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6. 有下列各式:①$(\frac{-2mn}{a^2b})^2$;②$-\frac{8m^4n^2}{a^5b} \cdot \frac{an}{bm^2}$;③$(\frac{2m}{-ab^2})^2 \cdot (\frac{nb}{a})^2$;④$\frac{2mn^2}{ab^2} ÷ \frac{a^3}{m}$.其中相等的两个式子是(
A.①②
B.①③
C.②③
D.③④
B
)A.①②
B.①③
C.②③
D.③④
答案
B
解析
①$(\frac{-2mn}{a^2b})^2=\frac{4m^2n^2}{a^4b^2}$
②$-\frac{8m^4n^2}{a^5b} \cdot \frac{an}{bm^2}=-\frac{8m^2n^3}{a^4b^2}$
③$(\frac{2m}{-ab^2})^2 \cdot (\frac{nb}{a})^2=\frac{4m^2}{a^2b^4} \cdot \frac{n^2b^2}{a^2}=\frac{4m^2n^2}{a^4b^2}$
④$\frac{2mn^2}{ab^2} ÷ \frac{a^3}{m}=\frac{2mn^2}{ab^2} \cdot \frac{m}{a^3}=\frac{2m^2n^2}{a^4b^2}$
相等的两个式子是①③
B
②$-\frac{8m^4n^2}{a^5b} \cdot \frac{an}{bm^2}=-\frac{8m^2n^3}{a^4b^2}$
③$(\frac{2m}{-ab^2})^2 \cdot (\frac{nb}{a})^2=\frac{4m^2}{a^2b^4} \cdot \frac{n^2b^2}{a^2}=\frac{4m^2n^2}{a^4b^2}$
④$\frac{2mn^2}{ab^2} ÷ \frac{a^3}{m}=\frac{2mn^2}{ab^2} \cdot \frac{m}{a^3}=\frac{2m^2n^2}{a^4b^2}$
相等的两个式子是①③
B
7. 计算$1 ÷ \frac{1 + m}{1 - m} \cdot (m^2 - 1)$的结果是(
A.$-m^2 - 2m - 1$
B.$-m^2 + 2m - 1$
C.$m^2 - 2m - 1$
D.$m^2 - 1$
B
)A.$-m^2 - 2m - 1$
B.$-m^2 + 2m - 1$
C.$m^2 - 2m - 1$
D.$m^2 - 1$
答案
B
解析
$1÷\frac{1+m}{1-m}\cdot(m^2 - 1)$
$=1\cdot\frac{1 - m}{1 + m}\cdot(m + 1)(m - 1)$
$=\frac{1 - m}{1 + m}\cdot(m + 1)(m - 1)$
$=(1 - m)(m - 1)$
$=-(m - 1)^2$
$=-(m^2 - 2m + 1)$
$=-m^2 + 2m - 1$
B
$=1\cdot\frac{1 - m}{1 + m}\cdot(m + 1)(m - 1)$
$=\frac{1 - m}{1 + m}\cdot(m + 1)(m - 1)$
$=(1 - m)(m - 1)$
$=-(m - 1)^2$
$=-(m^2 - 2m + 1)$
$=-m^2 + 2m - 1$
B
8. 已知$x^2 - 4x + 4与\vert y - 1\vert$互为相反数,则式子$(\frac{x}{y} - \frac{y}{x}) ÷ (x + y)$的值等于
$\frac{1}{2}$
.答案
$\frac{1}{2}$
解析
因为$x^2 - 4x + 4$与$\vert y - 1\vert$互为相反数，所以$x^2 - 4x + 4 + \vert y - 1\vert = 0$，即$(x - 2)^2 + \vert y - 1\vert = 0$。
因为$(x - 2)^2 \geq 0$，$\vert y - 1\vert \geq 0$，所以$x - 2 = 0$，$y - 1 = 0$，解得$x = 2$，$y = 1$。
$(\frac{x}{y} - \frac{y}{x}) ÷ (x + y)$
$= (\frac{x^2 - y^2}{xy}) ÷ (x + y)$
$= \frac{(x + y)(x - y)}{xy} × \frac{1}{x + y}$
$= \frac{x - y}{xy}$
将$x = 2$，$y = 1$代入，得$\frac{2 - 1}{2×1} = \frac{1}{2}$。
$\frac{1}{2}$
因为$(x - 2)^2 \geq 0$，$\vert y - 1\vert \geq 0$，所以$x - 2 = 0$，$y - 1 = 0$，解得$x = 2$，$y = 1$。
$(\frac{x}{y} - \frac{y}{x}) ÷ (x + y)$
$= (\frac{x^2 - y^2}{xy}) ÷ (x + y)$
$= \frac{(x + y)(x - y)}{xy} × \frac{1}{x + y}$
$= \frac{x - y}{xy}$
将$x = 2$，$y = 1$代入，得$\frac{2 - 1}{2×1} = \frac{1}{2}$。
$\frac{1}{2}$
9. 计算:(1)$(\frac{y^2}{-x^3})^4$;
(2)$\frac{m^2 - n^2}{(m - n)^2} \cdot (\frac{n - m}{mn})^2 ÷ \frac{m + n}{m}$.
(2)$\frac{m^2 - n^2}{(m - n)^2} \cdot (\frac{n - m}{mn})^2 ÷ \frac{m + n}{m}$.
答案
解:
(1)原式$=\frac{(y^{2})^{4}}{(-x^{3})^{4}}=\frac{y^{8}}{x^{12}}$.
(2)原式$=\frac{(m+n)(m-n)}{(m-n)^{2}}\cdot\frac{(n-m)^{2}}{m^{2}n^{2}}\cdot\frac{m}{m+n}=\frac{m-n}{mn^{2}}$.
(1)原式$=\frac{(y^{2})^{4}}{(-x^{3})^{4}}=\frac{y^{8}}{x^{12}}$.
(2)原式$=\frac{(m+n)(m-n)}{(m-n)^{2}}\cdot\frac{(n-m)^{2}}{m^{2}n^{2}}\cdot\frac{m}{m+n}=\frac{m-n}{mn^{2}}$.
10. 计算:(1)$(-\frac{a}{b})^2 \cdot (-\frac{b}{a})^3 ÷ (-ab^4)$;
(2)$(xy - x^2) ÷ \frac{x^2 - 2xy + y^2}{xy} \cdot \frac{x - y}{x^2}$.
(2)$(xy - x^2) ÷ \frac{x^2 - 2xy + y^2}{xy} \cdot \frac{x - y}{x^2}$.
答案
解:
(1)原式$=\frac{a^{2}}{b^{2}}\cdot(-\frac{b^{3}}{a^{3}})\cdot\frac{1}{-ab^{4}}=\frac{a^{2}}{b^{2}}\cdot\frac{b^{3}}{a^{3}}\cdot\frac{1}{ab^{4}}=\frac{1}{a^{2}b^{3}}$.
(2)原式$=-x(x-y)\cdot\frac{xy}{(x-y)^{2}}\cdot\frac{x-y}{x^{2}}=-y$.
(1)原式$=\frac{a^{2}}{b^{2}}\cdot(-\frac{b^{3}}{a^{3}})\cdot\frac{1}{-ab^{4}}=\frac{a^{2}}{b^{2}}\cdot\frac{b^{3}}{a^{3}}\cdot\frac{1}{ab^{4}}=\frac{1}{a^{2}b^{3}}$.
(2)原式$=-x(x-y)\cdot\frac{xy}{(x-y)^{2}}\cdot\frac{x-y}{x^{2}}=-y$.
11. 有这样一道题:计算$\frac{x^2 - 2x + 1}{x^2 - 1} ÷ \frac{x - 1}{x^2 + x} ÷ (\frac{1}{x})^3$的值,其中$x = 2$. 小明同学把$x = 2错抄为x = -2$,但是他计算的结果也是正确的,你说这是怎么回事?
答案
解:$\frac{x^{2}-2x+1}{x^{2}-1}÷\frac{x-1}{x^{2}+x}÷(\frac{1}{x})^{3}=\frac{(x-1)^{2}}{(x+1)(x-1)}\cdot\frac{x(x+1)}{x-1}\cdot x^{3}=x^{4}$,
∴当x=2或x=-2时,原式的值都等于16.
∴当x=2或x=-2时,原式的值都等于16.
12. 已知实数$x,y满足\vert x - 3\vert + y^2 - 4y + 4 = 0$,求式子$\frac{x^2 - y^2}{xy} \cdot \frac{1}{x^2 - 2xy + y^2} ÷ \frac{x}{x^2y - xy^2}$的值.
答案
解:$\frac{x^{2}-y^{2}}{xy}\cdot\frac{1}{x^{2}-2xy+y^{2}}÷\frac{x}{x^{2}y-xy^{2}}=\frac{(x+y)(x-y)}{xy}\cdot\frac{1}{(x-y)^{2}}\cdot\frac{xy(x-y)}{x}=\frac{x+y}{x}$.
∵$|x-3|+y^{2}-4y+4=0$,
∴$|x-3|+(y-2)^{2}=0$,
∴x=3,y=2,
∴原式$=\frac{3+2}{3}=\frac{5}{3}$.
∵$|x-3|+y^{2}-4y+4=0$,
∴$|x-3|+(y-2)^{2}=0$,
∴x=3,y=2,
∴原式$=\frac{3+2}{3}=\frac{5}{3}$.
