Câu hỏi của Nguyễn Thị Kim Ngân

Question

giúp mình nha

Cheewin · Accepted Answer

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 $Q=\dfrac{-9}{2}\Leftrightarrow x=\dfrac{3}{2}$Câu trả lời: 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 $Q=\dfrac{-9}{2}\Leftrightarrow x=\dfrac{3}{2}$

T.Thùy Ninh · Answer

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 $\left[{}\begin{matrix}x-\dfrac{1}{2}=0\x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\x=-3\end{matrix}\right.$Câu trả lời: 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 $\left[{}\begin{matrix}x-\dfrac{1}{2}=0\x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\x=-3\end{matrix}\right.$

Cheewin · Answer

Bài 3: $A=x-x^2$

$\Leftrightarrow A=-x^2+2.\dfrac{1}{2}x-\dfrac{1}{4}+\dfrac{1}{4}$

$\Leftrightarrow A=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}$

Max $A=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{2}$

b) $N=2x-2x^2-5$

$\Leftrightarrow N=-2x^2+2x-5$

$\Leftrightarrow N=-2.\left(x^2-x+\dfrac{5}{2}\right)$

$\Leftrightarrow N=-2.\left(x^2-2.\dfrac{1}{2}.x+\dfrac{1}{4}+\dfrac{9}{4}\right)$

$\Leftrightarrow N=-2.\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\le\dfrac{-9}{2}$

Max $N=\dfrac{-9}{2}\Leftrightarrow x=\dfrac{1}{2}$Câu trả lời: Bài 3: $A=x-x^2$