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\(2D=x^2-4xy+4y^2+x^2-12x+36+6y^2-36y+54+10\)\(2D=\left(x-2y\right)^2+\left(x-6\right)^2+6\left(y-3\right)^2+10\)
\(2D\ge10\) => D>=5 khi x=2y=6
\(F=3x^2+x+4=3\left(x^2+\dfrac{2x}{6}+\dfrac{1}{36}\right)+\dfrac{47}{12}\)
F=\(3\left(x+\dfrac{1}{6}\right)^2+\dfrac{47}{12}\ge\dfrac{47}{12}\) khi x=-1/6
\(2E=4x^2-4xy+y^2+y^2-4y+4+3996\)
\(2E=\left(2x-y\right)^2+\left(y-2\right)^2+3996\ge3996\)
E>=1998 khi 2x=y=2
bài 4;
\(B=-3x^2+x=-3\left(x^2-\dfrac{2x}{6}+\dfrac{1}{36}\right)+\dfrac{1}{12}\)
\(B=-3\left(x-\dfrac{1}{6}\right)^2+\dfrac{1}{12}\le\dfrac{1}{12}\)
khi x=1/6
bài 5:
\(a,\left(x+2\right)^2=0=>x=-2\)
\(b,\left(x-6\right)^2+\left(y+1\right)^2=0\rightarrow\left\{{}\begin{matrix}x=6\\y=-1\end{matrix}\right.\)
c,\(x^2+2y^2-2xy-2x+2=0\)
\(x^2-4xy+4y^2+x^2-4x+4=0\)
\(\left(x-2y\right)^2+\left(x-2\right)^2=0\rightarrow\left\{{}\begin{matrix}x=2y\\x=2\end{matrix}\right.\)
đây nhá bạn, khá tốn time của mình 
ừ thì mình sẽ giúp bạn mà câu a bạn viết sai đề nha
1/a)\(2x^2+3x-5=2x^2-2x+5x-5=2x\left(x-1\right)+5\left(x-1\right)=\left(2x+5\right)\left(x-1\right)\)
b)\(4x^2-3x-1=4x^2-4x+x-1=4x\left(x-1\right)+\left(x-1\right)=\left(4x+1\right)\left(x-1\right)\)
c)Sai đề: \(3x^2+6xy+3y^2-3z^2\)
\(=3\left(x^2+2xy+y^2-z^2\right)\)
\(=3\left[\left(x+y\right)^2-z^2\right]\)
\(=3\left(x+y+z\right)\left(x+y-z\right)\)
d)Sai đề:\(x^3-2x^2y+xy^2-9x=x\left(x-2xy+y^2-9\right)=x\left[\left(x-y\right)^2-9\right]=x\left(x-y+3\right)\left(x-y-3\right)\)
e)\(2x-2y-x^2+2xy-y^2=2\left(x-y\right)-\left(x-y\right)^2=\left(x-y\right)\left(2-x+y\right)\)
f)Hình như sai đề đúng không?
\(x^3-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x+y+1\right)\left(x+y-1\right)\)
2/a.\(7x-6x^2-2=0\)
\(\Leftrightarrow-\left(6x^2-3x-4x+2\right)=0\)
\(\Leftrightarrow3x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(x-1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}3x-2=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\x=1\end{matrix}\right.\)
b.\(16x-5x^2-3=0\)
\(\Leftrightarrow-\left(5x^2-15x-x+3\right)=0\)
\(\Leftrightarrow5x\left(x-3\right)-\left(x-3\right)=0\)
\(\Leftrightarrow\left(5x-1\right)\left(x-3\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}5x-1=0\\x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\x=3\end{matrix}\right.\)
c.\(2x^2+3x-5=0\)
\(\Leftrightarrow2x^2-2x+5x-5=0\)
\(\Leftrightarrow2x\left(x-1\right)+5\left(x-1\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}2x+5=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}=-2,5\\x=1\end{matrix}\right.\)
Bài 1:
a) Ta có: \(x-2y+x^2-4y^2\)
\(=\left(x-2y\right)+\left(x-2y\right)\left(x+2y\right)\)
\(=\left(x-2y\right)\left(1+x+2y\right)\)
b) Ta có: \(x^2-4x^2y^2+y^2+2xy\)
\(=\left(x^2+2xy+y^2\right)-4x^2y^2\)
\(=\left(x+y\right)^2-\left(2xy\right)^2\)
\(=\left(x+y+2xy\right)\left(x+y-2xy\right)\)
c) Ta có: \(x^6-x^4+2x^3+2x^2\)
\(=x^4\left(x^2-1\right)+2x^2\left(x+1\right)\)
\(=x^4\left(x+1\right)\left(x-1\right)+2x^2\left(x+1\right)\)
\(=\left(x+1\right)\left[x^4\left(x-1\right)+2x^2\right]\)
\(=\left(x+1\right)\left(x^5-x^4+2x^2\right)\)
\(=x^2\left(x+1\right)\left(x^3-x^2+2\right)\)
d) Ta có: \(x^3+3x^2+3x+1-8y^3\)
\(=\left(x^3+3x^2+3x+1\right)-\left(2y\right)^3\)
\(=\left(x+1\right)^3-\left(2y\right)^3\)
\(=\left(x+1-2y\right)\left[\left(x+1\right)^2+2y\left(x+1\right)+\left(2y\right)^2\right]\)
\(=\left(x+1-2y\right)\left(x^2+2x+1+2xy+2y+4y^2\right)\)
Bài 1: Thực hiện phép tính
a) 3x(2x2 - 5x + 9) = \(6x^3-15x^2+27x\)
b) 5x(x2-xy+1) = \(5x^3-5xy+5x\)
c) -2/3x2y(3xy-x2+y) = \(-2x^3y^2+\dfrac{2}{3}x^4y-\dfrac{2}{3}x^2y^2\)
2) Thực hiện phép tính
a) (5x-2y) (x2-xy+1) = \(5x^3+5x-7y-2x^3y+2xy^2\)
b) (x+3y)(x2-2xy+y) = \(x^3-x^2y+xy+6xy^2+y^2\)
c) (3x-5y) (4x+ 7y) = \(12x^2-xy-35y^2\)
Bài 3: Rút gọn các biểu thức sau(bằng cách khai triển hằng đẳng thức):
a) (x+y)2+(x-y)2
= \(x^2+2xy+y^2+x^2-2xy+y^2\)
= \(\left(x^2+x^2\right)+\left(2xy-2xy\right)+\left(y^2+y^2\right)\)
= \(2x^2+2y^2=2\left(x^2+y^2\right)\)
b) (x+2)(x-2)-(x-3)(x+1)
= \(x^2-4\) - \(\left(x^2-2x-3\right)\)= \(x^2-4-x^2+2x+3\)
= \(\left(x^2-x^2\right)+2x+\left(-4+3\right)\)=\(2x-1\)
c) (x-2)(x+2)-(x-2)2
=>\(x^2-4-\left(x^2-2.x.2+2^2\right)=x^2-4-x^2-4x+4=\left(x^2-x^2\right)+\left(-4+4\right)-4x=-4x\)
d) (2x+y)(4x2-2xy+y2)-(2x-y)(4x2+2xy+y2)
= \(8x^3+y^3-\left(8x^3-y^3\right)\)
= \(8x^3+y^3-8x^3+y^3\)
= \(\left(8x^3-8x^3\right)+\left(y^3+y^3\right)\)= \(2y^3\)