a) \(x^3+3x=x\left(x+3\right)\)

b) \(9x^2-6x=3x\left(3x-2\right)\)

c) \(5y^{10}+15y^6=5y^6\left(y^4+3\right)\)

d) \(x^4y-2x^2y^2+5xy=xy\left(x^3-2xy+5\right)\)

\(\left(x-2\right)\left(x^2-2x+4\right)\)

Ta có : x² - 6x + 5

= x² - x - 5x + 5 

= (x² - x) - (5x - 5)

= x(x - 1) - 5(x - 1)

= (x - 5)(x - 1)

Đọc đề

Tao có \(x^2-6x+5\)

=\(x^2-x-5x+5\)

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

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

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

Bài 2:

1)  \(x^2-4x+4=\left(x-2\right)^2\)

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

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

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

5) \(\dfrac{1}{25}x^2-64y^2=\left(\dfrac{1}{5}x-8y\right)\left(\dfrac{1}{5}x+8y\right)\)

6) \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)

Bài 2:

7) \(x^3+\dfrac{1}{27}=\left(x+\dfrac{1}{3}\right)\left(x^2+\dfrac{1}{3}x+\dfrac{1}{9}\right)\)

8) \(x^3+64=\left(x+4\right)\left(x^2+4x+16\right)\)

9) \(\left(a+b\right)^2-\left(2a-b\right)^2=\left(a+b+2a-b\right)\left(a+b-2a+b\right)=3a\left(-a+2b\right)\)

10) \(\left(a+b\right)^2-\left(a-b\right)^2=\left(a+b+a-b\right)\left(a+b-a+b\right)=2a\cdot2b=4ab\)

11) \(\left(a+b\right)^3+\left(a-b\right)^3=\left(a+b+a-b\right)\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)

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

\(=2a\left(3a^2+b^2\right)\)

12) \(\left(6x-1\right)^2-\left(3x+2\right)^2=\left(6x-1+3x+2\right)\left(6x-1-3x-2\right)=\left(9x+1\right)\left(3x-3\right)\)

1:

1: ,4x^2-6x=2x(2x-3)

2: 9x^3y^2+3x^2y^2=3x^2y^2(3x+1)

3: x^3+2x^2+3x=x(x^2+2x+3)

4: 2x^2-4x=2x(x-2)

5: 3x-6y=3(x-2y)

6: x^2-3x=x(x-3)

7: 6x^2y+4xy^2+2xy

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

8: 5x^2(x-2y)-15x(x-2y)

=(x-2y)(5x^2-15x)

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

9: =3(x-y)+5y(x-y)

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

10: =(x-1)(3x+5)

11: =2(2x-1)-3(2x-1)

=-(2x-1)

ak

x8 + -7x4 + -8 = 0 Reorder the terms: -8 + -7x4 + x8 = 0 Solving -8 + -7x4 + x8 = 0 Solving for variable 'x'. Factor a trinomial. (-1 + -1x4)(8 + -1x4) = 0

bn nói j vậy bạn

bn vui lòng làm ra từng bước cho mk dc ko???

2x⁴ - 3x³ - 7x² +6x+8

= 2x⁴ - 4x³ + x³ - 2x² - 5x² +10x - 4x +8

= 2x³.(x-2) +x².(x-2) - 5x.(x-2) - 4.(x-2)

= (x-2).(2x³ +x² - 5x -4)

= (x-2).(2x³ + 2x² - x² - x - 4x-4)

= (x-2).(x+2).(2x² -x -4)

....

Bó tay với hệ số bất định