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

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

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

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

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

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

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

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

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

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

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

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

Còn lại bn lm nốt nha! 

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) \(4x^2-y^2+4x+1\)

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

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

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

\(a,x^2\left(x-2\right)-4x+8\\ =\left(x^2-4\right)\left(x-2\right)\\ =\left(x-2\right)^2\left(x+2\right)\\ b,x^2+7xy+10y^2\\ =x^2+2xy+5xy+10y^2\\ =x\left(x+2y\right)+5y\left(x+2y\right)\\ =\left(x+5y\right)\left(x+2y\right)\)

1.A

2.C

3.B

4.C

a

c

b

c

(x-1)y^2-4(x-1)y

a) Ta có: \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2\)

\(=\left(4x^2-3x-18-4x^2-3x\right)\left(4x^2-3x-18+4x^2+3x\right)\)

\(=\left(-6x-18\right)\left(8x^2-18\right)\)

\(=-6\left(x+3\right)\cdot2\left(4x^2-9\right)\)

\(=-12\left(x+3\right)\left(2x-3\right)\left(2x+3\right)\)

b) Ta có: \(9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)

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

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

\(=-\left(x+3y+5\right)\left(7x+9y-1\right)\)

c) Ta có: \(-4x^2+12xy-9y^2+25\)

\(=-\left(4x^2-12xy+9y^2-25\right)\)

\(=-\left[\left(2x-3y\right)^2-25\right]\)

\(=-\left(2x-3y-5\right)\left(2x-3y+5\right)\)

d) Ta có: \(x^2-2xy+y^2-4m^2+4mn-n^2\)

\(=\left(x^2-2xy+y^2\right)-\left(4m^2-4mn+n^2\right)\)

\(=\left(x-y\right)^2-\left(2m-n\right)^2\)

\(=\left(x-y-2m+n\right)\left(x-y+2m-n\right)\)

a) (4x²-3x-18)²-(4x²+3x)²

=(4x²-3x-18-4x²-3x)(4x²-3x-18+4x²+3x)

=(-6x-18)(8x²-18)

=-48x³+108x-144x²+324

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)x^2-(a+b)x+ab

= x^2 - ax - bx + ab

= (x^2 - ax) - (bx - ab)

= x(x-a) - b(x-a)

= (x-b)(x-a) 

b)7x^3-3xyz-21x^2+9z

= 

c)4x+4y-x^2(x+y)

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

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

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

d) y^2+y-x^2+x

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

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

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

e)4x^2-2x-y^2-y

= [(2x)^2 - y^2] - (2x +y)

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

= (2x -y -1)(2x+y)

f)9x^2-25y^2-6x+10y

= 

ko biết làm

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

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

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