15x( x- y) - 25x + 25y 

= 15x(x-y) - 25 ( x - y)

= (15x - 25) ( x-y)

= 5 (3x - 5)(x-y)

a; x^8 + x^4 + 1 

= x^8 + 2x^4 + 1 - x^4 

= (x^4 + 1) - ( x^2)^2 

= (x^4 - x^2 + 1)( x^4 + x^2 + 1)

= ( x^4 - x^2 + 1)(x^4 +2x^2 + 1 - x^2)

= ( x^4 - x^2 + 1)[(x^2 + 1)^2 - (x)^2 ] 

= ( x^4 - x^2 + 1)( x^2 -x + 1)( x^2 +x + 1) 

Câu hỏi là j?

ko biet cau hoi

m)\(x^4-256=\left(x^2\right)^2-16^2\)

\(=\left(x^2+16\right)\left(x^2-16\right)\)

\(=\left(x^2+16\right)\left(x+4\right)\left(x-4\right)\)

h) \(15x\left(x-y\right)-25x+25y\)

\(=15x\left(x-y\right)-25\left(x-y\right)\)

\(=\left(x-y\right)\left(15x-25\right)\)

\(=5\left(x-y\right)\left(3x-5\right)\)

k) \(-y^2+\frac{1}{9}=\frac{1}{9}-y^2\)

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

\(=\left(y+\frac{1}{3}\right)\left(\frac{1}{3}-y\right)\)

Lời giải:

a) \(3x+3y-x^2-2xy-y^2\)

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

\(=3(x+y)-(x+y)^2=(x+y)(3-x-y)\)

b) \(15x^2-15xy-25x+25y\)

\(=(15x^2-15xy)-(25x-25y)\)

\(=15x(x-y)-25(x-y)=(x-y)(15x-25)\)

\(=5(x-y)(3x-5)\)

a) Ta có: \(27m\left(m+n\right)-m-n\)

\(=27m\left(m+n\right)-\left(m+n\right)\)

\(=\left(m+n\right)\left(27m-1\right)\)

b) Ta có: \(15x\left(x-y\right)-25x+25y\)

\(=15x\left(x-y\right)-25\left(x-y\right)\)

\(=5\left(x-y\right)\left(3x-5\right)\)

c) Ta có: \(12x^2-3xy+8xz-2yz\)

\(=3x\left(4x-y\right)+2z\left(4x-y\right)\)

\(=\left(4x-y\right)\left(3x+2z\right)\)

d) Ta có: \(x^3+x^2y-x^2z-xyz\)

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

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

Ta có: \(25x^2-2xy+\frac{1}{25}y^2\)

\(=\left(5x\right)^2-2\cdot5x\cdot\frac{1}{5}y+\left(\frac{1}{5}y\right)^2\)

\(=\left(5x-\frac{1}{5}y\right)^2\)(*)

Thay \(x=-\frac{1}{5}\) và y=-5 vào biểu thức (*), ta được:

\(\left[5\cdot\frac{-1}{5}-\frac{1}{5}\cdot\left(-5\right)\right]^2=\left(-1-5\right)^2=\left(-6\right)^2=36\)

Vậy: 36 là giá trị của biểu thức \(25x^2-2xy+\frac{1}{25}y^2\) tại \(x=-\frac{1}{5}\) và y=-5

a)7x² + 6x - 13

= 7x² - 7x + 13x - 13

= (7x² - 7x) + (13x - 13)

= 7x(x - 1) + 13(x - 1)

= (7x + 13)(x - 1)

b) 15(x - y) - 25x + 25y

= 15(x - y) - (25x - 25y)

= 15(x - y) - 25(x - y)

= (15 - 25)(x - y)

= -10(x - y)

c) 36 - 4a² + 20ab - 25b²

= 36 - (4a² - 20ab + 25b²)

= 6² - (2a - 5b)²

= (6 - 2a + 5b)(6 + 2a - 5b)

= (5b - 2a + 6)(2a - 5b + 6)

nhớ tik mik nha. Ak có j ko hiểu thì cứ bình luận ở dưới mik giải cho

a, \(7x^2+6x-13\)

=\(7x^2-7x+13x-13\)

=\(\left(7x^2-7x\right)+\left(13x-13\right)\)

=\(7x\left(x-1\right)+13\left(x-1\right)\)

=\(\left(x-1\right)\left(7x+13\right)\)

M= x(12z - 5y)

\(M=12xz-5xy=x\left(12z-5y\right)\)

\(N=64^2-49=64^2-7^2=\left(64-7\right)\left(64+7\right)\)

\(P=25y^2-16z^2=\left(5y\right)^2-\left(4z\right)^2=\left(5y-4z\right)\left(5y+4z\right)\)

\(Q=4-25y^2=2^2-\left(5y\right)^2=\left(2-5y\right)\left(2+5y\right)\)

\(R=25x^2-10x+1=\left(5x-1\right)^2\)

\(O=16-8y+y^2=\left(4-y\right)^2\)

\(X=1-6x+9x^2=\left(1-3x\right)^2\)