a/ ĐKXĐ: ....
\(\Leftrightarrow2x^2-5x-3+1-\sqrt{x-2}+1-\sqrt{4-x}=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+1\right)-\frac{x-3}{1+\sqrt{x-2}}+\frac{x-3}{1+\sqrt{4-x}}=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+1-\frac{1}{1+\sqrt{x-2}}+\frac{1}{1+\sqrt{4-x}}\right)=0\)
\(\Leftrightarrow x=3\)
b/ ĐKXĐ: ...
\(\Leftrightarrow3\left(\sqrt{1-x}+1\right)=\sqrt{x+3}+\sqrt{x}\)
Ta có \(-3\le x\le1\Rightarrow\sqrt{1-x}+1\ge1\Rightarrow VT\ge3\)
\(\left\{{}\begin{matrix}\sqrt{x+3}\le2\\\sqrt{x}\le1\end{matrix}\right.\) \(\Rightarrow VP\le3\)
Dấu "=" xảy ra khi và chỉ khi \(x=1\)
c/ ĐKXĐ: ....
\(\Leftrightarrow2x+12-6\sqrt{2x+3}+5-x-2\sqrt{4-x}=0\)
\(\Leftrightarrow\frac{\left(2x+12\right)^2-36\left(2x+3\right)}{\left(2x+12\right)^2+6\sqrt{2x+3}}+\frac{\left(5-x\right)^2-4\left(4-x\right)}{\left(5-x\right)^2+2\sqrt{4-x}}=0\)
\(\Leftrightarrow\frac{4\left(x-3\right)^2}{\left(2x+12\right)^2+6\sqrt{2x+3}}+\frac{\left(x-3\right)^2}{\left(5-x\right)^2+2\sqrt{4-x}}=0\)
\(\Leftrightarrow\left(x-3\right)^2\left(\frac{4}{\left(2x+12\right)^2+6\sqrt{2x+3}}+\frac{1}{\left(5-x\right)^2+2\sqrt{4-x}}\right)=0\)
\(\Rightarrow x=3\)
