8. 计算:
(1)$\frac{x^{2}}{(x + y)^{2}}+\frac{2y^{2}}{(y + x)^{2}}-\frac{y^{2}-2xy}{(-x - y)^{2}}$; (2)$\frac{x^{2}-4x + 4}{x^{2}-4}+\frac{x - 2}{x^{2}+2x}+2$;
(3)$\frac{a}{a + 2}+\frac{a}{2 - a}+\frac{3a - 2}{a^{2}-4}$; (4)$x + 2y+\frac{4y^{2}}{x - 2y}+\frac{4x^{2}y}{4y^{2}-x^{2}}$.
(1)$\frac{x^{2}}{(x + y)^{2}}+\frac{2y^{2}}{(y + x)^{2}}-\frac{y^{2}-2xy}{(-x - y)^{2}}$; (2)$\frac{x^{2}-4x + 4}{x^{2}-4}+\frac{x - 2}{x^{2}+2x}+2$;
(3)$\frac{a}{a + 2}+\frac{a}{2 - a}+\frac{3a - 2}{a^{2}-4}$; (4)$x + 2y+\frac{4y^{2}}{x - 2y}+\frac{4x^{2}y}{4y^{2}-x^{2}}$.
答案
解:(1)原式$=\frac{x^{2}+2y^{2}-y^{2}+2xy}{(x + y)^{2}}=\frac{(x + y)^{2}}{(x + y)^{2}}=1$.
(2)原式$=\frac{(x - 2)^{2}}{(x + 2)(x - 2)}+\frac{x - 2}{x(x + 2)}+2=\frac{x - 2}{x + 2}+\frac{x - 2}{x(x + 2)}+2=\frac{x(x - 2)}{x(x + 2)}+\frac{x - 2}{x(x + 2)}+\frac{2x(x + 2)}{x(x + 2)}=\frac{x^{2}-2x+x - 2+2x^{2}+4x}{x(x + 2)}=\frac{3x^{2}+3x - 2}{x(x + 2)}$.
(3)原式$=\frac{a(a - 2)-a(a + 2)+3a - 2}{(a + 2)(a - 2)}=\frac{-(a + 2)}{(a + 2)(a - 2)}=-\frac{1}{a - 2}$.
(4)原式$=\frac{x^{2}-4y^{2}+4y^{2}}{x - 2y}-\frac{4x^{2}y}{(x + 2y)(x - 2y)}=\frac{x^{2}(x + 2y)-4x^{2}y}{(x + 2y)(x - 2y)}=\frac{x^{2}(x - 2y)}{(x + 2y)(x - 2y)}=\frac{x^{2}}{x + 2y}$.
(2)原式$=\frac{(x - 2)^{2}}{(x + 2)(x - 2)}+\frac{x - 2}{x(x + 2)}+2=\frac{x - 2}{x + 2}+\frac{x - 2}{x(x + 2)}+2=\frac{x(x - 2)}{x(x + 2)}+\frac{x - 2}{x(x + 2)}+\frac{2x(x + 2)}{x(x + 2)}=\frac{x^{2}-2x+x - 2+2x^{2}+4x}{x(x + 2)}=\frac{3x^{2}+3x - 2}{x(x + 2)}$.
(3)原式$=\frac{a(a - 2)-a(a + 2)+3a - 2}{(a + 2)(a - 2)}=\frac{-(a + 2)}{(a + 2)(a - 2)}=-\frac{1}{a - 2}$.
(4)原式$=\frac{x^{2}-4y^{2}+4y^{2}}{x - 2y}-\frac{4x^{2}y}{(x + 2y)(x - 2y)}=\frac{x^{2}(x + 2y)-4x^{2}y}{(x + 2y)(x - 2y)}=\frac{x^{2}(x - 2y)}{(x + 2y)(x - 2y)}=\frac{x^{2}}{x + 2y}$.
9. 分式中,在分子、分母都是整式的情况下,如果分子的次数低于分母的次数,称这样的分式为真分式. 例如,分式$\frac{2}{x + 1}$,$\frac{4x^{2}}{x^{3}-3x}$是真分式. 如果分子的次数不低于分母的次数,称这样的分式为假分式. 例如,分式$\frac{x - 1}{x + 1}$,$\frac{x^{2}}{x - 1}$是假分式. 一个假分式可以化为一个整式与一个真分式的和(差)的形式. 例如,$\frac{x - 1}{x + 1}=\frac{(x + 1)-2}{x + 1}=1-\frac{2}{x + 1}$.
(1)将假分式$\frac{2x - 3}{x + 1}$化为一个整式与一个真分式的和(差)的形式;
(2)如果分式$\frac{x^{2}}{x - 1}$的值为整数,求$x$的整数值.
(1)将假分式$\frac{2x - 3}{x + 1}$化为一个整式与一个真分式的和(差)的形式;
(2)如果分式$\frac{x^{2}}{x - 1}$的值为整数,求$x$的整数值.
答案
解:(1)$\frac{2x - 3}{x + 1}=\frac{2(x + 1)-5}{x + 1}=2-\frac{5}{x + 1}$.
(2)$\frac{x^{2}}{x - 1}=\frac{x^{2}-1 + 1}{x - 1}=x + 1+\frac{1}{x - 1}$,
$\because$分式的值为整数,且$x$为整数,
$\therefore x - 1=\pm1$,$\therefore x = 2$或$x = 0$.
(2)$\frac{x^{2}}{x - 1}=\frac{x^{2}-1 + 1}{x - 1}=x + 1+\frac{1}{x - 1}$,
$\because$分式的值为整数,且$x$为整数,
$\therefore x - 1=\pm1$,$\therefore x = 2$或$x = 0$.
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