a)\(3x^2+6x-3=\sqrt{\frac{x+7}{3}}\)
Đk:\(x\ge-7\)
\(pt\Leftrightarrow9x^4+36x^3+18x^2-36x+9=\frac{x+7}{3}\)
\(\Leftrightarrow9x^4+36x^3+18x^2-36x+9-\frac{x+7}{3}=0\)
\(\Leftrightarrow\left(x^2+\frac{5x}{3}-\frac{4}{3}\right)\left(9x^2+21x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{\sqrt{69}+7}{6}\\x=\frac{\sqrt{73}-5}{6}\end{cases}}\) (thỏa)
b)\(2x^2+2x+1=\left(2x+3\right)\left(\sqrt{x^2+x+2}-1\right)\)
\(\Leftrightarrow2x^2+2x+1=\left(2x+3\right)\sqrt{x^2+x+2}-2x-3\)
\(\Leftrightarrow2x^2+4x+4=\left(2x+3\right)\sqrt{x^2+x+2}\)
\(\Leftrightarrow\frac{2x^2+4x+4}{2x+3}=\sqrt{x^2+x+2}\)
\(\Leftrightarrow\frac{2x^2+4x+4}{2x+3}-2x=\sqrt{x^2+x+2}-2x\)
\(\Leftrightarrow\frac{2x^2+4x+4}{2x+3}-2x=\frac{x^2+x+2-4x^2}{\sqrt{x^2+x+2}+2x}\)
\(\Leftrightarrow\frac{-2\left(x+2\right)\left(x-1\right)\left(3x+2\right)}{\left(2x+3\right)\left(3x+2\right)}=\frac{x^2+x+2-4x^2}{\sqrt{x^2+x+2}+2x}\)
\(\Leftrightarrow\frac{-2\left(x+2\right)\left(x-1\right)\left(3x+2\right)}{\left(2x+3\right)\left(3x+2\right)}=\frac{-\left(x-1\right)\left(3x+2\right)}{\sqrt{x^2+x+2}+2x}\)
\(\Leftrightarrow\frac{-2\left(x+2\right)\left(x-1\right)\left(3x+2\right)}{\left(2x+3\right)\left(3x+2\right)}-\frac{-\left(x-1\right)\left(3x+2\right)}{\sqrt{x^2+x+2}+2x}=0\)
\(\Leftrightarrow-\left(x-1\right)\left(3x+2\right)\left(\frac{2\left(x+2\right)}{\left(2x+3\right)\left(3x+2\right)}-\frac{1}{\sqrt{x^2+x+2}+2x}\right)=0\)
\(\Leftrightarrow x=1;x=-\frac{2}{3}\) (thỏa)
mình bảo là đưa về dạng \(A^2=B^2\)hoặc \(A^2+B^2=0\)cơ, giúp mình nhé
