\(ĐK:x\ge\frac{1}{2}\)
Bình phương 2 vế ta dc:
\(x^2+2x+2x-1+2\sqrt{\left(x^2+2x\right)\left(2x-1\right)}=3x^2+4x+1\)
\(\Leftrightarrow3x^2+4x+1-x^2-2x-2x+1=2\sqrt{\left(x^2+2x\right)\left(2x-1\right)}\)
\(\Leftrightarrow2x^2+2=2\sqrt{\left(x^2+2x\right)\left(2x-1\right)}\)
\(\Leftrightarrow x^2+1=\sqrt{\left(x^2+2x\right)\left(2x-1\right)}\)
\(\Rightarrow x^4+2x^2+1=2x^3+3x^2-2x\)
\(\Leftrightarrow x^4+2x^2+1-2x^3-3x^2+2x=0\)
\(\Leftrightarrow\left(x^2-x-1\right)^2=0\Leftrightarrow x^2-x-1=0\)
\(\Delta=\left(-1\right)^2-4.\left(-1\right)=5>0\)
\(\Rightarrow x_1=\frac{1+\sqrt{5}}{2}\left(TM\right);x_2=\frac{1-\sqrt{5}}{2}\left(loai\right)\)
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