\(5x^2+y^2+4xy-14x-6y+2016=4x^2+4xy+y^2-6\left(2x+y\right)+9+x^2+2x+1+2006\)

\(=\left(2x+y\right)^2-6xy+9+\left(x+1\right)^2+2006\)

\(=\left(2x+y-3\right)^2+\left(x+1\right)^2+2006\)

lập luận nha gtnn là 2006

5x^2+y^2+4xy-14x-6y+2016

=4x^2+x^2+y^2+y^2-y^2+4xy-14x-6y+9+49+1958

=4x^2+4xy+y^2+x^2-14x+49+y^2-6y+9-y^2+1958

=(4x^2+4xy+y^2)+(x^2-14x+49)+(y^2-6y+9)-y^2+1958

=(2x+y)^2+(x-7)^2+(y-3)^2-y^2+1958

Mà: + (2x+y)^2+(x-7)^2+(y-3)^2-y^2\(\ge\) 1958

Vậy GTNN là: 1958

phân tich M=(2x+y)² + (x-1)² - 6(2x+y) + 2024

   M= ( 2x + y - 3 )² + ( x- 1 )² + 2015

M >= 2015

Dấu = xảy ra khi 2x + y - 3 = 0 và x-1 =0

suy ra x = y = 1

vậy GTNN M= 2015 khi và chi khi x=y=1

a)

\(A=2x^2-3x+1=2\left(x^2-\frac{3}{2}x+\frac{9}{16}\right)-2.\frac{9}{16}+1=2\left(x-\frac{3}{4}\right)^2-\frac{1}{8}\ge-\frac{1}{8}\)

Vậy \(MinA=-\frac{1}{8}\Leftrightarrow\left(x-\frac{3}{4}\right)^2=0\Leftrightarrow x=\frac{3}{4}\)

b)

\(B=5x^2+y^2+10+4xy-15x-6y\)

\(=\left[\left(2x\right)^2+y^2-3^2+2.2x.y-2.y.3-2.2x.3\right]+\left(x^2-3x+\frac{9}{4}\right)+\frac{27}{4}\)

\(=\left(2x+y-3\right)^2+\left(x-\frac{3}{2}\right)^2+\frac{27}{4}\ge\frac{27}{4}\)

Vậy \(MinB=\frac{27}{4}\Leftrightarrow\hept{\begin{cases}\left(2x+y-3\right)^2=0\\\left(x-\frac{3}{2}\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}2x+y-3=0\\x-\frac{3}{2}=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{3}{2}\\y=0\end{cases}}}\)

A là -0,125

\(a,A=2x^2+9y^2-6xy-6x-12y+2049\)

\(=x^2-6xy+9y^2+x^2-10x+25+4x-12y+2024\)

\(=\left(x-3y\right)^2+\left(x-5\right)^2+4\left(x-3y\right)+2024\)

\(=\left(x-3y\right)^2+4\left(x-3y\right)+4+\left(x-5\right)^2+2020\)

\(=\left(x-3y+2\right)^2+\left(x-5\right)^2+2020\)

\(A_{min}=2020\Leftrightarrow\hept{\begin{cases}\left(x-3y+2\right)^2=0\\\left(x-5\right)^2=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x-3y+2=0\\x-5=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x-3y+2=0\\x=5\end{cases}\Rightarrow5-3y+2=0}\)

\(\Rightarrow3y=7\Leftrightarrow y=\frac{7}{3}\)

Vậy \(A_{min}=2020\Leftrightarrow\hept{\begin{cases}x=5\\y=\frac{7}{3}\end{cases}}\)

b tương tự nhé

a) Ta có: \(A=9x^2-12x+10\)

\(=\left(3x\right)^2-2\cdot3x\cdot2+4+6\)

\(=\left(3x-2\right)^2+6\)

Ta có: \(\left(3x-2\right)^2\ge0\forall x\)

\(\Rightarrow\left(3x-2\right)^2+6\ge6\forall x\)

Dấu '=' xảy ra khi \(3x-2=0\)

\(\Leftrightarrow3x=2\)

hay \(x=\frac{2}{3}\)

Vậy: Giá trị nhỏ nhất của biểu thức \(A=9x^2-12x+10\) là 6 khi \(x=\frac{2}{3}\)

bac hai thi bien doi ve tong binh phuong

\(A=\left(x^2-2.3x+9\right)+\left(y^2+2.\frac{5}{2}y+\frac{25}{4}\right)+\left(1-9-\frac{25}{4}\right)\)cu ep vao BP thua de ra ngoai

\(A=\left(x-3\right)^2+\left(y+\frac{5}{2}\right)^2+\left(1-9-\frac{25}{4}\right)\)

\(A\ge\left(1-9-\frac{25}{4}\right)\)co tinh de nguyen cac gia tri them bot de ban de hieu

dang thuc khi x=3; y=-5/2

Cảm ơn bạn nha...

\(H=\left[\left(y^2+2.\left(2x\right)y-2.3y-2.\left(2x\right)+\left(2x\right)^2+3^2\right)\right]+\left(x^2-2.10x+10^2\right)+\left(10-3^2-10^2\right)\)

\(H=\left(y+2x-3\right)^2+\left(x-10\right)^2+\left(10-3^2-10^2\right)\)

H>=10-9-100

dang thuc khi x=10

y=3-2x=3-20=-17