Question

[Nguyễn Thị Kim Ngân]

giúp mình nha

Answer 1

[Cheewin]

Câu 1)\(H=\left(x-y+z\right)^2+\left(z-y\right)^2+2\left(x-y+z\right)\left(y-z\right)\)

\(\Leftrightarrow H=\left(x-y+z+z-y\right)^2\)

\(\Leftrightarrow H=\left(x-2y+2z\right)^2\)

Câu 2: \(Q=2x^2-6x\)

\(\Leftrightarrow Q=2\left(x^2-2.\dfrac{3}{2}.x+\left(\dfrac{3}{2}\right)^2\right)-\dfrac{9}{2}\)

\(\Leftrightarrow Q=2.\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge\dfrac{-9}{2}\)

Min \(Q=\dfrac{-9}{2}\Leftrightarrow x=\dfrac{3}{2}\)

Answer 2

[T.Thùy Ninh]

2.

\(a,Q=2x^2-6x=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge\dfrac{-9}{2}\)Vậy \(Min_Q=\dfrac{-9}{2}\) khi \(x-\dfrac{3}{2}=0\Rightarrow x=\dfrac{3}{2}\)

\(b,M=x^2+y^2-x+6y+10=\left(x^2-x+\dfrac{1}{4}\right)+\left(y^2+6y+9\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

vậy \(Min_M=\dfrac{3}{4}\)khi \(\left[{}\begin{matrix}x-\dfrac{1}{2}=0\\x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-3\end{matrix}\right.\)