\(A=\left(x^2-2xy+y^2\right)+2\left(x-y\right)+1+x^2+6x+9+1978\)

\(=\left(x-y\right)^2+2\left(x-y\right)+1+\left(x+3\right)^2+1978\)

\(=\left(x-y+1\right)^2+\left(x+3\right)^2+1978\ge1978\)

\(A_{min}=1978\) khi \(\left\{{}\begin{matrix}x=-3\\y=-2\end{matrix}\right.\)

Giups mik vs

8x-2x^2+5=-2(x^2-4x+4)+8+5

=-2(x-2)^2+13

Vì (x-2)^2 lớn hơn hoặc bằng 0

=>-2(x-2)^2 bé hơn hoặc bằng 0

=>gtln=13

Lp 8 thì chịu oy 

gia tri lon nhat la 6

A=4-x²+3x

=-x²+3x+4

=\(-x^2+3x-\)\(\frac{9}{4}+\frac{25}{4}\)

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

\(=\frac{25}{4}-\left(x-\frac{3}{2}\right)^2\)

\(\Rightarrow-\left(x-\frac{3}{2}\right)^2\le0\) voi moi x

\(\Rightarrow-\left(x-\frac{3}{2}\right)^2\le\frac{25}{4}\)

Vay GTLN la : \(\frac{25}{4}\)

Dau "=" xay ra khi : \(x-\frac{3}{2}=0\Rightarrow x=\frac{3}{2}\)

mấy bạn chuyên toán giải giùm mk bài b) giùm ạ, mk đaq rất cần

1, a) 

Ta có:

\(x^2+2x+1=\left(x+1\right)^2\)

Thay x=99 vào ta có:

\(\left(99+1\right)^2=100^2=10000\)

b) Ta có:

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

Thay x=101 vào ta có:

\(\left(101-1\right)^3=100^3=1000000\)

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

\(=x^2-2xy+y^2+2x^2-3x+2\)

\(=\left(x-y\right)^2+2\left(x-\frac{3}{4}\right)^2+\frac{7}{8}\)

do\(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(x-\frac{3}{4}\right)^2\ge0\end{cases}\Rightarrow P\ge\frac{7}{8}}\)

\(\Rightarrow P_{min}=\frac{7}{8}\)đạt được khi \(x=y=\frac{3}{4}\)