\(\sqrt{x-2}+\sqrt{4-x}+\sqrt{2x-5}=2x^2-5x\)
ĐK:\(\dfrac{5}{2}\le x\le4\)
\(pt\Leftrightarrow\sqrt{x-2}-1+\sqrt{4-x}-1+\sqrt{2x-5}-1=2x^2-5x-3\)
\(\Leftrightarrow\dfrac{x-2-1}{\sqrt{x-2}+1}+\dfrac{4-x-1}{\sqrt{4-x}+1}+\dfrac{2x-5-1}{\sqrt{2x-5}+1}=\left(x-3\right)\left(2x+1\right)\)
\(\Leftrightarrow\dfrac{x-3}{\sqrt{x-2}+1}-\dfrac{x-3}{\sqrt{4-x}+1}+\dfrac{2\left(x-3\right)}{\sqrt{2x-5}+1}-\left(x-3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(\dfrac{1}{\sqrt{x-2}+1}-\dfrac{1}{\sqrt{4-x}+1}+\dfrac{2}{\sqrt{2x-5}+1}-2x-1\right)=0\)
Dễ thấy: \(\dfrac{1}{\sqrt{x-2}+1}-\dfrac{1}{\sqrt{4-x}+1}+\dfrac{2}{\sqrt{2x-5}+1}-2x-1< 0\forall\dfrac{5}{2}\le x\le4\)
\(\Rightarrow x-3=0\Rightarrow x=3\)
Em có cách khác:D Mặc dù ko chắc nhưng ngứa tay nên đăng:v
ĐK:\(4\ge x\ge\frac{5}{2}\)
\(PT\Leftrightarrow2x^2-8x+6+\sqrt{x-2}\left(\sqrt{x-2}-1\right)+\left(1-\sqrt{4-x}\right)+\sqrt{2x-5}\left(\sqrt{2x-5}-1\right)=0\)
\(\Leftrightarrow2\left(x-3\right)\left(x-1\right)+\sqrt{x-2}\left(\frac{x-3}{\sqrt{x-2}+1}\right) +\frac{x-3}{\sqrt{4-x}+1}+\sqrt{2x-5}\left(\frac{2\left(x-3\right)}{\sqrt{2x-5}+1}\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left[2\left(x-1\right)+\frac{\sqrt{x-2}}{\sqrt{x-2}+1}+\frac{1}{\sqrt{4-x}+1}+\frac{2\sqrt{2x-5}}{\sqrt{2x-5}+1}\right]=0\)
Dễ thấy cái ngoặc to luôn >0 với mọi \(4\ge x\ge\frac{5}{2}\)
Cái này có vẻ xử lí cái ngoặc to nhanh hơn nhỉ:D
