Question

[Nguyễn Kim Oanh]

Có ai lm đc bài này k ạ.. giúp mik vs... cảm mơn trc nhé.

\(\sqrt{x+2-3\sqrt{2x-5}}+\sqrt{x-2-\sqrt{2x-5}}=\sqrt{8}\)

Answer 1

[Phương An]

\(\sqrt{x+2-3\sqrt{2x-5}}+\sqrt{x-2-\sqrt{2x-5}}=\sqrt{8}\)

\(\Leftrightarrow\sqrt{2x+4-6\sqrt{2x-5}}+\sqrt{2x-4-2\sqrt{2x-5}}=4\)

\(\Leftrightarrow\sqrt{\left(\sqrt{2x-5}-3\right)^2}+\sqrt{\left(\sqrt{2x-5}-1\right)^2}=4\)

\(\Leftrightarrow\left|\sqrt{2x-5}-3\right|+\left|\sqrt{2x-5}-1\right|=4\) (1)

(~ ~ ~) Với \(\dfrac{5}{2}\le x< 3\)

\(\left(1\right)\Leftrightarrow4-2\sqrt{2x-5}=4\)

\(\Leftrightarrow\sqrt{2x-5}=0\)

\(\Leftrightarrow x=\dfrac{5}{2}\) (nhận)

(~ ~ ~) Với \(3\le x\le7\)

=> pt vô nghiệm

(~ ~ ~) Với 7 < x

\(\left(1\right)\Leftrightarrow2\sqrt{2x-5}-4=4\)

\(\Leftrightarrow4\left(2x-5\right)=64\)

\(\Leftrightarrow x=\dfrac{64+20}{8}\)

\(\Leftrightarrow x=\dfrac{21}{2}\) (nhận)

Vậy \(x\in\left\{\dfrac{5}{2};\dfrac{21}{2}\right\}\)