trả lời
mik đăng cho hay thôi chứ bt làm rồi
cảm ơn bn
\(\sqrt{3x^2+24x+22}=2x+1\) (\(x\ge\frac{-1}{2}\) )
Binh phuong hai ve ta duoc:
\(\left(\sqrt{3x^2+24x+22}\right)^2=\left(2x+1\right)^2\)
\(\Leftrightarrow3x^2+24x+22=4x^2+4x+1\)
\(\Leftrightarrow x^2-20x-21=0\)
\(\Leftrightarrow x^2+x-21x-21=0\)
\(\Leftrightarrow x\left(x+1\right)-21\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-21\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-21=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=21\end{cases}}}\) (nhan)
Vay x = -1 hoac x = 21
