a, Đặt \(A=\left(y+1\right)\left(y-2\right)\left(y-3\right)\left(y-6\right)=\left(y+1\right)\left(y-6\right)\left(y-2\right)\left(y-3\right)\)
\(=\left(y^2-6y+y-6\right)\left(y^2-3y-2y+6\right)\)
\(=\left(y^2-5y-6\right)\left(y^2-5y+6\right)\)
Đặt \(y^2-5y=x\), ta có:
\(A=\left(x-6\right)\left(x+6\right)=x^2-36=\left(y^2-5y\right)^2-36\ge-36\)
Dấu "=" xảy ra <=> \(y^2-5y=0\Leftrightarrow\left[{}\begin{matrix}y=0\\y=5\end{matrix}\right.\)
Vậy Amin = -36 khi y = 0 hoặc y = 5
b, Đặt \(B=\left(y+2\right)\left(y-3\right)\left(y-4\right)\left(y-9\right)=\left(y+2\right)\left(y-9\right)\left(y-3\right)\left(y-4\right)\)
\(=\left(y^2-9y+2y-18\right)\left(y^2-4y-3y+12\right)\)
\(=\left(y^2-7y-18\right)\left(y^2-7y+12\right)\)
Đặt \(y^2-7y-3=t\),ta có:
\(B=\left(t-15\right)\left(t+15\right)=t^2-225=\left(y^2-7y-3\right)^2-225\ge-225\)
Dấu "="xảy ra <=> \(y^2-7y-3=0\Leftrightarrow\left[{}\begin{matrix}y=\dfrac{7+\sqrt{61}}{2}\\y=\dfrac{7-\sqrt{61}}{2}\end{matrix}\right.\)
Vậy...
