Đúng đề không bạn cái đó phân tích nhân tử được hả

4x^2 - 5x + 9

= 4x^2 + 4x - 9x + 9

= 4x( x-1) -9(x-1)

= (4x-9)(x-1)

h nha

thanks

nếu nhóm 4x²+4x thì tại sao lại thành 4x(x-1)

\(4x^2-5x-6=4x^2-8x+3x-6\)

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

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

5x² + 10xy - 4x - 8y

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

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

=(5x² + 10xy) - (4x + 8y) = 5x.(x + 2y) - 4.(x + 2y) = (5x - 4).(x + 2y)

\(5x^2+10xy-4x-8y=\left(5x^2+10xy\right)-\left(4x+8y\right)=5x\left(x+2y\right)-4\left(x+2y\right)=\left(5x-4\right)\left(x+2y\right)\)

b. 5x + 5y - 3x - 3y

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

= 2x + 2y

= 2(x + y)

#)Giải :

\(x^3-2x-4\)

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

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

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

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

\(x^4+2x^3+5x^2+4x-12\)

\(=x^4+x^3+6x^2+x^3+x^2+6x-2x^2-2x-12\)

\(=x^2\left(x^2+x+6\right)+x\left(x^2+x+6\right)-2\left(x^2+x+6\right)\)

\(=\left(x^2+x+6\right)\left(x^2+x-2\right)\)

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

Câu 1.

Đoán được nghiệm là 2.Ta giải như sau:

\(x^3-2x-4\)

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

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

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

cảm ơn nha!

b: \(=x\left(x^4-5x^2+4\right)\)

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

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

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

a: Đa thức này không phân tích được bạn nhé

b: \(5x^2+6xy+y^2\)

\(=5x^2+5xy+xy+y^2\)

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

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

c: \(x^2-7x+10=\left(x-2\right)\left(x-5\right)\)

d: \(4x^2+12x+9-y^2\)

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

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

a) Ta có: \(8x+4x^2-12xy\)

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

b) Ta có: \(5x^3-10x^2+5x\)

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

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

c) Ta có: \(x^3+x^2y-xy^2-y^3\)

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

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

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

d) Ta có: \(x^2-8x-9\)

\(=x^2-9x+x-9\)

\(=\left(x-9\right)\left(x+1\right)\)

a. `8x+4x^2-12xy=4x(2+x-3y)`

b) `5x^3-10x^2+5x=5x(x^2-2x+1)`

c) `x^3+x^2y-xy^2-y^3=x^2(x+y)-y^2(x+y)=(x+y)(x^2-y^2)=(x+y)^2 (x-y)`

d) `x^2-8x-9=(x^2-2.x.4+4^2)-25=(x-4)^2-5^2=(x+1)(x-9)`

\(4x\left(x+1\right)^2-5x^2\left(x+1\right)-4\left(x+1\right) =\left(x+1\right)\left[4x\left(x+1\right)-5x^2-4\right]=\left(x+1\right)\left(4x^2+4x-5x^2-4\right)=\left(x+1\right)\left(-x^2+4x-4\right)=-\left(x+1\right)\left(x-2\right)^2\)

\(4x\left(x+1\right)^2-5x^2\left(x+1\right)-4\left(x+1\right)\)

\(=4x\left(x^2+2x+1\right)-5x^2\left(x+1\right)-4\left(x+1\right)\)

\(=4x^3+8x^2+4x-5x^3-5x-4x-4\)

\(=-x^3+8x^2-5x-4\)