\(\sqrt{2+3x}-3\sqrt{8+12x}+\sqrt{50+75x}=5\left(đk:x\ge-\dfrac{2}{3}\right)\)
\(\Leftrightarrow\sqrt{2+3x}-6\sqrt{2+3x}+5\sqrt{2+3x}=5\)
\(\Leftrightarrow0=5\left(VLý\right)\)
Vậy \(S=\varnothing\)
\(\sqrt{2+3x}-3\sqrt{8+12x}+\sqrt{50+75x}=5\left(x\ge-\dfrac{2}{3}\right)\\ \Leftrightarrow\sqrt{2+3x}-6\sqrt{2+3x}+5\sqrt{2+3x}=5\\ \Leftrightarrow0\sqrt{2+3x}=5\\ \Leftrightarrow x\in\varnothing\)
g: Ta có: \(\sqrt{3x+2}-3\sqrt{12x+8}+\sqrt{75x+50}=5\)
\(\Leftrightarrow\sqrt{3x+2}-6\sqrt{3x+2}+5\sqrt{3x+2}=5\)
\(\Leftrightarrow0\sqrt{3x+2}=5\)(vô lý)
