Đặt \(A=x^2-4xy+5y^2+10x-22y+28\)

\(=x^2-4xy+10x+5y^2-22y+28\)

\(=x^2-x\left(4y-10\right)+5y^2-22y+28\)

\(=x^2-2.x.\frac{4y-10}{2}+\left(\frac{4y-10}{2}\right)^2+5y^2-22y-\left(\frac{4y-10}{2}\right)^2+28\)

\(=\left(x-\frac{4y-10}{2}\right)^2+5y^2-22y-\frac{16y^2-80y+100}{4}+28\)

\(=\left(x-\frac{4y-10}{2}\right)^2+5y^2-22y-4y^2+20y-25+28\)

\(=\left(x-\frac{4y-10}{2}\right)^2+y^2-2y+3=\left(x-\frac{4y-10}{2}\right)^2+y^2-2.y.1+1^2+2\)

\(=\left(x-\frac{4y-10}{2}\right)^2+\left(y-1\right)^2+2\)

Vì \(\left(x-\frac{4y-10}{2}\right)^2\ge0;\left(y-1\right)^2\ge0=>\left(x-\frac{4y-10}{2}\right)^2+\left(y-1\right)^2\ge0\)

\(=>\left(x-\frac{4y-10}{2}\right)^2+\left(y-1\right)^2+2\ge2\) (với mọi x,y)

Dấu "=" xảy ra \(< =>\hept{\begin{cases}\left(x-\frac{4y-10}{2}\right)^2=0\\\left(y-1\right)^2=0\end{cases}}< =>\hept{\begin{cases}x-\frac{4y-10}{2}=0\\y=1\end{cases}}< =>\hept{\begin{cases}x-\frac{4-10}{2}=0\\y=1\end{cases}}\)

\(< =>\hept{\begin{cases}x=-3\\y=1\end{cases}}\)

Vậy MInA=2 khi x=-3;y=1

Amin=2

\(4-2x^2\)

\(=2^2-2x^2\)

\(2^2-2x^2\ge0\)

Rồi làm tiếp nha 

BT trên không có GTNN ,bn nên xem lại đề

Chứng minh đẳng thức:

1) xét vế trái (a+b)(a-b)=a²-ab+ab-b² =a²-b²=vế phải

2) xét vt (a+b)(a²-ab+b²) =a³-a²b+ab²+a²b-ab²+b³ =a³+b³=vp

3) (a-b)(a²+ab+b²)=a³+a²b+ab²-a²b-ab²-b³ =a³- b³ =vp

4) (a+b)²=(a+b)(a+b)=a²+ab+ab+b² =a²+2ab+b²=vp

5) (a-b)² =(a-b)(a-b)=a²-ab-ab+b² =a²-2ab+b²=vp

6) (a+b)³ =(a+b)(a+b)(a+b)=(a²+2ab+b²)(a+b) = a³+2a²b+ab²+a²b+2ab²+b³= a³+3a²b+3ab²+b³=vp

7)(a-b)³=(a-b)(a-b)(a-b)=(a²-2ab+b²)(a-b) = a³-2a²b+ab²-a²b+2ab²-b³ =a³-3a²b+3ab²-b³=vp

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

     \(=1\)

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

Ta có : \(x+y-2=0\Rightarrow x+2=-y\)

\(\Rightarrow N=-x^2y-xy^2+2xy+2\)

     \(N=-xy\left(x+y-2\right)+2=2\)

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

= -2( x² +2x/2 +1/4) +5 + 1/2

= -2(x+1/2)² + 5,5

GTLN = 5,5

mk làm tắt giỏi toán moi hieu

\(-2x^2+x+5=-2\left(x^2-\frac{x}{2}\right)+5=-2\left(x^2-2.x.\frac{1}{4}+\frac{1}{16}\right)+\frac{1}{8}+5=-2\left(x-\frac{1}{4}\right)^2+\frac{41}{8}\le\frac{41}{8}\)Do đó Max = \(\frac{41}{8}\Leftrightarrow x=\frac{1}{4}\)

Kp nha,e học lớp 1