\(\frac{3^{10}\cdot\left(-5\right)^{21}}{\left(-5\right)^{20}\cdot3^{12}}=\)
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\(\frac{5.\left(2^2.3^2\right)^9.\left(2^2\right)^6-2.\left(2^2.3\right)^{14}.3^4}{5.2^{28}.3^{18}+7.2^{29}.3^{18}}\)
\(=\frac{5.2^{18}.3^{18}.2^{12}-2.2^{28}.3^{14}.3^4}{2^{28}.3^{18}.\left(5+7.2\right)}\)
\(=\frac{5.2^{30}.3^{18}-2^{29}.3^{18}}{2^{28}.3^{18}.19}=\frac{2^{28}.3^{18}.\left(5.4-2\right)}{2^{28}.3^{18}.19}\)
\(=\frac{5.4-2}{19}=\frac{18}{19}\)
a)
Điều kiện: $x\ne-2,-5,-10,-17.$
$\dfrac3{(x+2)(x+5)}+\dfrac5{(x+5)(x+10)}+\dfrac7{(x+10)(x+17)}=\dfrac{x}{(x+2)(x+17)}$
$\Leftrightarrow\dfrac1{x+2}-\dfrac1{x+5}+\dfrac1{x+5}-\dfrac1{x+10}+\dfrac1{x+10}-\dfrac1{x+17}=\dfrac{x}{(x+2)(x+17)}$
$\Leftrightarrow\dfrac1{x+2}-\dfrac1{x+17}=\dfrac{x}{(x+2)(x+17)}$
$\Leftrightarrow\dfrac{15}{(x+2)(x+17)}=\dfrac{x}{(x+2)(x+17)}$
$\Leftrightarrow x=15.$
b)
Điều kiện: $x\ne1,3,8,20.$
$\dfrac2{(x-1)(x-3)}+\dfrac5{(x-3)(x-8)}+\dfrac{12}{(x-8)(x-20)}-\dfrac1{x-20}=-\dfrac34$
$\Leftrightarrow\left(\dfrac1{x-3}-\dfrac1{x-1}\right)+\left(\dfrac1{x-8}-\dfrac1{x-3}\right)+\left(\dfrac1{x-20}-\dfrac1{x-8}\right)-\dfrac1{x-20}=-\dfrac34$
$\Leftrightarrow-\dfrac1{x-1}=-\dfrac34$
$\Leftrightarrow\dfrac1{x-1}=\dfrac34$
$\Leftrightarrow4=3(x-1)$
$\Leftrightarrow3x=7$
$\Leftrightarrow x=\dfrac73.$
$\textbf{a)}$
$\dfrac{x-1}{2}+\dfrac{x-2}{5}=\dfrac14+\dfrac{x-7}{10}$
$\Leftrightarrow\dfrac{5(x-1)+2(x-2)}{10}=\dfrac14+\dfrac{x-7}{10}$
$\Leftrightarrow\dfrac{7x-9}{10}=\dfrac14+\dfrac{x-7}{10}$
$\Leftrightarrow14x-18=5+2x-14$
$\Leftrightarrow12x=9$
$\Leftrightarrow x=\dfrac34.$
$\textbf{b)}$
$\dfrac{3-2}{2x-3}=\dfrac25+\dfrac1{2x-3}-\dfrac32$
$\Leftrightarrow1-\dfrac2{2x-3}=-\dfrac{11}{10}+\dfrac1{2x-3}$
$\Leftrightarrow\dfrac{21}{10}=\dfrac3{2x-3}$
$\Leftrightarrow21(2x-3)=30$
$\Leftrightarrow42x=93$
$\Leftrightarrow x=\dfrac{31}{14}.$


\(\frac{3^{10}.\left(-5\right)^{20}.\left(-5\right)}{3^{10}.3^2.\left(-5\right)^{20}}=\frac{-5}{3^2}=\frac{-5}{9}\)
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