1/Tự chép lại đb nha :v

 =a² - 9b²+2ab+3a²-8b²-12ab+6ab-3b²-2a²+ab

= 2a²-3ab-20b²

= (2a²+5ab) - (8ab+20b²)

= a(2a+5b) - 4b(2a+5b)

=(2a+5b)(a-4b)

câu 2 tương tự nhé :)

Bài 1: 

a: Ta có: \(\left(6x+3\right)-\left(2x-5\right)\left(2x+1\right)\)

\(=\left(2x+1\right)\left(3-2x+5\right)\)

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

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

b) Ta có: \(\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)

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

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

\(=\left(3x-2\right)\left(3x-6\right)\)

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

Bài 2: 

a: Ta có: \(\left(a-b\right)\left(a+2b\right)-\left(b-a\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)

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

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

\(=\left(a-b\right)\left(2a-4b\right)\)

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

f: Ta có: \(x^2-6xy+9y^2+4x-12y\)

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

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

whatterfice

1. \(4x^2-17xy+13y^2=4x^2-4xy-13xy+13y^2=4x\left(x-y\right)-13y\left(x-y\right)=\left(x-y\right)\left(4x-13y\right)\)

2. \(2x\left(x-5\right)-x\left(3+2x\right)=26\Leftrightarrow2x^2-10x-3x-2x^2=26\Leftrightarrow-13x=26\Leftrightarrow x=-2\)

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

\(\Leftrightarrow\left(2a-3b\right)^2-2\left(2a-3b\right)\left(2b-3a\right)+\left(2b-3a\right)^2=\left(2a-3b-2b+3a\right)^2=\left(5a-5b\right)^2\)

\(=25\left(a-b\right)^2=25\cdot100=2500\)

Mấy câu dễ mình làm trước nhé. Mấy câu khó hơn mình trình bày sau :)

1) 2x² - 5xy - 3y² = 2x² + xy - 6xy - 3y² = x( 2x + y ) - 3y( 2x + y ) = ( 2x + y )( x - 3y )

2) 7x² + 3xy - 10y² = 7x² - 7xy + 10xy - 10y² = 7x( x - y ) + 10y( x - y ) = ( x - y )( 7x + 10y )

3) x² + 5x - 2 = ( x² + 5x + 25/4 ) - 33/4 = ( x + 5/2 )² - \(\left(\frac{\sqrt{33}}{2}\right)^2\)= \(\left(x+\frac{5}{2}-\frac{\sqrt{33}}{2}\right)\left(x+\frac{5}{2}+\frac{\sqrt{33}}{2}\right)\)

6) x⁴ + 324 = ( x⁴ + 36x² + 324 ) - 36x² = ( x² + 18 )² - ( 6x )² = ( x² - 6x + 18 )( x² + 6x + 18 )

4) x⁸ + x⁷ + 1

= x⁸ + x⁷ + x⁶ - x⁶ + 1

= x⁶( x² + x + 1 ) - ( x⁶ - 1 )

= x⁶( x² + x + 1 ) - ( x³ - 1 )( x³ + 1 )

= x⁶( x² + x + 1 ) - ( x - 1 )( x² + x + 1 )( x³ + 1 )

= ( x² + x + 1 )( x⁶ - ( x - 1 )( x³ + 1 ) ]

= ( x² + x + 1 )( x⁶ - x⁴ + x³ - x + 1 )

5) x⁷ + x⁵ + 1

= x⁷ + x⁶ - x⁶ + x⁵ + 1

= x⁵( x² + x + 1 ) - ( x⁶ - 1 )

= x⁵( x² + x + 1 ) - ( x³ - 1 )( x³ + 1 )

= x⁵( x² + x + 1 ) - ( x - 1 )( x² + x + 1 )( x³ + 1 )

= ( x² + x + 1 )[ x⁵ - ( x - 1 )( x³ + 1 ) ]

= ( x² + x + 1 )( x⁵ - x⁴ + x³ - x + 1 )

7) x⁵ - 5x³ + 4x

= x⁵ - x³ - 4x³ + 4x

= x³( x² - 1 ) - 4x( x² - 1 )

= ( x² - 1 )( x³ - 4x )

= ( x - 1 )( x + 1 )x( x² - 4 )

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

8) Xin hàng :)

e) Ta có: \(a^3-a^2-a+1\)

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

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

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

f) Ta có: \(x^3-2xy-x^2y+2y^2\)

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

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

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

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

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

d) Đề sai ko ???

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

f) \(x^3-2xy-x^2y+2y^2=x^2\left(x-y\right)-2y\left(x-y\right)=\left(x-y\right)\left(x^2-2y\right)\)

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

\(b,=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)

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

\(d,=2\left(x^2+2xy+y^2-4z^2\right)=2\left(\left(x+y\right)^2-4z^2\right)=2\left(x+y-2z\right)\left(x+y+2z\right)\)

\(e,=a^2\left(a-1\right)-\left(a-1\right)=\left(a-1\right)\left(a^2-1\right)\)

\(f,=x^2\left(x-y\right)-2y\left(x-y\right)=\left(x^2-2y\right)\left(x-y\right)\)

Phân tích đa thức thành nhân tử
a) (1-2x)(1+2x)-x(x+2)(x-2)

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

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

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

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

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

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

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

\(=a\left(a^3+6a^2b+12ab^2+8b^3\right)-b\left(8a^3+12a^2b+6ab^2+b^3\right)\)

\(=a^4+6a^3b+12a^2b^2+8b^3a-8a^3b-12a^2b^2+6ab^3-b^4\)

\(=a^4+6a^3b+8b^3a-8a^3b-6ab^3-b^4\)

\(=\left(a^4-b^4\right)+\left(6a^3b-6ab^3\right)+\left(8b^3a-8a^3b\right)\)

\(=\left(a-b\right)\left(a^3+a^2b+ab^2+b^3\right)+6ab\left(a^2-b^2\right)+8ab\left(b^2-a^2\right)\)

\(=\left(a-b\right)\left(a^3+a^2b+ab^2+b^3\right)+6ab\left(a-b\right)\left(a+b\right)-8ab\left(a-b\right)\left(a+b\right)\)

\(=\left(a-b\right)\left(a^3+a^2b+ab^2+b^3+6a^2b+6ab^2-8a^2b-8ab^2\right)\)

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

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

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

\(a^3\left(c-b^2\right)+b^3\left(a-c^2\right)+c^3\left(b-a^2\right)+abc\left(abc-1\right)\)

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

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

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

\(=\left(c-b^2\right)\left[a^2\left(a-c^2\right)-b\left(a-c^2\right)\right]\)

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

ủa phần a mình phân tích rồi mà bạn hu hu