Đề bài \(ĐK\left(x\ge-\frac{3}{2}\right)\)
\(=>\left(x-3\right)^2+\left(\sqrt{2x+3}-3\right)^2=0\)
mà \(\left(x-3\right)^2+\left(\sqrt{2x+3}-3\right)^2\ge0\)
dấu = xảy ra khi x=3 (chọn )
zậy...
:V cách khác
Ta có:
\(x^2-4x+21=6\sqrt{2x+3}\left(x\ge-\frac{3}{2}\right)\)
\(\Leftrightarrow x^2-4x+21-18=6\left(\sqrt{2x+3}-3\right)\)
\(\Leftrightarrow x^2-4x+3=6\cdot\frac{2x-6}{\sqrt{2x+3}+3}\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)-\frac{12\left(x-3\right)}{\sqrt{2x+3}+3}=0\)
\(\Leftrightarrow\left(x-3\right)\left[x-1-\frac{12}{\sqrt{2x+3}+3}\right]=0\)
:V
