[Toán 8] Phương trình

C

chaobanhao

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H

hanh7a2002123

$\frac{x - 1}{x} - \frac{1}{x + 1} = \frac{2x - 1}{x^2 + x}$
ĐKXĐ: $x\neq 0; x\neq -1$
\Leftrightarrow $\frac{(x-1)(x+1)}{x(x+1)}-\frac{x}{x(x+1)}=\frac{2x-1}{x^2+x}$
\Leftrightarrow $\frac{x^2-1}{x^2+x}-\frac{x}{x^2+x}=\frac{2x-1}{x^2+x}$
\Leftrightarrow $\frac{x^2-x-1}{x^2+x}=\frac{2x-1}{x^2+x}$
Vì $x\neq 0$
\Rightarrow $x^2-x-1=2x-1$
\Leftrightarrow $x^2=3x$
\Leftrightarrow $ x=3$ ( tm đkxđ)
 
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