Question

giải phương trình

a) \(\sqrt{2x^2-\sqrt{2}x+\frac{1}{4}}=\sqrt{2}x\)

b) \(\sqrt{4x+8}+\frac{1}{3}\sqrt{9x+18}=3\sqrt{\frac{x+2}{4}}+\sqrt{2}\)

Answer 1

b) \(\sqrt{4x+8}+\frac{1}{3}\sqrt{9x+18}=3\sqrt{\frac{x+2}{4}}+\sqrt{2}\)

⇔ \(2\sqrt{x+2}+\frac{1}{3}\cdot3\sqrt{x+2}=\frac{3\sqrt{x+2}}{2}+\sqrt{2}\)

⇔ \(3\sqrt{x+2}-\frac{3\sqrt{x+2}}{2}=\sqrt{2}\)

⇔ \(\frac{3\sqrt{x+2}}{2}=\sqrt{2}\)

⇔ \(\frac{3}{2}=\frac{\sqrt{2}}{\sqrt{x-2}}\)

⇔ \(\sqrt{\frac{9}{4}}=\sqrt{\frac{2}{x+2}}\)

⇔ \(\frac{2}{x+2}=\frac{9}{4}\)

⇔ \(x+2=\frac{8}{9}\)

⇔ \(x=\frac{8}{9}-2=-\frac{10}{9}\)

Answer 2

a) \(\sqrt{2x^2-\sqrt{2}x+\frac{1}{4}}=\sqrt{2}x\)

⇔ \(2x^2-\sqrt{2}x+\frac{1}{4}=2x^2\)

⇔ \(-\sqrt{2}x+\frac{1}{4}=0\)

⇔ \(\sqrt{2}x=\frac{1}{4}\)

⇔ \(x=\frac{\sqrt{2}}{8}\)