a/ ĐK: \(x\ge-\frac{1}{2}\)
\(\Leftrightarrow4x^2-4x+1=\left(2x+1\right)^2\)
\(\Leftrightarrow4x^2-4x+1=4x^2+4x+1\)
\(\Rightarrow x=0\)
b/ ĐKXĐ: \(0\le x\le5\)
\(\Leftrightarrow\sqrt{x+3}+\sqrt{5-x}=2\left(x-2\sqrt{x}+1\right)+4\)
\(\Leftrightarrow\sqrt{x+3}+\sqrt{5-x}=2\left(\sqrt{x}-1\right)^2+4\)
Ta có \(VT\le\sqrt{\left(1+1\right)\left(x+3+5-x\right)}=4\)
\(\left(\sqrt{x}-1\right)^2\ge0\Rightarrow VP\ge4\)
\(\Rightarrow VT\le VP\)
Dấu "=" xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}x+3=5-x\\\sqrt{x}-1=0\end{matrix}\right.\) \(\Rightarrow x=1\)
a) ĐK: x ≥\(\frac{-1}{2}\)
\(\sqrt{4x^2-4x+1}=2x+1\)
⇔ \(4x^2-4x+1=\left(2x+1\right)^2\)
⇔\(4x^2-4x+1=4x^2+4x+1\)
\(\Leftrightarrow x=0\)
