toán 8 giải phương trình

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nhokdangyeu01

$\frac{x+1}{x-1}+\frac{1}{x}=\frac{-1}{x^2-1}$ ĐKXĐ: x#-1;0;1
\Leftrightarrow $\frac{x(x+1)^2+(x+1)(x-1)}{x(x+1)(x-1)}=\frac{-x}{x(x-1)(x+1)}$
\Leftrightarrow $x^3+2x^2+x+x^2-1=-x$
\Leftrightarrow $x^3+3x^2+2x-1=0$
\Leftrightarrow $x=\sqrt[3]{\frac{1}{2}+\sqrt[]{\frac{23}{108}}}+\sqrt[3]{\frac{1}{2}-\sqrt[]{\frac{23}{108}}}-1$
 
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