3. 已知$x - y = xy$,则$\frac{1}{x}-\frac{1}{y}=$________。
答案
-1
4. 计算:
(1)$\frac{-3a^{2}}{b} \cdot \frac{ab^{2}}{-a^{3}b} \cdot (-\frac{6b}{a^{2}})$;
(2)$\frac{x + y}{x^{4}-y^{4}} \div \frac{1}{x^{2}+y^{2}}$。
(1)$\frac{-3a^{2}}{b} \cdot \frac{ab^{2}}{-a^{3}b} \cdot (-\frac{6b}{a^{2}})$;
(2)$\frac{x + y}{x^{4}-y^{4}} \div \frac{1}{x^{2}+y^{2}}$。
答案
(1)$-\frac{18b}{a^{2}}$ (2)$\frac{1}{x - y}$
1. 填空:
(1)$6x^{2}y^{4} \cdot \frac{1}{2xy^{2}}=$________;
(2)$\frac{2}{xy} \div \frac{4}{x^{2}y}=$________;
(3)$\frac{2x - y}{x} \cdot \frac{y}{y - 2x}=$________;
(4)$(\frac{2}{-3x})^{3}=$________;
(5)$-a^{3} \div \frac{a^{2}}{b}=$________;
(6)$\frac{a^{2}-4}{a^{2}+2a + 1} \cdot \frac{a + 1}{a + 2}=$________。
(1)$6x^{2}y^{4} \cdot \frac{1}{2xy^{2}}=$________;
(2)$\frac{2}{xy} \div \frac{4}{x^{2}y}=$________;
(3)$\frac{2x - y}{x} \cdot \frac{y}{y - 2x}=$________;
(4)$(\frac{2}{-3x})^{3}=$________;
(5)$-a^{3} \div \frac{a^{2}}{b}=$________;
(6)$\frac{a^{2}-4}{a^{2}+2a + 1} \cdot \frac{a + 1}{a + 2}=$________。
答案
(1)$3xy^{2}$ (2)$\frac{x}{2}$ (3)$-\frac{y}{x}$ (4)$-\frac{8}{27x^{3}}$ (5)$-ab$ (6)$\frac{a - 2}{a + 1}$
2. 计算:
(1)$\frac{5a - 10}{9a^{3}b} \cdot \frac{6ab}{a^{2}-4}$;
(2)$(-12x^{4}y)^{2} \div (-\frac{3x^{2}}{y})^{3}$;
(3)$(a - 5) \cdot \frac{25 - a^{2}}{a^{2}-10a + 25}$;
(4)$\frac{2x - 4}{x^{2}+3x} \div \frac{x - 2}{x^{2}+6x + 9}$。
(1)$\frac{5a - 10}{9a^{3}b} \cdot \frac{6ab}{a^{2}-4}$;
(2)$(-12x^{4}y)^{2} \div (-\frac{3x^{2}}{y})^{3}$;
(3)$(a - 5) \cdot \frac{25 - a^{2}}{a^{2}-10a + 25}$;
(4)$\frac{2x - 4}{x^{2}+3x} \div \frac{x - 2}{x^{2}+6x + 9}$。
答案
(1)$\frac{10}{3a^{3}+6a^{2}}$ (2)$-\frac{16x^{2}y^{5}}{3}$ (3)$-a - 5$ (4)$\frac{2x + 6}{x}$
3. 已知$a + b + c = 0$,且$abc \neq 0$。证明:$a(\frac{1}{b}+\frac{1}{c})+b(\frac{1}{c}+\frac{1}{a})+c(\frac{1}{a}+\frac{1}{b})+3 = 0$。
答案
$a(\frac{1}{b}+\frac{1}{c})+b(\frac{1}{c}+\frac{1}{a})+c(\frac{1}{a}+\frac{1}{b})=\frac{a}{b}+\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}+\frac{c}{b}=\frac{b + c}{a}+\frac{a + c}{b}+\frac{a + b}{c}=-1 - 1 - 1=-3$,从而得证
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