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ĐK \(x\ge-5\)
Ta có: \(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow\sqrt{x+5}\left(\sqrt{4}-3+\dfrac{4}{3}\sqrt{9}\right)=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Rightarrow x=\left(\dfrac{6}{3}\right)^2-5=-1\)(TMĐK)
Vậy x=-1
\(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow\sqrt{2^2\left(x+5\right)}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45=6}\)\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow\sqrt{x+5}=2\)
\(\Leftrightarrow x+5=4\)
\(\Leftrightarrow x=-1\)
vậy có tập nghiệm S là -1
điều kiện xác định \(x\ge-5\)
\(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\\ \Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\\ \Leftrightarrow3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=2\\ \Leftrightarrow x+5=4\\ \Leftrightarrow x=-1\left(TMĐK\right)\)
vậy x=-1
