Chọn đáp án đúng: Cho \(A=x^2-y^2+2y-1\)
\(A=x^2-y^2+2y-1=x^2-\left(y^2-2y+1\right)\)
\(=x^2-\left(y-1\right)^2=\left(x+y-1\right)\left(x-y+1\right)\)
Tìm min B biết \(B=x^2-8x+20\)
Trả lời: minB = khi x =
\(B=x^2-8x+20=\left(x^2-8x+16\right)+4=\left(x-4\right)^2+4\)
Do \(\left(x-4\right)^2\ge0\forall x\Rightarrow\left(x-4\right)^2+4\ge4\forall x\)
Vậy min B = 4 khi x = 4.
Tìm max B biết \(B=15-4x-x^2\)
\(B=15-4x-4x^2=-\left(x^2+4x-15\right)\)
\(=-\left[\left(x^2+4x+4\right)-19\right]=-\left[\left(x+2\right)^2-19\right]\)
Do \(\left(x+2\right)^2\ge0\Rightarrow\left(x+2\right)^2-19\ge-19\Rightarrow-\left[\left(x+1\right)^2-19\right]\le19\)
Vậy max B = 19 khi x = -2.
Phân tích đa thức thành nhân tử: \(A=x^3-x^2y-4x+4y\)
\(A=x^3-x^2y-4x+4y=x^2\left(x-y\right)-4\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2-4\right)=\left(x-y\right)\left(x+2\right)\left(x-2\right)\)