Bài 1:

\(1,Sửa:x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\\ 2,=6\left(x^2+2xy+y^2\right)=6\left(x+y\right)^2\\ 3,=2y\left(y^2+4y+4\right)=2y\left(y+2\right)^2\\ 4,=5\left(x^2-2xy+y^2\right)=5\left(x-y\right)^2\)

Bài 2:

\(1,=x\left(x^2-64\right)=x\left(x-8\right)\left(x+8\right)\\ 2,=2y\left(4x^2-9\right)=2y\left(2x-3\right)\left(2x+3\right)\\ 3,=3\left(x^3-1\right)=3\left(x-1\right)\left(x^2+x+1\right)\)

Bài 3:

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

Bài 1;

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

2) \(6x^2+12xy+6y^2=6\left(x^2+2xy+y^2\right)=6\left(x+y\right)^2\)

3) \(2y^3+8y^3+8y=10y^3+8y=2y\left(5y^2+4\right)\)

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

Bài 2:

1) \(x^3-64x=x\left(x^2-64\right)=x\left(x-8\right)\left(x+8\right)\)

2) \(8x^2y-18y=2y\left(4x^2-9\right)=2y\left(2x-3\right)\left(2x+3\right)\)

3) \(24x^3-3=3\left(8x^3-1\right)=3\left(2x-1\right)\left(4x^2+2x+1\right)\)

Bài 3:

1) \(5x^2+10x+5-5y^2=5\left(x^2+2x+1-y^2\right)=5\left[\left(x+1\right)^2-y\right]=5\left(x-y+1\right)\left(x+y+1\right)\)

2) \(3x^3-6x^2+3x-12xy^2=3x\left(x^2-2x+1-4y^2\right)=3x\left[\left(x-1\right)^2-\left(2y\right)^2\right]=3x\left(x-2y-1\right)\left(x+2y-1\right)\)

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

4) \(2x^3-2xy^2-8x^2+8xy=2x\left(x^2-y^2-4x+4y\right)=2x\left[\left(x-y\right)\left(x+y\right)-4\left(x-y\right)\right]=2x\left(x-y\right)\left(x+y-4\right)\)

a) =( x²+y²-5)²-[2(xy-2)]²

⁼( x²+y²-5)²- (2xy-4)²

=(x²+y²-5+2xy-4)(x²+y²-5-2xy+4)

=[(x+y)²-9][(x-y)²-1]

phân tích tiếp HĐT 2 ở 2 thừa số

đề sai rùi phải là : \(36\left(x-y\right)^2-25\left(2x-1\right)^2\)

\(=>\left[6\left(x-y\right)\right]^2-\left[5\left(2x-1\right)\right]^2=\left[6\left(x-y\right)-5\left(2x-1\right)\right]\left[6\left(x-y\right)+5\left(2x-1\right)\right]\)

\(=>\left(6x-6y-10x+5\right)\left(6x-6y+10x-5\right)=\left(5-4x-6y\right)\left(16x-6y-5\right)\)

Áp dụng HDT : x^2 -y^2 =(x-y) (x+y)

Ủng hộ = 1 cái t i c k nha cảm ơn

Phải \(2x^4\) thì mới phân tích được c hứu?

x^8+x^4+1=x^8-x^2+x^4-x+x^2+x+1=x^2(x^6-1)+x(x^3-1)+x^2+x+1=x^2(x^3-1)(x^3+1)+x(x^3-1)+x^2+x+1=x^2(x^3+1)(x-1)(x^2+x+1)+x(x-1)(x^2+x+1)+x^2+x+1=(x^2+x+1)[x^2(x^3+1)(x-1)+x(x-1)+1)]

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

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

bài 2 :

0,25x³+x²+x=0

<=>0,25x³+0,5x²+0,5x²+x=0

<=>0,25x²(x+2)+0,5x(x+2)=0

<=>(x+2)(0,25x²+0,5x)=0

<=>(x+2)x(0,25x+0,5)=0

<=>x+2=0 hoặc x=0 hoặc 0,25x+0,5=0

=>x=-2 hoặc x=0 hoặc x=-2

vậy x=0 hoặc x=-2

36(x-y)²-25(2x-y)²

= 36(x-y)² - 100(x-y)²

=(36-100)(x-y)²

= -64(x-y)²

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)

\(a^{2b}-b^2.\left(x+y\right)^2=\left(a^b\right)^2-\left(b\left(x+y\right)\right)^2=\left(a^b+bx+by\right)\left(a^b-bx-by\right)\)

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

TL:

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

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

\(=3x.\left(x+2\right)\) 

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

~Study well~ :)