\(1)4x^2-25+\left(2x+7\right).\left(5.2x\right)\)

\(=\left(2x\right)^2-5^2-\left(2x+7\right).\left(2x-5\right)\)

\(=\left(2x.5\right)\left(2x+5\right).\left(2x+7\right)\left(2x-5\right)\)

\(=\left(2x-5\right)\left(2x+5-2x+7\right)\)

\(=\left(2x-5\right).12\)

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

\(=3(x+4)-\left(x+4\right)\)

\(=\left(3-x\right)\left(x+4\right)\)

\(3)5x^2-2y^2-10x+10y\)

\(=5\left(x^2-y^2\right)-10\left(x-4\right)\)

\(=5\left(x-y\right)\left(x+y\right)-10\left(x-y\right)\)

\(=\left(x-y\right)[5(x+y)-10]\)

Còn lại bn lm nốt nha! 

`#3107.101107`

`x^2 - y^2 + 10x - 10y`

`= (x^2 - y^2) + (10x - 10y)`

`= (x - y)(x + y) + 10(x - y)`

`= (x + y + 10)(x - y)`

_____

Sử dụng HĐT:

`A^2 - B^2 = (A - B)(A + B).`

2x^2-5x-7

=2x^2+2x-7x-7

=2x.(x+!) -7.(x+1)

(2x-7).(x+1)

x^2-y^2-5x+5y

=(x-y).(x+y)-5.(x-y)

=(x-y-5).(x+y)

5x^3-5x^2.y-10x^2+10xy

=5x^2.(x-y)-10x.(x-y)

=(5x^2-10x).(x-y)

=5x.(x-2).(x-y)

a)x²−2x−4y²−4ya)x²-2x-4y²-4y

=x²−2x−4y²−4y+2xy−2xy=x²-2x-4y²-4y+2xy-2xy

=(x²−2xy−2x)+(2xy−4y²−4y)=(x²-2xy-2x)+(2xy-4y²-4y)

=x(x−2y−2)+2y(x−2y−2)=x(x-2y-2)+2y(x-2y-2)

=(x+2y)(x−2y−2)=(x+2y)(x-2y-2)

b)x4+2x³−4x−4b)x4+2x³-4x-4

=x4+2x³+2x²−2x²−4x−4=x4+2x³+2x²-2x²-4x-4

=(x4+2x³+2x²)−(2x²+4x+4)=(x4+2x³+2x²)-(2x²+4x+4)

=x²(x²+2x+2)−2(x²+2x+2)=x²(x²+2x+2)-2(x²+2x+2)

=(x²−2)(x²+2x+2)=(x²-2)(x²+2x+2)

c)x³+2x²y−x−2yc)x³+2x²y-x-2y

=x²(x+2y)−(x+2y)=x²(x+2y)-(x+2y)

=(x²−1)(x+2y)=(x²-1)(x+2y)

=(x+1)(x−1)(x+2y)=(x+1)(x-1)(x+2y)

d)3x²−3y²−2(x−y)²d)3x²-3y²-2(x-y)²

=3(x²−y²)−2(x−y)²=3(x²-y²)-2(x-y)²

=3(x+y)(x−y)−2(x−y)²=3(x+y)(x-y)-2(x-y)²

=(x−y)[3(x+y)−2(x−y)]=(x-y)[3(x+y)-2(x-y)]

=(x−y)(3x+3y−2x+2y)=(x-y)(3x+3y-2x+2y)

=(x−y)(x+5y)=(x-y)(x+5y)

e)x³−4x²−9x+36e)x³-4x²-9x+36

=(x³−4x²)−(9x−36)=(x³-4x²)-(9x-36)

=x²(x−4)−9(x−4)=x²(x-4)-9(x-4)

=(x−4)(x²−9)=(x-4)(x²-9)

=(x−4)(x²−3²)=(x-4)(x²-3²)

=(x−4)(x+3)(x−3)=(x-4)(x+3)(x-3)

f)x²−y²−2x−2yf)x²-y²-2x-2y

=(x²−y²)−(2x+2y)=(x²-y²)-(2x+2y)

=(x+y)(x−y)−2(x+y)=(x+y)(x-y)-2(x+y)

=(x+y)(x−y−2)

hok tốt nhé

k đi

a) \(5x^2+5xy-x-y\)

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

\(=\left(x+y\right)\left(5x-1\right)\)

b) \(5x^2-10y+5y^2-20z^2\)

\(=5.\left(x^2-2y+y^2-4z^2\right)\)

Đề sai ở đâu đó.

c) \(4x^2-y^2+4x+1\)

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

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

\(=\left(2x+y+1\right)\left(2x-y+1\right)\)

\(x^2-xy-10x+10y\)

\(=x\left(x-y\right)-10\left(x-y\right)\)

\(=\left(x-y\right)\left(x-10\right)\)

a)Ta có:  \(x\left(x-y\right)+5x-5y=x\left(x-y\right)+\left(5x-5y\right)\)

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

b) \(x^2-y^2+10x+25=\left(x^2+10x+25\right)-y^2=\left(x+5\right)^2-y^2\)

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

\(x^2-4x+5y^2-10y+9=0\\ \Leftrightarrow\left(x^2-4x+4\right)+\left(5y^2-10y+5\right)=0\\ \Leftrightarrow\left(x-2\right)^2+5\left(y^2-2y+1\right)=0\\ \Leftrightarrow\left(x-2\right)^2+5\left(y-1\right)^2=0\)

Vì \(\left(x-2\right)^2\ge0;5\left(y-1\right)^2\ge0\) mà \(\left(x-2\right)^2+5\left(y-1\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x-2\right)^2=0\\5\left(y-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)