Sửa lại câu h

h) \(x^2-4x+1=0\)

<=> \(\left(x^2-4x+4\right)-3=0\)

<=> \(\left(x-2\right)^2=3\)

<=> \(\left[{}\begin{matrix}x-2=\sqrt{3}\\x+2=-\sqrt{3}\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=\sqrt{3}+2\\x=2-\sqrt{3}\end{matrix}\right.\)

Vậy nghiệm của PT là S = \(\left\{2+\sqrt{3};2-\sqrt{3}\right\}\)

Nhân ra là ok!

f) \(4x^2-12x+9=0\)

<=> \(\left(2x-3\right)^2\) = 0

<=> \(2x-3=0\)

<=> \(2x=3\) <=> \(x=\dfrac{3}{2}\)

Vậy ...............

g) \(3x^2+7x+2=0\)

<=> \(\left(3x^2+6x\right)+\left(x+2\right)=0\)

<=> \(3x\left(x+2\right)+\left(x+2\right)=0\)

<=> \(\left(x+2\right)\left(3x+1\right)=0\)

<=> \(\left[{}\begin{matrix}x=-2\\x=\dfrac{-1}{3}\end{matrix}\right.\)

Vậy ........................

h) \(x^2-4x+1=0\)

<=> \(\left(x^2-4x+4\right)-3=0\)

<=> \(\left(x-2\right)^2=3\)

<=> \(\left[{}\begin{matrix}x+2=\sqrt{3}\\x+2=-\sqrt{3}\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=\sqrt{3}-2\\x=-\sqrt{3}-2\end{matrix}\right.\)

Vậy .........................

i) \(2x^2-6x+1=0\)

<=> \(2\left(x^2-3x+2,25\right)-3,5=0\)

<=> \(\left(x-1,5\right)^2=1,75\)

<=> \(\left[{}\begin{matrix}x-1,5=\sqrt{1,75}\\x-1,5=-\sqrt{1,75}\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=\sqrt{1,75}+1,5\\x=-\sqrt{1,75}+1,5\end{matrix}\right.\)

Vậy ...................

j) \(3x^2+4x-4=0\)

<=> \(\left(3x^2+6x\right)-\left(2x+4\right)=0\)

<=> \(3x\left(x+2\right)-2\left(x+2\right)\) = 0

<=> \(\left(x+2\right)\left(3x-2\right)=0\)

<=> \(\left[{}\begin{matrix}x=-2\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy ....................................