Question

Giải phương trình:

a) \(\sqrt{4x^2-4x+1}=2x+1\)

b) \(\sqrt{x+3}+4\sqrt{x}-2x=6-\sqrt{5-x}\)

Answer 1

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:

\(\left\{{}\begin{matrix}x+3=5-x\\\sqrt{x}-1=0\end{matrix}\right.\) \(\Rightarrow x=1\)

Answer 2

a) ĐK: x ≥\(\frac{-1}{2}\)
\(\sqrt{4x^2-4x+1}=2x+1\)

⇔ \(4x^2-4x+1=\left(2x+1\right)^2\)

⇔\(4x^2-4x+1=4x^2+4x+1\)

\(\Leftrightarrow x=0\)