Bài 2
Ta có :
\(3y^2-12=0\)
\(3y^2=0+12\)
\(3y^2=12\)
\(y^2=12:3\)
\(y^2=4\)
\(\Rightarrow y=\pm2\)
b) \(\left|x+1\right|+2=0\)
\(\left|x+1\right|=0+2\)
\(\Rightarrow\orbr{\begin{cases}x+1=2\\x+1=-2\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}}\)
\(N=\frac{3}{2x^2+6}\)
Ta có: \(x^2\ge0\Rightarrow2x^2+6\ge6\)
\(\Rightarrow N_{Max}=\frac{3}{2x^2+6}=\frac{3}{6}=1,5\)
\(\Leftrightarrow2x^2+6=6\Leftrightarrow x=0\)
\(3y^2-12=0\)
\(3y^2=0+12=12\)
\(y^2=12:3=4\)
Vậy x = -2 hoặc 2
\(\left|x+1\right|+2=0\)
Ta có : \(\left|x+1\right|\ge0\Rightarrow\left|x+1\right|+2\ge2\)
=> Phương trình vô nghiệm '^'
#Sel