a) \(\sqrt{\frac{3x-1}{x+2}}\) = 2
⇔ \(\left(\text{}\sqrt{\frac{3x-1}{x+2}}\right)^2\) = 22
⇔ \(\frac{3x-1}{x+2}\) = 4
⇔ 3x - 1 = 4.(x+2)
⇔ 4x + 8 - 3x +1 = 0
⇔ x + 9 = 0
Vậy: x = -9
b) \(\frac{\sqrt{5x-7}}{\sqrt{2x-1}}\) = 1
⇔ \(\left(\text{}\sqrt{\frac{5x-7}{2x-1}}\right)^2\) = 12
⇔ \(\frac{5x-7}{2x-1}\) = 1
⇔ 5x - 7 = 1(2x -1)
⇔ 5x - 7 - 2x +1 = 0
⇔ 3x - 6 = 0
⇔ 3x = 6
Vậy: x = 2
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