g) Ta có: \(2\sqrt{9x+9}-\dfrac{1}{4}\sqrt{16x+16}-5\sqrt{x+1}=-6-\sqrt{4x+4}\)
\(\Leftrightarrow6\sqrt{x+1}-\sqrt{x+1}-5\sqrt{x+1}+2\sqrt{x+1}=-6\)
\(\Leftrightarrow\sqrt{x+1}=-3\)(Vô lý)
Vậy: \(x\in\varnothing\)
h) Ta có: \(\sqrt{5-x^2}=x-1\)
\(\Leftrightarrow5-x^2=x^2-2x+1\)
\(\Leftrightarrow x^2-2x+1-5+x^2=0\)
\(\Leftrightarrow2x^2-2x-4=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow x=2\)
i) Ta có: \(\dfrac{1}{3}\sqrt{9x+9}-2\sqrt{x+1}+8\sqrt{\dfrac{4x+4}{25}}=11\)
\(\Leftrightarrow\sqrt{x+1}-2\sqrt{x+1}+8\cdot\dfrac{2}{5}\sqrt{x+1}=11\)
\(\Leftrightarrow\dfrac{11}{5}\cdot\sqrt{x+1}=11\)
\(\Leftrightarrow\sqrt{x+1}=5\)
\(\Leftrightarrow x+1=25\)
hay x=24
k) Ta có: \(\sqrt{x^2+x+1}=x+1\)
\(\Leftrightarrow x^2+x+1=x^2+2x+1\)
\(\Leftrightarrow x=0\)
