Question

Giải các phương trình tích sau:

a) 9(x-3)^2= 4(x+2)^2

b) (x+1)^2= 4(x^2-2x+1)^2

Answer 1

a) \(9\left(x-3\right)^2=4\left(x+2\right)^2\Leftrightarrow9\left(x^2-6x+9\right)=4\left(x^2+4x+4\right)\)

\(\Leftrightarrow9x^2-54x+81=4x^2+16x+16\)

\(\Leftrightarrow9x^2-54x+81-4x^2-16x-16=0\Leftrightarrow5x^2-70x+65=0\)

\(\Leftrightarrow5x^2-5x-65x+65=0\Leftrightarrow5x\left(x-1\right)-65\left(x-1\right)=0\)

\(\Leftrightarrow\left(5x-65\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}5x-65=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}5x=65\\x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{65}{5}=13\\x=1\end{matrix}\right.\) vậy \(x=13;x=1\)

Answer 2

a. \(9\left(x-3\right)^2=4\left(x+2\right)^2\Leftrightarrow\left(3\left(x-3\right)\right)^2-\left(2\left(x+2\right)\right)^2=0\Leftrightarrow\left(3x-9+2x+4\right)\left(3x-9-2x-4\right)=0\Leftrightarrow\left(5x-5\right)\left(x-13\right)=0\Leftrightarrow\left[{}\begin{matrix}5x-5=0\\x-13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=13\end{matrix}\right.\)

b. \(\left(x+1\right)^2=4\left(x^2-2x+1\right)^2\Leftrightarrow\left(x+1\right)^2=4\left(\left(x-1\right)^2\right)^2\Leftrightarrow\left(x+1\right)^2-\left(2\left(x-1\right)^2\right)^2=0\Leftrightarrow\left(x+1+2x^2-4x+2\right)\left(x+1-2x^2+4x-2\right)=0\Leftrightarrow\left(2x^2-3x+3\right)\left(-2x^2+5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}2x^2-3x+3=0\\-2x^2+5x-1=0\end{matrix}\right.\Leftrightarrow x=\dfrac{5\pm\sqrt{17}}{4}\)

Vì \(2x^2-3x+3>0\)