Question

Tìm x

3(4x-5)^2-8x+10=0

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Answer 1

\(3\left(4x-5\right)^2-8x+10=0\)

<=> \(3\left(16x^2-40x+25\right)-8x+10=0\)

<=> \(48x^2-120x+75-8x+10=0\)

<=> \(48x^2-128x+85=0\)

<=> \(48x^2-68x-60x-85=0\)

<=> \(48x\left(x-\frac{17}{12}\right)-60\left(x-\frac{17}{12}\right)=0\)

<=> \(\left(48x-60\right)\left(x-\frac{17}{12}\right)=0\)

<=> \(\hept{\begin{cases}48x-60=0\\x-\frac{17}{12}=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=\frac{5}{4}\\x=\frac{17}{12}\end{cases}}\)