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第92页

信息发布者：

解$：(1)$原式$=\frac {(a + b)^2}{a²b²}÷\frac {b^2-a^2}{a^2b^2}$

$=\frac {(a + b)^2}{a^2b^2}×\frac {a^2b^2}{(b + a)(b - a)}$

$=\frac {a + b}{b - a}$

解：$(2)$原式$=(\frac {3x^2}{4y})^2\ \mathrm {·}\frac {2y}{3x} + \frac {x^2}{2y^2} \div \frac {2y^2}{x}$

$ =\frac {9x^4}{16y^2}\ \mathrm {·}\frac {2y}{3x} + \frac {x^2}{2y^2}\ \mathrm {·}\frac {x}{2y^2}$

$ =\frac {3x^3}{8y} + \frac {x^3}{4y^4}$

$ =\frac {3x^3y^3 + 2x^3}{8y^4}$

$ =\frac {x^3(3y^3 + 2)}{8y^4}$

解：$(3)$原式$=(\frac {x}{x + y} + \frac {2y}{x + y})\ \mathrm {·}\frac {xy}{x + 2y} \div (\frac {1}{x} + \frac {1}{y})$

$ =\frac {x + 2y}{x + y}\ \mathrm {·}\frac {xy}{x + 2y} \div \frac {x + y}{xy}$

$ =\frac {xy}{x + y}\ \mathrm {·}\frac {xy}{x + y}$

$ =\frac {x^2y^2}{(x + y)^2}$

解：$(4)$原式$=(\frac {a + b}{a - b})^2\ \mathrm {·}\frac {2a - 2b}{3a + 3b} - \frac {a^2}{a^2 - b^2} \div \frac {a}{b}$

$ =\frac {(a + b)^2}{(a - b)^2}\ \mathrm {·}\frac {2(a - b)}{3(a + b)} - \frac {a^2}{(a - b)(a + b)}\ \mathrm {·}\frac {b}{a}$

$ =\frac {2(a + b)}{3(a - b)} - \frac {ab}{(a - b)(a + b)}$

$ =\frac {2(a + b)^2 - 3ab}{3(a - b)(a + b)}$

$ =\frac {2a^2 + 4ab + 2b^2 - 3ab}{3(a^2 - b^2)}$

$ =\frac {2a^2 + ab + 2b^2}{3(a^2 - b^2)}$

解: 方法一：原式$=(\frac {3x}{x - 2} - \frac {x}{x + 2})\ \mathrm {·}\frac {4 - x^2}{x}$

$ =[\frac {3x(x + 2) - x(x - 2)}{(x - 2)(x + 2)}] ·\frac {-(x^2 - 4)}{x}$

$ =\frac {3x^2 + 6x - x^2 + 2x}{x^2 - 4}\ \mathrm {·}\frac {-(x - 2)(x + 2)}{x}$

$ =\frac {2x^2 + 8x}{(x - 2)(x + 2)}\ \mathrm {·}\frac {-(x - 2)(x + 2)}{x}$

$ =2x(x + 4)\ \mathrm {·}\frac {-1}{x}$

$ =-2(x + 4) $

$= -2x - 8$

方法二：原式$=\frac {3x}{x - 2}\ \mathrm {·}\frac {4 - x^2}{x} - \frac {x}{x + 2}\ \mathrm {·}\frac {4 - x^2}{x}$

$ =3\ \mathrm {·}\frac {-(x^2 - 4)}{x - 2} - \frac {-(x^2 - 4)}{x + 2}$

$ =3\ \mathrm {·}\frac {-(x - 2)(x + 2)}{x - 2} - \frac {-(x - 2)(x + 2)}{x + 2}$

$ =-3(x + 2) + (x - 2)$

$ =-3x - 6 + x - 2 $

$= -2x - 8$

解:(1)原式$=\frac {p^2}{mnp}+\frac {\mathrm {m^2}}{mnp}+\frac {n^2}{mnp}$

$=\frac {p^2+\mathrm {m^2}+n^2}{mnp}$

解$：(2)$

$\begin {aligned}&\frac {c - a}{(a - b)(b - c)}-\frac {a - b}{(b - c)(c - a)}+\frac {b - c}{(c - a)(a - b)}\\=&\frac {-(a - c)^2}{(a - b)(b - c)(c - a)}-\frac {(a - b)^2}{(b - c)(c - a)(a - b)}+\frac {(b - c)^2}{(c - a)(a - b)(b - c)}\\=&\frac {-(c - a)^2-(a - b)^2+(b - c)^2}{(a - b)(b - c)(c - a)}\\=&\frac {-(a - b)^2-(b - c)^2-(c - a)^2}{(a - b)(b - c)(c - a)}\end {aligned}$

解:$\frac {b}{a}-\frac {a}{b}=\frac {b^2-a^2}{ab}=\frac {(b - a)(b + a)}{ab}$

$∵a＞b＞0$

$∴b - a＜0$，$a + b＞0$，$ab＞0$

$∴\frac {(b - a)(b + a)}{ab}＜0$

$∴\frac {b}{a}＜\frac {a}{b}$

答：$\frac {b}{a}＜\frac {a}{b}$。