Đáp án:
Giải thích các bước giải:
`B=(\frac{21}{x^{2}-9}-\frac{x-4}{3-x}-\frac{x-1}{3+x}):(1-\frac{1}{x+3})` (ĐKXĐ: $x\neq3;-3$)
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`=(\frac{21}{x^{2}-9}+\frac{x-4}{x-3}-\frac{x-1}{x+3}):\frac{x+2}{x+3}`
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`=(\frac{21}{(x+3).(x-3)}+\frac{(x+3).(x-4)}{(x+3).(x-3)}-\frac{(x-1).(x-3)}{(x+3).(x-3)}).\frac{x+3}{x+2}`
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`=\frac{6+3x}{(x+3).(x-3)}.\frac{x+3}{x+2}`
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`=\frac{3.(2+x)}{(x+3).(x-3)}\frac{x+3}{x+2}`
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`=\frac{3}{x-3}`
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`b)|2x+1|=5`
$⇔$\(\left[ \begin{array}{l}2x+1=5\\2x+1=-5\end{array} \right.\)
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$⇔$\(\left[ \begin{array}{l}2x=4\\2x=-6\end{array} \right.\)
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$⇔$\(\left[ \begin{array}{l}x=2\\x=-3(loại)\end{array} \right.\)
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Thay vào $B$ ta có:
$B=\dfrac{3}{2-3}$
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$=\dfrac{3}{-1}=-3$
