Question

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Answer 1

[HT.Phong (9A5)]

\(c)\left(2x-3\right)^2-\left(x-1\right)=0\\ \Leftrightarrow\left(4x^2-12x+9\right)-\left(x-1\right)=0\\ \Leftrightarrow4x^2-12x+9-x+1=0\\ \Leftrightarrow4x^2-13x+10=0\\ \Leftrightarrow\left(4x^2-8x\right)-\left(5x-10\right)=0\\ \Leftrightarrow4x\left(x-2\right)-5\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(4x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\4x-5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{4}\end{matrix}\right.\)

\(d)\left(3x-2\right)\cdot\left(3x+2\right)-9\left(x-1\right)\cdot x=0\\ \Leftrightarrow\left(3x\right)^2-2^2-9\left(x-1\right)x=0\\ \Leftrightarrow9x^2-4-9x^2+9x=0\\ \Leftrightarrow9x-4=0\\ \Leftrightarrow9x=4\\ \Leftrightarrow x=\dfrac{4}{9}\)