$有意义,则$x$需满足的条件是.

4.$x≠19$

5.(大庆市)方程$\frac{1}{2x+3}+\frac{1}{x}=0$的解为.

5.$x=-1$

6.先化简,再求值:
(1)(长春市)$\frac{x^3}{x-2}-\frac{2x^2}{x-2}$,其中$x=\sqrt{2}$;
(2)(青海省)$( \frac{1}{y}-\frac{1}{x} )÷( \frac{x}{y}-\frac{y}{x} )$,其中$x=2-y$.

6.解:(1)原式$=\frac{x^3-2x^2}{x-2}=\frac{x^2(x-2)}{x-2}=x^2$,
当$x=\sqrt{2}$时,原式$=(\sqrt{2})^2=2$.
(2)$(\frac{1}{y}-\frac{1}{x})÷(\frac{x}{y}-\frac{y}{x})$
$=(\frac{x}{xy}-\frac{y}{xy})÷(\frac{x^2}{xy}-\frac{y^2}{xy})$
$=\frac{x-y}{xy}÷\frac{x^2-y^2}{xy}$
$=\frac{x-y}{xy}×\frac{xy}{x^2-y^2}$
$=\frac{x-y}{xy}×\frac{xy}{(x+y)(x-y)}$
$=\frac{1}{x+y}$.
$\because x=2-y$,$\therefore x+y=2$,
$\therefore$原式$=\frac{1}{x+y}=\frac{1}{2}$.