Câu a:

\(x^2-y^2-x+3y-2=\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}\right)-\left(y^2-2.y.\frac{3}{2}+\frac{9}{4}\right)\)

\(< =>\left(x-\frac{1}{2}\right)^2-\left(y-\frac{3}{2}\right)^2\)

\(< =>\left(x-\frac{1}{2}+y-\frac{3}{2}\right)\left(x-\frac{1}{2}-y+\frac{3}{2}\right)=\left(x+y-2\right)\left(x-y+1\right)\)

a) \(x^2+4y^2+4xy\)

\(=x^2+2.x.2y+\left(2y\right)^2\)

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

b) \(\left(x+y\right)^2-\left(x-y\right)^2\)

\(=\left(x+y-x+y\right)\left(x+y+x-y\right)\)

\(=2y.2x\)

\(=4xy\)

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

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

a) \(x^6-y^6=\left(x^2\right)^3-\left(y^2\right)^3\)

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

\(P=x^2-6xy+9y^2=\left(x-3y\right)^2\)

(Áp dụng 7 hằng đẳng thức đáng nhớ)

a) 3x^2 y - 6xy^2 = 3xy ( x - 2y)

b) 9 - ( x-  y)^2 = ( 3 )^2 - ( x- y)^2

                        = ( 3 -x + y )( 3 + x + y ) 

a/ \(3x^2y-6xy^2\)\(=3xy\left(x-2y\right)\) ( đây là p²   đặt nhân tử chung )

b/9-(x -y )²    =( 3 -x +y ) ( 3 + x+y )   ( dùng hđt số 3 để giải )

\(3x^2y-6xy^2\)

\(=3xy\left(x-2y\right)\)

\(9-\left(x-y\right)^2\)

\(=\left(3-x+y\right)\left(3+x-y\right)\)

\(2x^2y-4xy^2+6xy=2xy\left(x-2y+3\right)\)

2x²y - 4xy² + 6xy

= 2xy . ( x - 2y + 3 )

\(2x^2y-4xy^2+6xy=2xy\cdot\left(x-2y+3\right)\)

a) \(2x^3+x^2-4x-12\)

\(=2x^3-4x^2+5x^2-10x+6x-12\)

\(=2x^2\left(x-2\right)+5x\left(x-2\right)+6\left(x-2\right)\)

\(=\left(x-2\right)\left(2x^2+5x+6\right)\)

b) \(5x^2+6xy+y^2\)

\(=5x^2+5xy+xy+y^2\)

\(=5x\left(x+y\right)+y\left(x+y\right)\)

\(=\left(x+y\right)\left(5x+y\right)\)

a,7(x+y)

b,2xy(x-3y)

chuc hk tot.

\(a,x^2+6x+9\)

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

\(b,10x-25-x^2\)

\(=-\left(x^2-10x+25\right)\)

\(=-\left(x-5\right)^2\)

\(c,8x^3-\frac{1}{8}\)

\(=8x^3-\left(\frac{1}{2}\right)^3\)

\(=\left(8x-\frac{1}{2}\right)\left(64x^2+4x+\frac{1}{4}\right)\)

\(d,8x^3+12x^2+6xy^2+y^3\)

\(=2\left(4x^3+6x^2+3xy^2+\frac{1}{2}y^3\right)\)

hok tốt!

ai tra loi dung minh cho

Điệp viên 007 sai c

c, \(8x^3-\frac{1}{8}=\left(2x\right)^3-\left(\frac{1}{2}\right)^3=\left(2x-\frac{1}{2}\right)\left(4x^2+x+\frac{1}{4}\right)\)

Ta có

a,    x²-x-y²-y

=x²-y²-(x+y)

=(x-y)(x+y) - (x+y)

=(x+y)(x-y-1)

b,   x²-2xy+y²-z²

=(x-y)²-z²

=(x-y-z)(x-y+z)

con bai 32, 33 neu ban tra loi duoc minh h them

nhanh nha