Question

\(x-3\sqrt{x}-4=0\)

\(\sqrt{2x-1}+\sqrt{x-1}=5\)

\(x^2+2x+7=3\sqrt{x^2+1)\left(x+3\right)}\)

Answer 1

a: \(\Leftrightarrow\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)=0\)

=>căn x-4=0

=>x=16

b: \(\Leftrightarrow2x-1+x-1+2\sqrt{\left(2x-1\right)\left(x-1\right)}=25\)

\(\Leftrightarrow2\sqrt{\left(2x-1\right)\left(x-1\right)}=25-3x+2=-3x+23\)

\(\Leftrightarrow4\left(2x-1\right)\left(x-1\right)=\left(-3x+23\right)^2\)

\(\Leftrightarrow4\left(2x^2-3x+1\right)=\left(3x-23\right)^2\)

\(\Leftrightarrow9x^2-138x+529-8x^2+12x-4=0\)

=>x^2-126x+525=0

hay \(x\in\left\{121.69;4.31\right\}\)