\(A=-x^2+2xy-4y^2+2x+10y-3\)

\(=-x^2+2xy-y^2+2x-2y-1-3y^2+12y-12+10\)

\(=-\left(x^2-2xy+y^2-2x+2y+1\right)-3\left(y^2-4y+4\right)+10\)

\(=-\left(x-y-1\right)^2-3\left(y-2\right)^2+10< =10\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-y-1=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=y+1=3\end{matrix}\right.\)

\(B=-4x^2-5y^2+8xy+10y+12\)

\(=-4x^2+8xy-4y^2-y^2+10y-25+37\)

\(=-4\left(x^2-2xy+y^2\right)-\left(y^2-10y+25\right)+37\)

\(=-4\left(x-y\right)^2-\left(y-5\right)^2+37< =37\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-y=0\\y-5=0\end{matrix}\right.\)

=>x=y=5

\(A=-x^2+2xy-4y^2+2x+10y-3\)

\(=10-\left(x^2+y^2+1-2xy-2x+2y\right)-3\left(y^2-4y+4\right)\)

\(=10-\left(x-y-1\right)^2-3\left(y-2\right)^2\le10\)

Vậy \(MaxA=10\), đạt được khi và chỉ khi \(\left\{{}\begin{matrix}x-y-1=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)

Lời giải:
$-A=x^2-2xy+4y^2-2x-10y+3$

$=(x^2-2xy+y^2)+3y^2-2x-10y+3$

$=(x-y)^2-2(x-y)+3y^2-12y+3$

$=(x-y)^2-2(x-y)+1+3(y^2-4y+4)-10$

$=(x-y+1)^2+3(y-2)^2-10\geq 0+0-10=-10$

$\Rightarrow A\leq 10$

Vậy $A_{\max}=10$. Giá trị này đạt tại $x-y+1=y-2=0$

$\Leftrightarrow y=2; x=1$

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:

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

Trần Việt Hoàng !!! Em xem lại đề nhé! Cô nghĩ là M= - x^2+2xy-4y^2+2x+10y-8

\(6M=-6x^2+12xy-24y^2+12x+60y-48\)

\(=(-4x^2+12xy+9y^2)+(-2x^2+12x)+(-15y^2+60y)-48\)

\(=-(2x-3y)^2-2(x^2-6x+9)-15(y^2-4y+4)+30\)

\(=-(2x-3y)^2-2(x-3)^2-15(y-2)^2+30\le30\)

Dấu " = " xảy ra khi : 2x - 3y = 0 ; x - 3 = 0 , y - 2 = 0 => \(\hept{\begin{cases}x=3\\y=2\end{cases}}\)

Vậy GTLN của M là \(\frac{30}{8}=5\)tại x = 3 , y = 2

Chúc bạn học tốt :>

1) \(a^2+\frac{1}{a^2}=14\Leftrightarrow a^2+\frac{1}{a^2}+2a.\frac{1}{a}=16\Leftrightarrow\left(a+\frac{1}{a}\right)^2=16\Rightarrow a+\frac{1}{a}=4\)

\(\Rightarrow\left(a+\frac{1}{a}\right)\left(a^2+\frac{1}{a^2}\right)=a^3+\frac{1}{a}+a+\frac{1}{a^3}=a^3+4+\frac{1}{a^3}=4.14=56\)

\(\Rightarrow a^3+\frac{1}{a^3}=52\)

Ta có : \(\left(a^2+\frac{1}{a^2}\right)\left(a^3+\frac{1}{a^3}\right)=a^5+\frac{1}{a}+a+\frac{1}{a^5}=a^5+4+\frac{1}{a^5}=14.52\)

\(\Rightarrow a^5+\frac{1}{a^5}=14.52-4=724\)

2) \(A=2xy-x^2-4y^2+2x+10y-2000\)

\(=\left(-x^2+2xy-y^2\right)+\left(2x-2y\right)+\left(-3y^2+12y-12\right)-1988\)

\(=-\left(x-y\right)^2+2\left(x-y\right)-1-3\left(y^2-4y+4\right)-1987\)

\(=-\left(x-y-1\right)^2-3\left(y-2\right)^2-1987\le-1987\forall x;y\) có GTLN là 2013

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

Vậy \(A_{max}=-1987\) tại \(x=3;y=2\)

Giups mik vs