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

\(=x^2\left(x-1\right)-2x\cdot\left(x-1\right)\)

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

5: \(y^2\left(x^2+y\right)-z\cdot x^2-zy\)

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

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

7: \(5\left(x+y\right)^2+15\left(x+y\right)\)

\(=5\left(x+y\right)\cdot\left(x+y\right)+5\left(x+y\right)\cdot3\)

=5(x+y)(x+y+3)

9: \(7x\left(y-4\right)^2-\left(4-y\right)^3\)

\(=7x\left(y-4\right)^2+\left(y-4\right)^3\)

\(=\left(y-4\right)^2\left(7x+y-4\right)\)

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

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

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

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

2: \(5x\left(x-2\right)-3x^2\left(x-2\right)\)

\(=\left(x-2\right)\left(5x-3x^2\right)=x\left(5-3x\right)\left(x-2\right)\)

4: 3x(x-5y)-2y(5y-x)

=3x(x-5y)+2y(x-5y)

=(x-5y)(3x+2y)

6: b(a-c)+5c-5a

=b(a-c)-5(a-c)

=(a-c)(b-5)

8: 9x(x-y)-10(y-x)^2

=9x(x-y)-10(x-y)^2

=(x-y)(9x-10x+10y)

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

10: \(\left(a-b\right)^2-\left(a+b\right)\left(b-a\right)\)

\(=\left(a-b\right)^2+\left(a+b\right)\left(a-b\right)\)

=(a-b)(a-b+a+b)

=2a(a-b)

12: 2x(x-3)+y(x-3)+(3-x)

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

=(x-3)(2x+y-1)

thoi minh luoi lam minh ko giai het duoc dau

- Đề bài bài 4 nhầm nha. 

- Phải là : 19^x + 5^y + 1980z = 1975^430 + 2004

a) => y+42+2y= -12-14+2y

y+2y-2y = -12-14-42

y= -68

b) => 15+y-5-5y= -12-5y

y-5y+5y= -12-15+5

y = -22

c) => 2y+5-8y+21= -3-5y-2

2y-8y+5y= -3-2-5-21

-y= -31=>y=31

d)=> -13+3y+23= -120+y

3y-y= -120+13-23

2y= -130=>y= -65

e) => -21+32+5y= 16+4y

5y-4y= 16+21-32

y= 5

bài 1

a)y-(-42-2y) = (-12) - 14 +2y

y +42 + 2y = -12 -14 +2y

3y + 42 = -26 +2y

y = -68

b)15-(-y+5)-5y=-(12+5y+2)

15+y-5-5y=-12-5y-2

10-4y=-14-5y

-4y+5y=-14-10=-24

c)2y-(-5+8y-21)=-3-(5y+2)

2y+5-8y+21=-3y-5y-2

-6y+26=-8y-2

-6y+8y=-2-26

2y=-28

y=-28/2=-14

Bài 2:

a: =6(15-5)=6*10=60

b: =9(-3+23)=9*20=180

c: =11(-10+210)=11*200=2200

d: =125*4+125*4=125*8=1000

\(a,=\left(x+y\right)\left(y+z\right)\\ b,=x\left(x^2+2x+1\right)=x\left(x+1\right)^2\\ c,=\left(x-y\right)\left(x+y\right)+\left(x-y\right)=\left(x+y+1\right)\left(x-y\right)\\ d,= \left(2x-5\right)\left(2x+5\right)\\ e,=\left(4y-3\right)\left(4y+3\right)\)

a: ta có: \(\dfrac{2x-5}{7x-1}=\dfrac{4x+3}{14x-9}\)

\(\Leftrightarrow\left(2x-5\right)\left(14x-9\right)=\left(7x-1\right)\left(4x+3\right)\)

\(\Leftrightarrow28x^2-18x-70x+45=28x^2+21x-4x-3\)

=>-88x+45=17x-3

=>-105x=-48

hay x=16/35

b: Sửa đề: \(\dfrac{x}{4}=\dfrac{y}{9}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{4}=\dfrac{y}{9}=\dfrac{x-y}{4-9}=\dfrac{105}{-5}=-21\)

Do đó: x=-84; y=-189

c: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{2x-5y}{2\cdot3-5\cdot4}=\dfrac{56}{-14}=-4\)

Do đó:x=-12; y=-16

e: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x^2}{2}=\dfrac{y^2}{3}=\dfrac{x^2+y^2}{2+3}=\dfrac{125}{5}=25\)

Do đó: \(x^2=50;y^2=75\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{5\sqrt{2};-5\sqrt{2}\right\}\\y\in\left\{5\sqrt{3};-5\sqrt{3}\right\}\end{matrix}\right.\)

a) = (x + 3)² - y² = (x + 3 - y)(x + 3 + y)

b) = x²(x - 3) -4(x - 3) = (x - 3)(x² - 4) = (x - 3)(x - 2)(x + 2)

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

d) Nhầm đề. tui sửa lại x³ + y³ + 2x² - 2xy + 2y²

= x³ + y³ + 2(x² - xy + y²) = (x + y)(x² - xy + y²) + 2(x² - xy + y²) = (x² - xy + y²)(x + y + 2)

e) = x⁴ - x³ - x³ + x² - x² + x + x - 1 = x³(x - 1) - x²(x - 1) - x(x - 1) + x - 1 = (x - 1)(x³ - x² - x + 1) = (x - 1)(x - 1)(x² - 1) = (x - 1)³(x + 1)

f) = x³ - 3x² - x² + 3x + 9x - 27 = x²(x - 3) - x(x - 3) + 9(x - 3) = (x-3)(x² - x + 9)

g) chắc là 3xyz 

= x²y + xy² + y²z + yz² + x²z + xz² + 3xyz = x²y + xy² + xyz + y²z + yz² + xyz + x²z + xz² + xyz = (x + y + z)(xy + yz + xz)

h) = 2³ -(3x)³ = (2 - 3x)(4 + 6x + 9x²)

i) = (x + y - x + y)(x + y + x - y) = 2y*2x = 4xy

k) = (x³ - y³)(x³ + y³) = (x - y)(x² + xy +y2)(x + y)(x² - xy +y2).