Bài 2:
a: \(=-9y^2+3yz=-3y\left(3y-z\right)\)
b: \(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
Bài 1.
a) \(A=25x^2+3y^2-10x+11=\left(25x^2-10x+1\right)+3y^2+10\)
\(=\left(5x-1\right)^2+3y^2+10\)
Do \(\left(5x-1\right)^2\ge0;y^2\ge0\Rightarrow A\ge10\)
Dấu = xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}5x-1=0\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=0\end{matrix}\right.\)
Vậy \(A_{min}=10\) đạt được khi \(x=\dfrac{1}{5};y=0\).
b) \(B=\left(x-3\right)^2+\left(x-11\right)^2=x^2-6x+9+x^2-22x+121\)
\(=2x^2-28x+130=2\left(x^2-14x+49\right)+32=2\left(x-7\right)^2+32\)
Do \(\left(x-7\right)^2\ge0\Rightarrow B\ge32\)
Dấu = xảy ra \(\Leftrightarrow x-7=0\Leftrightarrow x=7\)
Vậy \(B_{min}=32\) đạt được khi \(x=7\)