\(A=2x^2+9y^2-6xy-6x-12y+2004\)

\(=\left(9y^2-6xy-12y\right)+2x^2-6x+2004\)

\(=\left[9y^2-6y\left(x+2\right)+\left(x+2\right)^2\right]+2x^2-6x+2004\)\(=\left(3y-x-2\right)^2+2x^2-6x+2004-x^2-4x-4\)\(=\left(3y-x-2\right)^2+\left(x^2-10x+25\right)+1979\)

\(=\left(3y-x-2\right)^2+\left(x-5\right)^2+1979\)

Với mọi giá trị của x;y ta có:

\(\left(3y-x-2\right)^2\ge0;\left(x-5\right)^2\ge0\)

\(\Rightarrow\left(3y-x-2\right)^2+\left(x-5\right)^2+1979\ge1979\)

Vậy Min A = 1979

Để A = 1979 thì \(\left\{{}\begin{matrix}3y-x-2=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3y-5-2=0\\x=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3y=7\\x=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{7}{3}\\x=5\end{matrix}\right.\)

avt xấu

fan Tùg Sơn

Like cho củ cà rốt --- Không vì 1 lí do nào cả>.<. Yêu đươg đau đầu-_-

vg.

\(a,x^4+4y^4=\left(x^4+4x^2y^2+4y^4\right)-4x^2y^2\)

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

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

\(b,81x^4+4y^4=\left(81x^4+36x^2y^2+4y^4\right)-36x^2y^2\)

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

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

\(x^{10}+x^8+1\)

\(=\left(x^{10}-x\right)+\left(x^8-x^2\right)+\left(x^2+x+1\right)\)

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

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

\(=x\left(x^3-1\right)\left(x^6+x^3+1\right)+x^2\left(x^3-1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\)\(=x\left(x-1\right)\left(x^2+x+1\right)\left(x^6+x^3+1\right)+x^2\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\)\(=\left(x^2+x+1\right)\left[x\left(x-1\right)\left(x^6+x^3+1\right)+x^2\left(x-1\right)\left(x^3+1\right)+1\right]\)

hình thì bn tự vẽ nhan

ta có góc A =84 độ=>góc B+góc C =96độ.

=>góc B /2=góc C /2

=>góc IBC+góc ICB=96/2=48độ

xét tam giác BIC có:góc BIC=180độ -góc IBC-góc ICB=180-48=132độ

tick cho mik nhan

\(E=5x^2+y^2-4xy+4x-8y+1\)

\(=\left(y^2-4xy-8y\right)+5x^2+4x+1\)

\(=\left[y^2-2y\left(2x+4\right)+\left(2x+4\right)^2\right]+5x^2+4x+1-\left(2x+4\right)^2\)\(=\left(y-2x-4\right)^2+5x^2+4x+1-4x^2-16x-16\)\(=\left(y-2x-4\right)^2+\left(x^2-12x+36\right)-51\)

\(=\left(y-2x-4\right)^2+\left(x-6\right)^2-51\)

Với mọi giá trị của x ta có:

\(\left(y-2x-4\right)^2\ge0;\left(x-6\right)^2\ge0\)

\(\Rightarrow\left(y-2x-4\right)^2+\left(x-6\right)^2-51\ge-51\)

Vậy Min E = -51

Để E = 51 thì \(\left\{{}\begin{matrix}y-2x-4=0\\x-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y-12-4=0\\x=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=16\\x=6\end{matrix}\right.\)