Question

\(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)   tìm x

 

Answer 1

Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)

\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)

\(\Leftrightarrow3\sqrt{x+5}=6\)

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

\(\Leftrightarrow x+5=4\)

hay x=-1

Answer 2

\(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\cdot\sqrt{9x+45}=6\left(x\ge-5\right)\)

\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+\dfrac{4}{3}\cdot3\cdot\sqrt{x+5}=6\)

\(\Leftrightarrow3\cdot\sqrt{x+5}=6\)

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

\(\Leftrightarrow x+5=2^2=4\)

\(\Leftrightarrow x=-1\left(N\right)\)