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

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

= (2 - x)(x + y)

2/ 5x² - 5xy -10x + 10y

= 5x(x - y) - 10(x - y)

= (5x - 10(x - y)

3/ 4x² + 8xy - 3x - 6y

= 4x(x + 2y) - 3(x + 2y)

= (4x - 3)(x + 2y)

1) 2x + 2y - x(x + y) 

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

= (2 - x)(x + y)

2) 5x² - 5xy - 10x + 10y 

= 5x(x - y) - 10(x - y)

= (5x - 10)(x - y)

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

3) 4x² + 8xy - 3x - 6y  

= 4x(x + 2y) - 3(x + 2y)

= (4x - 3)(x + 2y)

4) 2x² + 2y² - x²z + z - y²z - 2 

= 2(x² + y² - z(x² + y²) - (2 - z)

= (2 - z)(x² + y²) - (2 - z)

= (2 - z)(x² + y²)

5) x² + xy - 5x - 5y

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

= (x - 5)(x + y)

6) x(2x - 7) - 4x + 14 

= x(2x - 7) - 2(2x - 7) 

= (x - 2)(2x - 7)

7)x² - 3x + xy - 3y  

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

= (x - 3)(x + y)

5/ x² + xy - 5x - 5y 

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

= (x - 5)(x + y)

6/ x(2x - 7) - 4x + 14

= 2x² - 7x - 4x + 14

= (2x² - 4x) - (7x - 14)

= 2x(x - 2) -7(x - 2)

= (2x - 7)(x - 2)

7/ x² - 3x + xy - 3y

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

= (x + y)(x - 3) 

a) \(2x\left(x-7\right)-5y\left(x-7\right)=\left(x-7\right)\left(2x-5y\right)\)

b) \(5x^3y+10x^2y+5xy=5xy\left(x^2+2x+1\right)=5xy\left(x+1\right)^2\)

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

d) \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1=\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)

\(=\left(x^2+3x\right)^2+2\left(x^2+3x\right)+1=\left(x^2+3x+1\right)^2\)

a: \(=\left(x-7\right)\left(2x-5y\right)\)

b: \(=5xy\left(x^2+2x+1\right)=5xy\left(x+1\right)^2\)

a) \(=\left(2x-5y\right)\left(x-7\right)\)

b) \(=5xy\left(x^2+2x+1\right)=5xy\left(x+1\right)^2\)

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

\(B=x^5y^2+\dfrac{1}{2}x^5y^2-6xy+1=\dfrac{3}{2}x^5y^2-6xy+1\)

\(B=-\dfrac{1}{7}x^2y+x^5y^2-xy+\dfrac{1}{2}x^5y^2-5xy+\dfrac{1}{7}x^2y+2021^0\\ =\left(-\dfrac{1}{7}x^2y+\dfrac{1}{7}x^2y\right)+\left(x^5y^2+\dfrac{1}{2}x^5y^2\right)-\left(xy+5xy\right)+1\\ =0+\dfrac{3}{2}x^5y^2-6xy+1\\ =\dfrac{3}{2}x^5y^2-6xy+1\)

Bài 1:

\(a,6x^2-15x^3y\\ b,=-\dfrac{2}{3}x^2y^3+\dfrac{2}{3}x^4y-\dfrac{8}{3}xy\)

Bài 2:

\(a,=20x^3-10x^2+5x-20x^3+10x^2+4x=9x\\ b,=3x^2-6x-5x+5x^2-8x^2+24=24-11x\\ c,=x^5+x^3-2x^3-2x=x^5-x^3-2x\)

câu d của bài 2 là của bài 1 nha mình để nhầm chỗ huhu

Đáp án:

 a.3x³−5x²+7xa.3x³−5x²+7x

b.−4x²y−10x²y+2xyb.−4x²y−10x²y+2xy

c.−x³+2x²+29x+20c.−x³+2x²+29x+20

d.2x⁴−3x³+2x²+3x−4d.2x⁴−3x³+2x²+3x−4

e.x²−4y²e.x²−4y²

h.2x²−6x+13h.2x²−6x+13

g.3xy⁴−12y²+2x²yg.3xy⁴−12y²+2x²y 

f.−2x²y³+y−3f.−2x²y³+y−3

Giải thích các bước giải:

 a.3x.(x²−5x+7)a.3x.(x²−5x+7)

=3x³−5x²+7x=3x³−5x²+7x

b.−2xy.(2x³+5x−1)b.−2xy.(2x³+5x−1)

=−4x⁴y−10xy²+2xy=−4x⁴y−10xy²+2xy

c.(x+4).(−x²+6x+5)c.(x+4).(−x²+6x+5)

=−x³+6x²+5x−4x²+24x+20=−x³+6x²+5x−4x²+24x+20

=−x³+2x²+29x+20=−x³+2x²+29x+20

d.(x²−1).(2x²−3x+4)d.(x²−1).(2x²−3x+4)

=2x⁴−3x³+4x²−2x²+3x−4=2x⁴−3x³+4x²−2x²+3x−4

=2x⁴−3x³+2x2+3x−4=2x⁴−3x³+2x2+3x−4

$e.(x+2y).(x-2y)$

=x²−(2y)²=x²−(2y)²

=x²−4y²=x²−4y²

h.(3x−1)²−7(x²+2)h.(3x−1)²−7(x²+2)

=9x²−6x+1−7x²−14=9x²−6x+1−7x²−14

=2x²−6x+13=2x²−6x+13

g.(6x²g.(6x²y⁵−xy³+4x³y²):2xy−xy³+4x³y²):2xy

=3xy⁴−12y²+2x²y=3xy⁴−12y²+2x²y 

f.(−12x³y⁴+6xy²−18xy):6xyf.(−12x³y⁴+6xy²−18xy):6xy

=−2x³y³+y−3

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

\(\Leftrightarrow x^2+2-y\left(x-1\right)=0\)

\(\Leftrightarrow x^2-1-y\left(x-1\right)+3=0\)

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

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

...

b ) \(3xy-5x-2y=3\)

\(\Leftrightarrow9xy-15x-6y=9\)

\(\Leftrightarrow9xy-15x-6y+10=19\)

\(\Leftrightarrow3y\left(3x-2\right)-5\left(3x-2\right)=19\)

\(\Leftrightarrow\left(3y-5\right)\left(3x-2\right)=19\)

...

c ) \(x^2-10xy-11y^2=13\)

\(\Leftrightarrow x^2-11xy+xy-11y^2=13\)

\(\Leftrightarrow x\left(x-11y\right)+y\left(x-11y\right)=13\)

\(\Leftrightarrow\left(x+y\right)\left(x-11y\right)=13\)

...

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

\(\Leftrightarrow xy-2-2x-3y=0\)

\(\Leftrightarrow y\left(x-3\right)-2\left(x-3\right)-8=0\)

\(\Leftrightarrow\left(y-2\right)\left(x-3\right)=8\)

...

e ) \(5xy+x+2y=7\)

\(\Leftrightarrow5xy+x+2y-7=0\)

\(\Leftrightarrow5x\left(y+\dfrac{1}{5}\right)+2\left(y+\dfrac{1}{5}\right)-\dfrac{37}{5}=0\)

\(\Leftrightarrow\left(5x+2\right)\left(y+\dfrac{1}{5}\right)=\dfrac{37}{5}\)

\(\Leftrightarrow\left(5x+2\right)\left(5y+1\right)=37\)

...

P/s : Vì bài dài nên việc tìm x , y ( lập bảng ) bạn tự làm nhé

Thanks