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8. 计算$(-18)÷[(+\frac {1}{3})-(-\frac {1}{9})]$的结果为______.
答案
$-\frac{81}{2}$
解析
$(-18)÷\left[\left(+\frac{1}{3}\right)-\left(-\frac{1}{9}\right)\right]$
$=(-18)÷\left(\frac{1}{3}+\frac{1}{9}\right)$
$=(-18)÷\left(\frac{3}{9}+\frac{1}{9}\right)$
$=(-18)÷\frac{4}{9}$
$=(-18)×\frac{9}{4}$
$=-\frac{162}{4}$
$=-\frac{81}{2}$
$=(-18)÷\left(\frac{1}{3}+\frac{1}{9}\right)$
$=(-18)÷\left(\frac{3}{9}+\frac{1}{9}\right)$
$=(-18)÷\frac{4}{9}$
$=(-18)×\frac{9}{4}$
$=-\frac{162}{4}$
$=-\frac{81}{2}$
9. 计算：$90^{\circ }-65^{\circ }14'15''= $______.
答案
$24^{\circ}45'45''$
解析
$90^{\circ}-65^{\circ}14'15''$
$=89^{\circ}59'60''-65^{\circ}14'15''$
$=(89^{\circ}-65^{\circ})+(59'-14')+(60''-15'')$
$=24^{\circ}+45'+45''$
$=24^{\circ}45'45''$
$=89^{\circ}59'60''-65^{\circ}14'15''$
$=(89^{\circ}-65^{\circ})+(59'-14')+(60''-15'')$
$=24^{\circ}+45'+45''$
$=24^{\circ}45'45''$
10. 当$x= 4,y= \frac {1}{5}$时,代数式$3x^{2}y-(-2x^{2}y)$的值为______.
答案
16
解析
$3x^{2}y - (-2x^{2}y)$
$=3x^{2}y + 2x^{2}y$
$=5x^{2}y$
当$x = 4$，$y=\frac{1}{5}$时，
$5x^{2}y = 5×4^{2}×\frac{1}{5}$
$=5×16×\frac{1}{5}$
$=16$
16
$=3x^{2}y + 2x^{2}y$
$=5x^{2}y$
当$x = 4$，$y=\frac{1}{5}$时，
$5x^{2}y = 5×4^{2}×\frac{1}{5}$
$=5×16×\frac{1}{5}$
$=16$
16
11. (32分)计算：
(1)$|-2\frac {1}{5}-(-2\frac {1}{4})|-(-1\frac {3}{4})$ (2)$[\frac {3}{5}-(\frac {2}{3}-\frac {1}{4})÷\frac {5}{6}]÷\frac {1}{10}$
(3)$3x-[5x-4(2x-1)]$ (4)$-3(a^{2}-bc+b^{2})-2(-a^{2}+bc-2b^{2})$
(1)$|-2\frac {1}{5}-(-2\frac {1}{4})|-(-1\frac {3}{4})$ (2)$[\frac {3}{5}-(\frac {2}{3}-\frac {1}{4})÷\frac {5}{6}]÷\frac {1}{10}$
(3)$3x-[5x-4(2x-1)]$ (4)$-3(a^{2}-bc+b^{2})-2(-a^{2}+bc-2b^{2})$
答案
（1）$\frac{9}{5}$ （2）1 （3）$6x-4$ （4）$-a^{2}+bc+b^{2}$
解析
(1) $|-2\frac{1}{5}-(-2\frac{1}{4})|-(-1\frac{3}{4})$
$=|-\frac{11}{5}+\frac{9}{4}|+\frac{7}{4}$
$=|-\frac{44}{20}+\frac{45}{20}|+\frac{7}{4}$
$=|\frac{1}{20}|+\frac{7}{4}$
$=\frac{1}{20}+\frac{35}{20}$
$=\frac{36}{20}$
$=\frac{9}{5}$
(2) $[\frac{3}{5}-(\frac{2}{3}-\frac{1}{4})÷\frac{5}{6}]÷\frac{1}{10}$
$=[\frac{3}{5}-(\frac{8}{12}-\frac{3}{12})×\frac{6}{5}]×10$
$=[\frac{3}{5}-\frac{5}{12}×\frac{6}{5}]×10$
$=[\frac{3}{5}-\frac{1}{2}]×10$
$=(\frac{6}{10}-\frac{5}{10})×10$
$=\frac{1}{10}×10$
$=1$
(3) $3x-[5x-4(2x-1)]$
$=3x-(5x-8x+4)$
$=3x-(-3x+4)$
$=3x+3x-4$
$=6x-4$
(4) $-3(a^{2}-bc+b^{2})-2(-a^{2}+bc-2b^{2})$
$=-3a^{2}+3bc-3b^{2}+2a^{2}-2bc+4b^{2}$
$=(-3a^{2}+2a^{2})+(3bc-2bc)+(-3b^{2}+4b^{2})$
$=-a^{2}+bc+b^{2}$
12. (12分)求代数式$2(2x-3y)-(-4x+2y+1)$的值,其中$x-y= 2$.
答案
原式$=8(x-y)-1$. 当$x-y=2$时,原式$=15$
解析
原式$=4x-6y+4x-2y-1$
$=8x-8y-1$
$=8(x-y)-1$
当$x-y=2$时，原式$=8×2-1=15$
$=8x-8y-1$
$=8(x-y)-1$
当$x-y=2$时，原式$=8×2-1=15$
13. (16分)符号“f”表示一种运算,它对一些数的运算如下：$f(1)= 1+\frac {2}{1},f(2)= 1+\frac {2}{2},f(3)= 1+\frac {2}{3},f(4)= 1+\frac {2}{4},...$
(1)利用以上运算的规律写出$f(n)= $______(n为正整数)；
(2)计算：$f(1)\cdot f(2)\cdot f(3)\cdot ... \cdot f(100)$.
(1)利用以上运算的规律写出$f(n)= $______(n为正整数)；
(2)计算：$f(1)\cdot f(2)\cdot f(3)\cdot ... \cdot f(100)$.
答案
（1）$1+\frac{2}{n}$ （2）$f(1)\cdot f(2)\cdot f(3)\cdot\cdots\cdot f(100)=\left(1+\frac{2}{1}\right)×\left(1+\frac{2}{2}\right)×\left(1+\frac{2}{3}\right)×\cdots×\left(1+\frac{2}{100}\right)=\frac{3}{1}×\frac{4}{2}×\frac{5}{3}×\cdots×\frac{102}{100}=\frac{101×102}{1×2}=5151$
