Question

Tìm x: a, (x-1)(x+1)-2x=0
           b, (2x-3)(x-1)=\(^{2x^2}\)
           c, (\(\frac{1}{2}x+3\)) ( \(\frac{2}{3}-x\)) = \(\frac{x^2}{2}+1\)

Answer 1

b) \(\left(2x-3\right)\left(x-1\right)=2x^2\)

<=> \(2x^2-5x+3=2x^2\)

<=> x=3/5

c) \(\left(\frac{1}{2}x+3\right)\left(\frac{2}{3}-x\right)=\frac{x^2}{2}+1\)

<=> \(\frac{1}{3x}-\frac{1}{2}x^2+2-3x=\frac{x^2}{2}+1\)

<=> \(x^2+\frac{8}{3}x-1=0\)

<=> \(\left[\begin{array}{nghiempt}x=-3\\x=\frac{1}{3}\end{array}\right.\)

Answer 2

Tìm x: a, (x-1)(x+1)-2x=0

<=> \(x^2-1-2x=0\)

<=> \(x=1\pm\sqrt{2}\)

KL: có 2 nghiệm ...
           b, (2x-3)(x-1)=2x²

^{<=> \(2x^2-5x+3=2x^2\)}

^{<=> \(x=\frac{3}{5}\)}

            ^{c) \(\left(\frac{1}{2}x+3\right)\left(\frac{2}{3}-x\right)=\frac{x^2}{2}+1\)}

^{<=> \(\frac{1}{3}x-\frac{1}{2}x^2+2-3x=\frac{x^2}{2}+1\)}

^{<=> \(x^2+\frac{8}{3}x-1=0\)}

^{<=> \(\left[\begin{array}{nghiempt}x=-3\\x=\frac{1}{3}\end{array}\right.\)}

^{KL: có 2 nghiệm ..}