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a) \(5x^2-12xy+9y^2-4x+4=\left(4x^2-12xy+9y^2\right)+x^2-4x+4=\left(2x-3y\right)^2+\left(x-2\right)^2\ge0\)
b) \(-x^2-2y^2+12x-4y+7=-\left(x^2-12x+36\right)-2\left(y^2+2y+1\right)+45=-\left(x-6\right)^2-2\left(y+1\right)^2+45\le45\)
c)\(4y^2+10x^2+12xy+6x+7=\left(4y^2+12xy+9x^2\right)+x^2+6x+9-2=\left(2y+3x\right)^2+\left(x+3\right)^2-2\ge-2\)
d) \(3-10x^2-4xy-4y^2=3-\left(4y^2+4xy+x^2\right)-9x^2=-\left(2y+x\right)^2-9x^2+3\le3\)
e)\(x^2-5x+y^2-xy-4y+16=\left(\frac{1}{2}x^2-xy+\frac{1}{2}y^2\right)+\frac{1}{2}\left(x^2-10x+25\right)+\frac{1}{2}\left(y^2-8y+16\right)-\frac{9}{2}=\frac{1}{2}\left(x-y\right)^2+\frac{1}{2}\left(x-5\right)^2+\frac{1}{2}\left(y-4\right)^2-\frac{9}{2}\ge-\frac{9}{2}\)Phần e) mới nghĩ đk v, tui biết đáp án sao do k xảy ra dấu bằng
\(a.x^2+x+\frac{1}{4}=x^2+2.x.\frac{1}{4}+\left(\frac{1}{4}\right)^2\)
\(=\left(x+\frac{1}{4}\right)^2\)
b) \(x^2+12xy+36xy^2=x^2+2.x.y+y^2\)
\(a.x^2+x+\frac{1}{4}=x^2+2.x.\frac{1}{4}+\left(\frac{1}{4}\right)^2\)
\(=\left(x+\frac{1}{4}\right)^2\)
b) \(x^2+12xy+36xy^2=x^2+2.x.y+y^2\)
\(a.x^2+x+\frac{1}{4}=x^2+2.x.\frac{1}{4}+\left(\frac{1}{4}\right)^2\)
\(=\left(x+\frac{1}{4}\right)^2\)
b) \(x^2+12xy+36xy^2=x^2+2.x.y+y^2\)
\(a.x^2+x+\frac{1}{4}=x^2+2.x.\frac{1}{4}+\left(\frac{1}{4}\right)^2\)
\(=\left(x+\frac{1}{4}\right)^2\)
b) \(x^2+12xy+36xy^2=x^2+2.x.y+y^2\)
\(=\left(x+y\right)^2\) c) \(4x^2-12xy+9y^2=\left(2x\right)^2-2.2x.3y+\left(3y\right)^2\) \(=\left(2x-3y\right)^2\)d) \(x^2-2x+4=x^2-2.x.4+4^2\)
\(=\left(x-4\right)^2\)
e) \(25x^2+4y^2-20xy=\left(5x\right)^2-2.5x.2y+\left(2y\right)^2\)
\(=\left(5x-2y\right)^2\)
^...^ 
^_^ Bài làm có gì ko hiểu bạn cứ hỏi nhé ^_^
mạng của mk bị lỗi bạn xem cái phần cuối cùng nhé xl bạn nhiều vì mạng của mk bị lỗi 
bình phương tổng chứ
b, B= x^2+ 2xy+y^2 +4y+4
= x^2+2xy+y^2+y^2+4y+4
=(x+y)^2+(y+2)^2
c, C= 2x^2+6xy+9y^2+2x+1
= x^2+6xy+9y^2+x^2+2x+1
= (x+3)^2+(x+1)^2
d, D= x(x+2) +(x+1)(x+3) +2
= x^2+2x+x^2+3x+x+3+2
= x^2+2x+1+x^2+4x+4
= (x+1)^2+(x+2)^2
e, E= x^2-2xy+2y^2+2y+1
= x^2-2xy+y^2+y^2+2y+1
= (x-y)^2+(y+1)^2
f, F= 4x^2-12xy+10y^2+4y+4
=4x^2-12xy+9y^2+y^2+4y+4
=(2x-3y)^2+(y+2)^2
g, G=2x^2+4xy+4y^2+4x+4
=x^2+4xy+4y^2+x^2+4x+4
=(x+2y)^2+(x+2)^2
Xong r.... dài quá...mới hè lớp 7 nên có j bỏ qua ak
a. \(9x^2+25-12xy+5y^2-10y\)
\(=\left(9x^2-12xy+4y^2\right)+\left(25+y^2-10y\right)\)
\(=9\left(x^2-\frac{4xy}{3}+\frac{4y^2}{9}\right)+\left(5-y\right)^2\)
\(=9\left(x-\frac{2y}{3}\right)^2+\left(5-y\right)^2\)
a) \(4x^2-4xy+y^2-9\)
\(=\left(2x-y\right)^2-3^2\)
\(=\left(2x-y+3\right)\left(2x-y-3\right)\)
b) \(x^2-36+4xy+4y^2\)
\(=\left(4y^2+4xy+x^2\right)-36\)
\(=\left(2y+x\right)^2-6^2\)
\(=\left(2y+x+6\right)\left(2y+x-6\right)\)
c) \(9x^2-12xy-25+4y^2\)
\(=\left(9x^2-12xy+4y^2\right)-25\)
\(=\left(3x-2y\right)^2-5^2\)
\(=\left(3x-2y+5\right)\left(3x-2y-5\right)\)
d) \(25x^2+10x-4y^2+1\)
\(=\left(25x^2+10x+1\right)-4y^2\)
\(=\left(5x+1\right)^2-\left(2y\right)^2\)
\(=\left(5x+2y+1\right)\left(5x-2y+1\right)\)
a)\(6x-9-x^2\)
\(=-\left(x^2+6x+9\right)\)
\(=-\left(x+3\right)^2\)
b)\(x^2+4y^2+4xy\)
\(=\left(x+2y\right)^2\)
c)\(x^2+8x+16\)
\(=\left(x+4\right)^2\)
d)\(9x^2-12xy+4y^2\)
\(=\left(3x-2y\right)^2\)
e)\(-25x^2y^2+10xy-1\)
\(=-\left(25x^2y^2-10xy+1\right)\)
\(=-\left(5xy-1\right)^2\)
f)\(4x^2-4x+1\)
\(=\left(2x-1\right)^2\)
j)\(x^2+6x+9\)
\(=\left(x+3\right)^2\)
h)\(9x^2-6x+1\)
\(=\left(3x-1\right)^2\)
#H
a, 6x - 9 - x2 = - x2 + 6x - 9 = - (x2 - 6x + 9) = - (x - 3)2
b, x2 + 4y2 + 4xy = x2 + 2. x . 2y + (2y)2 = (x + 2y)2
c, x2 + 8x + 16 = x2 + 2 . x . 4 + 42 = (x + 4)2
d, 9x2 - 12xy + 4y2 = (3x)2 - 2 . 3x . 2y + (2y)2 = (3x - 2y)2
e, - 25x2y2 + 10xy - 1 = - (25x2y2 - 10xy + 1) = - [(5xy)2 - 2 . 5xy + 1] = - (5xy - 1)2
f, 4x2 - 4x + 1 = (2x)2 - 2 . 2x + 1 = (2x - 1)2
j, x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
h, 9x2 - 6x + 1 = (3x)2 - 2 . 3x + 1 = (3x - 1)2
a) \(x^2-6x+9=x^2-2.3.x+3^2=\left(x-3\right)^2\)
b)\(x^2+4x+4=x^2+2.2.x+2^2=\left(x+2\right)^2\)
c)\(4x^2+4x+1=\left(2x\right)^2+2.2x.1+1^2=\left(2x+1\right)^2\)
d)\(4x^2+12xy+9y^2=\left(2x\right)^2+2.2x.3y+\left(3y\right)^2=\left(2x+3y\right)^2\)
e)\(x^2-8x+16=x^2-2.4.x+4^2=\left(x-4\right)^2\)
a) x2 -6x +9 = (x-3)2
b) x2+4x +4= (x+2)2
c) 4x2+4x+1= (2x+1)2
d) 4x2+12xy+9y2 = (2x+3y)2
e) x2-8x+16 = (x-4)2
Đây chính là hằng đẳng thức nhé bn....
a, \(4x^2-4x+1=\left(2x-1\right)^2\)
b, \(x^2+4xy+4y^2=\left(x+2y\right)^2\)
c, \(4x^2+4xy+y^2=\left(2x+y\right)^2\)
d, \(x^2+12xy+36y^2=\left(x+6y\right)^2\)
e, \(x^2-12xy+36y^2=\left(x-6y\right)^2\)
a, \(4x^2-4x+1\)
\(=4x^2-2x-2x+1=2x.\left(2x-1\right)-\left(2x-1\right)\)
\(=\left(2x-1\right)^2\)
b, \(x^2+4xy+4y^2\)
\(=x^2+2xy+2xy+4y^2\)
\(=x.\left(x+2y\right)+2y.\left(x+2y\right)\)
\(=\left(x+2y\right)^2\)
Chúc bạn học tốt!!! (bạn nhờ mình giải chi tiết bài này á)