(3x+2)(1-2x)=4x2-1
\(\Leftrightarrow\)3x-6x2+2-4x = 4x2-1
\(\Leftrightarrow\)3x-6x2-4x-4x2 = -1-2
\(\Leftrightarrow\)-10x2-x = -3
\(\Leftrightarrow-10x^2-x+3=0\)
\(\Leftrightarrow-10x^2-6x+5x+3=0\)
\(\Leftrightarrow-\left(10x^2+6x\right)+\left(5x+3\right)=0\)
\(\Leftrightarrow-2x\left(5x+3\right)+\left(5x+3\right)=0\)
\(\Leftrightarrow\)(5x+3)(-2x+1) = 0
\(\Leftrightarrow\left[{}\begin{matrix}5x+3=0\\-2x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-3}{5}\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy pt có tập nghiệm S = \(\left\{\frac{-3}{5};\frac{1}{2}\right\}\)
\(\Leftrightarrow-10x^{^{ }}^2-6x+5x+3=0\)
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