K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

NV
23 tháng 12 2020

\(C=\left(x^2+4y^2+25-4xy+10x-20y\right)+\left(y^2-2y+1\right)+2\)

\(C=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)

\(C_{min}=2\) khi \(\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)

26 tháng 12 2021

\(a,A=\left(x^2-4xy+4y^2\right)+10\left(x-2y\right)+25+\left(y^2-2y+1\right)+2\\ A=\left(x-2y\right)^2+10\left(x-2y\right)+5+\left(y-1\right)^2+2\\ A=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=2y-5\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)

\(b,\Leftrightarrow3x^3+10x^2-5+n=\left(3x+1\right)\cdot a\left(x\right)\)

Thay \(x=-\dfrac{1}{3}\Leftrightarrow3\left(-\dfrac{1}{27}\right)+10\cdot\dfrac{1}{9}-5+n=0\)

\(\Leftrightarrow-\dfrac{1}{9}+\dfrac{10}{9}-5+n=0\\ \Leftrightarrow-4+n=0\Leftrightarrow n=4\)

\(c,\Leftrightarrow2n^2-4n+5n-10+3⋮n-2\\ \Leftrightarrow2n\left(n-2\right)+5\left(n-2\right)+3⋮n-2\\ \Leftrightarrow n-2\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow n\in\left\{-1;1;3;5\right\}\)

22 tháng 12 2021

\(A=\left(x^2-6x+9\right)+2=\left(x-3\right)^2+2\ge2\\ A_{min}=2\Leftrightarrow x=3\\ B=2\left(x^2-10x+25\right)+51=2\left(x-5\right)^2+51\ge51\\ B_{min}=51\Leftrightarrow x=5\\ C=\left[\left(x^2-4xy+4y^2\right)+10\left(x-2y\right)+25\right]+\left(y^2-2y+1\right)+2\\ C=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\\ C_{min}=2\Leftrightarrow\left\{{}\begin{matrix}x-2y+5=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2y-5=2-5=-3\\y=1\end{matrix}\right.\)

22 tháng 12 2021

a) \(A=\left(x^2-6x+9\right)+2=\left(x-3\right)^2+2\ge2\)

\(minA=2\Leftrightarrow x=3\)

b) \(B=2\left(x^2-10x+25\right)+51=2\left(x-5\right)^2+51\ge51\)

\(minB=51\Leftrightarrow x=5\)

c) \(C=\left[x^2-2x\left(2y-5\right)+\left(2y-5\right)^2\right]+\left(y^2-2y+1\right)+2=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)

\(minC=2\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)

22 tháng 9 2020

A = x2 - 4xy + 5y2 + 10x - 22y + 2044

= ( x2 - 4xy + 4y2 + 10x - 20y + 25 ) + ( y2 - 2y + 1 ) + 2018

= [ ( x2 - 4xy + 4y2 ) + ( 10x - 20y ) + 25 ] + ( y - 1 )2 + 2018

= [ ( x - 2y )2 + 2( x - 2y ).5 + 52 ] + ( y - 1 )2 + 2018

= ( x - 2y + 5 )2 + ( y - 1 )2 + 2018 ≥ 2018 ∀ x, y

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-2y+5=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)

=> MinA = 2018 <=> x = -3 ; y = 1

4 tháng 8 2023

a) \(M=x^2-3x+10\)

\(M=x^2-2\cdot\dfrac{3}{2}\cdot x+\dfrac{9}{4}+\dfrac{31}{4}\)

\(M=\left(x^2-2\cdot\dfrac{3}{2}\cdot x+\dfrac{9}{4}\right)+\dfrac{31}{4}\)

\(M=\left(x-\dfrac{3}{2}\right)^2+\dfrac{31}{4}\)

Mà: \(\left(x-\dfrac{3}{2}\right)^2\ge0\) nên: \(M=\left(x-\dfrac{3}{2}\right)^2+\dfrac{31}{4}\ge\dfrac{31}{4}\)

Dấu "=" xảy ra 

\(\left(x-\dfrac{3}{2}\right)^2+\dfrac{31}{4}=\dfrac{31}{4}\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=0\)

\(\Leftrightarrow x-\dfrac{3}{2}=0\Leftrightarrow x=\dfrac{3}{2}\)

Vậy: \(M_{min}=\dfrac{31}{4}\) với \(x=\dfrac{3}{2}\)

b) \(N=2x^2+5y^2+4xy+8x-4y-100\)

\(N=x^2+x^2+4y^2+y^2+4xy+8x-4y-120+16+4\)

\(N=\left(x^2+4xy+4y^2\right)+\left(x^2+8x+16\right)+\left(y^2-4y+4\right)-120\)

\(N=\left(x+2y\right)^2+\left(x+4\right)^2+\left(y-2\right)^2-120\)

Mà:

\(\left\{{}\begin{matrix}\left(x+2y\right)^2\ge0\\\left(x+4\right)^2\ge0\\\left(y-2\right)^2\ge0\end{matrix}\right.\) nên \(N=\left(x+2y\right)^2+\left(x+4\right)^2+\left(y-2\right)^2-120\ge120\)

Dấu "=" xảy ra:

\(\left\{{}\begin{matrix}\left(x+2y\right)^2=0\\\left(x+4\right)^2=0\\\left(y-2\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-4+2y=0\\x=-4\\y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=-4\\y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=2\end{matrix}\right.\)

Vậy: \(N_{min}=120\) khi \(\left\{{}\begin{matrix}x=-4\\y=2\end{matrix}\right.\)

4 tháng 8 2023

a

\(M=x^2-3x+10=x^2-2.\dfrac{3}{2}.x+\dfrac{9}{4}+\dfrac{31}{4}\\ =\left(x-\dfrac{3}{2}\right)^2+\dfrac{31}{4}\ge\dfrac{31}{4}\)

Min M \(=\dfrac{31}{4}\) khi và chỉ khi \(x=\dfrac{3}{2}\)

18 tháng 7 2021

có vài chỗ ko thấy

 

22 tháng 7 2021

A = x^2 + 5y^2 + 4xy - 2y - 3 

= x^2 + 4xy + 4y^2 + y^2 - 2y + 1 - 4

= ( x + 2y )^2 + ( y - 1 )^2 - 4 >= -4 

Dấu ''='' xảy ra khi y = 1 ; x = -2 

Vậy GTNN A là -4 khi x = -2 ; y = 1