a)\(y^2+8y-20=0\)
\(\Leftrightarrow y^2+2\cdot y\cdot4+16-16-20=0\)
\(\Leftrightarrow\left(y+4\right)^2-36=0\)
\(\Leftrightarrow\left(y+4\right)^2=36\)
\(\Leftrightarrow y+4=\pm6\)
\(\Leftrightarrow y=2\)hoặc \(y=-10\)
Vậy.....
b)\(x^2+7x=0\)
\(\Leftrightarrow x\left(x+7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\end{matrix}\right.\)
Vậy .....
c)\(2y^2-5y=0\)
\(\Leftrightarrow y\left(2y-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}y=0\\2y-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=0\\y=\dfrac{5}{2}\end{matrix}\right.\)
Vậy ......
d)\(y^2-5y^2+4=0\)
\(\Leftrightarrow-4y^2+4=0\)
\(\Leftrightarrow-4\left(y^2-4\right)=0\)
\(\Leftrightarrow-4\left(y+4\right)\left(y-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}y=-4\\y=4\end{matrix}\right.\)
Vậy....
2) Bạn thực hiện phép chia đi
Cuối cùng có:
Để (x2+3x+a)\(⋮\)(x+1) thì a-2=0=>a=2
Chúc bạn học tốt
