Question

\(\sqrt{2x^2+3x+1}+\sqrt{1-3x}=2\sqrt{x^2+1}\)

Answer 1

Đặt \(\sqrt{1-3x}=a;\sqrt{x^2+1}=b\left(b>0;a\ge0\right)\)

\(\sqrt{2x^2+3x+1}=\sqrt{2\left(x^2+1\right)+\left(3x-1\right)}=\sqrt{2b^2-a^2}\)

\(\Leftrightarrow\sqrt{2b^2-a^2}+a=2b\)

\(\Leftrightarrow\sqrt{2b^2-a^2}=2b-a\) (2b ≥ a)

Bình phương lên:

\(2b^2-a^2=4b^2-4ab+a^2\)

\(\Leftrightarrow2b^2+2a^2-4ab=0\)

\(\Leftrightarrow a^2+b^2-2ab=0\)

\(\Leftrightarrow\left(a-b\right)^2=0\)

Tự giải tiếp đc ko ạ ??