tim min,max neu co:
a , A=x2+4x-6
b, B=|2x-1|+|2x-7|-10
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=x^2-2x+50\)
\(A=x^2-2x+1+49\)
\(A=\left(x-1\right)^2+49\ge49\)
Dấu "=" xảy ra khi:
\(x=1\)
\(B=12x-x^2\)
\(B=-x^2+12x\)
\(B=-x^2+12x-36+36\)
\(B=-\left(x^2-12x+36\right)+36\)
\(B=-\left(x-6\right)^2+36\le36\)
Dấu "=" xảy ra khi:
\(x=6\)
\(C=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)\)
\(C=\left[\left(x+1\right)\left(x-6\right)\right]\left[\left(x-2\right)\left(x-3\right)\right]\)
\(C=\left[x\left(x-6\right)+1\left(x-6\right)\right]\left[x\left(x-3\right)-2\left(x-3\right)\right]\)
\(C=\left(x^2-6x+x-6\right)\left(x^2-3x-2x+6\right)\)
\(C=\left(x^2-5x-6\right)\left(x^2-5x+6\right)\)
\(C=\left(x^2-5x\right)^2-36\ge-36\)
Dấu "=" xảy ra khi:
\(x^2-5x=0\)
\(\Rightarrow x\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
Không có max
`a)sqrt{x^2-2x+5}`
`=sqrt{x^2-2x+1+4}`
`=sqrt{(x-1)^2+4}`
Vì `(x-1)^2>=0`
`=>(x-1)^2+4>=4`
`=>sqrt{(x-1)^2+4}>=sqrt4=2`
Dấu "=" xảy ra khi `x=1.`
`b)2+sqrt{x^2-4x+5}`
`=2+sqrt{x^2-4x+4+1}`
`=2+sqrt{(x-2)^2+1}`
Vì `(x-2)^2>=0`
`=>(x-2)^2+1>=1`
`=>sqrt{(x-2)^2+1}>=1`
`=>sqrt{(x-2)^2+1}+2>=3`
Dấu "=" xảy ra khi `x=2`
Lời giải:
$A=-x^2+2x+2xy-4y^2-10y-3$
$-A=x^2-2x-2xy+4y^2+10y+3$
$=(x^2-2xy+y^2)+3y^2-2x+10y+3$
$=(x-y)^2-2(x-y)+1+(3y^2+8y+\frac{16}{3})-\frac{10}{3}$
$=(x-y-1)^2+3(y+\frac{4}{3})^2-\frac{10}{3}\geq 0+3.0-\frac{10}{3}=\frac{-10}{3}$
$\Rightarrow A\leq \frac{10}{3}$
Vậy $A_{\max}=\frac{10}{3}$
Giá trị này đạt tại $x-y-1=y+\frac{4}{3}$
$\Leftrightarrow (x,y)=(\frac{-1}{3}, \frac{-4}{3})$
B = 2\(x^2\) - 4\(x\) - 8
B = 2(\(x^2\) - 2\(x\) + 4) - 16
B = 2(\(x-2\))2 - 16
Vì (\(x-2\))2 ≥ 0 ∀ \(x\) ⇒ 2(\(x-2\))2 ≥ 0 ∀ \(x\)
⇒ 2(\(x-2\))2 - 16 ≥ -16 ∀ \(x\)
Dấu bằng xảy ra khi (\(x-2\))2 = 0 ⇒ \(x-2=0\) ⇒ \(x=2\)
Vậy Bmin = -16 khi \(x=2\)
Tìm min của C biết:
C = \(x^2\) - 2\(xy\) + 2y2 + 2\(x\) - 10y + 17
C = (\(x^2\) - 2\(xy\) + y2) + 2(\(x\) - y) + y2 - 8y + 16 + 1
C = (\(x\) - y)2 + 2(\(x\) - y) + 1 + (y2 - 8y + 16)
C = (\(x-y+1\))2 + (y - 4)2
Vì (\(x\) - y + 1)2 ≥ 0 ∀ \(x;y\); (y - 4)2 ≥ 0 ∀ y
Dấu bằng xảy ra khi: \(\left\{{}\begin{matrix}x-y+1=0\\y-4=0\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x-y+1=0\\y=4\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}x-4+1=0\\y=4\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=-1+4\\y=4\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\)
Vậy Cmin = 0 khi (\(x;y\)) = (3; 4)
a, \(A=x^2+2\cdot\frac{1}{2}x+\frac{1}{4}-\frac{9}{4}=\left(x+\frac{1}{2}\right)^2-\frac{9}{4}\)
=> \(A\ge-\frac{9}{4}\) dấu = xảy ra khi : \(x=\frac{-1}{2}\)
\(x^2+4x-6=\left(x+2\right)^2-10\ge0-10=-10\Rightarrow A_{min}=-10\Leftrightarrow x=-2\)
\(B=|2x-1|+|7-2x|-10\ge|2x-1+7-2x|-10=6-10=-4.\Rightarrow B_{min}=-4\Leftrightarrow\frac{1}{2}\le x\le\frac{7}{2}\)