Đáp án:
a, $\sqrt[]{x^2-2x+1}$ `-2(x-2) = 3`
⇔ $\sqrt[]{(x-1)^2}$ `-2(x-2) = 3`
`⇔ |x-1|-(2x-4) = 3`
`⇒ TH1 : (x-1) - (2x-4) = 3`
`⇔ x-1 - 2x+4 = 3`
`⇔ -x+3 = 3`
`⇔ -x = 0`
`⇔ x = 0` `
TH2 : `-(x-1) - (2x-4) = 3`
`⇔ 1-x-2x+4 = 3`
`⇔ -3x+5 = 3`
`⇔ -3x = -2`
`⇔ x = 2/3`
b, $\frac{\sqrt[]{4x^2-1} }{\sqrt[]{2x-1}}$ `=3`
⇔ $\frac{\sqrt[]{(2x-1)(2x+1)} }{\sqrt[]{2x-1}}$ `=3`
⇔ $\sqrt[]{\frac{(2x-1)(2x+1)}{2x-1}}$ `=3`
⇔ $\sqrt[]{2x+1}$ `=3`
`⇔ 2x+1 = 9`
`⇔ 2x = 8`
`⇔ x = 4`
c, $\frac{\sqrt[]{4x^2+4x+1}}{x-2}$ `=3`
`⇔ $\frac{\sqrt[]{(2x+1)^2}}{x-2}$ `=3`
⇔ $\frac{|2x+1|}{x-2}$ `=3`
`⇔ (2x+1)/(x-2) = 3`
`⇔ 2x+1 = 3(x-2)`
`⇔ 2x+1 = 3x-6`
`⇔ -x = -7`
`⇔ x = 7`
