Question

Giải phương trình \(\sqrt{x^2-x+3}+\sqrt{x^2+x+4}=7\)

Answer 1

\(\sqrt{x^2-x+3}+\sqrt{x^2+x+4}=7\)

\(\Leftrightarrow x^2-x+3+x^2+x+4+2\sqrt{\left(x^2-x+3\right)\left(x^2+x+4\right)}=49\)

\(\Leftrightarrow2x^2+4+2\sqrt{x^4+6x^2-x+12}=49\)

\(\Leftrightarrow x^2+\sqrt{x^4+6x^2-x+12}=21\)

\(\Leftrightarrow x^4+6x^2-x+12=\left(21-x^2\right)^2\)

\(\Leftrightarrow x^4+6x^2-x+12=x^4-42x^2+441\)

\(\Leftrightarrow48x^2-x-429=0\)

\(\Leftrightarrow\left(x-3\right)\left(48x+143\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{-143}{48}\end{matrix}\right.\)( thỏa )

Vậy....