\(A=x^3-15x^4+16x^3-29x^2\)

\(A=\left(x^3+16x^3\right)-15x^4-29x^2\)

\(A=17x^3-15x^4-29x^2\)

\(A=-15x^4+17x^3-29x^2\)

\(A=-x^2\left(15x^2-17x+29\right)\)

\(2x^4-8x^3-7x^3+28x^2+7x^2-28x-2x+8\\ =2x^3\left(x-4\right)-7x^2\left(x-4\right)+7x\left(x-4\right)-2\left(x-4\right)\\ =\left(x-4\right)\left(2x^3-7x^2+7x-2\right)\\ =\left(x-4\right)\left(2x^3-4x^2-3x^2+6x+x-2\right)\\ =\left(x-4\right)\left[2x^2\left(x-2\right)-3x\left(x-2\right)+\left(x-2\right)\right]\\ =\left(x-4\right)\left(x-2\right)\left(2x^2-2x-x+1\right)\\ =\left(x-4\right)\left(x-2\right)\left(2x-1\right)\left(x-1\right)\)

a)15x^2y-20xy^2-25xy=5xy(3x^2-4y-5)
b) x^2-9-2xy+y^2=(x^2-2xy+y^2)-9=(x-y)^2-3^2=(x-y-3)(x-y+3)
c)2x^2-5x-7=2x^2 -7x+2x-7 = x(2x-7)+(2x-7) =(2x-7)(x+1)

a. x² - 11 = \(x^2-\left(\sqrt{11}\right)^2=\left(x-\sqrt{11}\right)\left(x+\sqrt{11}\right)\)

b. x² - 5 = \(x^2-\left(\sqrt{5}\right)^2=\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)\)

c. x² - 7 = \(x^2-\left(\sqrt{7}\right)^2=\left(x+\sqrt{7}\right)\left(x-\sqrt{7}\right)\)

bạn ơi  bạn viết sai đề có đề nào viết  thế này đâu?????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????

Uchiha Itachi

Trả lời:

1, 15x + 15y = 15 ( x + y )

2, 8x - 12y = 4 ( 2x - 3y )

3, xy - x = x ( y - 1 )

4, x² + x = x ( x + 1 )

5, 3x²y - 8xy² = xy ( 3x - 8y )

6, 6x - 12xy - 18x² = 6x ( 1 - 2y - 3x )

1) 15x + 15y = 15(x + y)

2) 8x - 12y = 4(2x - 3y)

3) xy - x = x(y - 1)

4) x² + x = x(x + 1) 

5) 3x²y - 8xy² = xy(3x - 8y)

6) 6x - 12xy - 18x² = 6x(1 - 2y - 3x) 

1.\(15x+15y=15\left(x+y\right)\)

2.\(8x-12y=4\left(2x-3y\right)\)

3.\(xy-x=x\left(y-1\right)\)

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

5.\(3x^{2y}-8xy^2\)hay là \(\left(3x\right)^{2y}-\left(8xy\right)^2\)??

6.\(6x-12xy-18x^2=6x\left(1-2y-3x\right)\)

   6x^2 + 15x - 36

= 6x^2 + 24x - 9x - 36

= (6x^2 + 24x) - (9x + 36)

= 6x(x + 4) - 9(x+4)

= (6x - 9) (X + 4)

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

\(6x^2+15-36\)

\(=3\left(2x^2+5x-12\right)\)

\(=3\left(2x^2+8x-3x-12\right)\)

\(=3\left[2x\left(x+4\right)-3\left(x+4\right)\right]\)

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

\(-8x^3+1=1^3-\left(2x\right)^3=\left(1-2x\right)\left(1+2x+4x^2\right)\)

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

3x(x² - 5)

ta có :

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