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零五網(wǎng) 全部參考答案 啟東中學(xué)作業(yè)本 2025年啟東中學(xué)作業(yè)本八年級數(shù)學(xué)上冊人教版 第41頁解析答案
1. 分式乘方的運算法則:
分式乘方要把分子、分母分別乘方
,字母表達式為
$\left(\frac{a}\right)^n=\frac{a^n}{b^n}$
.
答案:分式乘方要把分子、分母分別乘方 $\left(\frac{a}\right)^n=\frac{a^n}{b^n}$
2. 分式混合運算的順序:
先乘方,再乘除,最后算加減
.
答案:先乘方,再乘除,最后算加減
1. 化簡$a÷b\cdot \frac {1}$的結(jié)果是 (
A
)
A.$\frac {a}{b^{2}}$
B.$a$
C.$ab^{2}$
D.$ab$
答案:A
解析:
解:$a÷b\cdot \frac{1} = a \cdot \frac{1} \cdot \frac{1} = \frac{a}{b^2}$
答案:A
2. 計算:$(-ab)÷\frac {2b^{2}}{a}=$
$-\frac{a^2}{2b}$
; $\frac {a}{6b}\cdot (\frac {3b}{a})^{2}=$
$\frac{3b}{2a}$
; $x^{2}÷(\frac {2x}{y})^{2}=$
$\frac{y^2}{4}$
.
答案:$-\frac{a^2}{2b}$ $\frac{3b}{2a}$ $\frac{y^2}{4}$
解析:
$(-ab)÷\frac {2b^{2}}{a}=-ab\cdot \frac {a}{2b^{2}}=-\frac {a^{2}}{2b}$;
$\frac {a}{6b}\cdot (\frac {3b}{a})^{2}=\frac {a}{6b}\cdot \frac {9b^{2}}{a^{2}}=\frac {3b}{2a}$;
$x^{2}÷(\frac {2x}{y})^{2}=x^{2}÷\frac {4x^{2}}{y^{2}}=x^{2}\cdot \frac {y^{2}}{4x^{2}}=\frac {y^{2}}{4}$
$-\frac {a^{2}}{2b}$;$\frac {3b}{2a}$;$\frac {y^{2}}{4}$
3. 計算:
(1)$\frac {4a^{2}b}{3cd^{2}}\cdot \frac {5c^{2}d}{4ab^{2}}÷\frac {2abc}{3d}$;
(2)$\frac {81-a^{2}}{a^{2}+6a+9}÷\frac {a-9}{2a+6}\cdot \frac {a+3}{a+9}$;
(3)$\frac {x^{2}}{y}÷\frac {-y}{x}\cdot (\frac {y}{x})^{2}$.
答案:解:(1)原式$=\frac{4a^2b}{3cd^2}\cdot\frac{5c^2d}{4ab^2}\cdot\frac{3d}{2abc}=\frac{5}{2b^2}$.
(2)原式$=-\frac{(a+9)(a-9)}{(a+3)^2}\cdot\frac{2(a+3)}{a-9}\cdot\frac{a+3}{a+9}=-2$.
(3)原式$=\frac{x^2}{y}\cdot\frac{-x}{y}\cdot\frac{y^2}{x^2}=-x$.
4. 計算:
(1)$(\frac {10c}{-a^{2}b})^{3}$;
(2)$(\frac {-3a}{4b^{2}})^{2}÷6a^{2}b$;
(3)$(\frac {y^{3}}{x})^{2}\cdot (\frac {-x}{2y^{2}})^{3}$.
答案:解:(1)$\left(\frac{10c}{-a^2b}\right)^3=-\frac{1000c^3}{a^6b^3}$.
(2)$\left(\frac{-3a}{4b^2}\right)^2÷6a^2b=\frac{9a^2}{16b^4}\cdot\frac{1}{6a^2b}=\frac{3}{32b^5}$.
(3)$\left(\frac{y^3}{x}\right)^2\cdot\left(\frac{-x}{2y^2}\right)^3=\frac{y^6}{x^2}\cdot\left(-\frac{x^3}{8y^6}\right)=-\frac{x}{8}$.
5. 計算下列各式:
(1)$(\frac {-a^{2}})\cdot (\frac {c}{-a})^{2}÷(-\frac {c^{2}})$;
(2)$(\frac {a^{2}b}{-cd})^{3}÷\frac {2a}{d^{3}}\cdot (\frac {c}{2a})^{2}$.
答案:解:(1)原式$=\frac{a^2}\cdot\frac{c^2}{a^2}\cdot\frac{c^2}=1$.
(2)原式$=\frac{a^6b^2}{-c^3d^3}\cdot\frac{d^3}{2a}\cdot\frac{c^2}{4a^2}=\frac{a^6b^2c^2d^3}{-8a^3c^3d^3}=-\frac{a^3b^2}{8c}$.
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