Ta có

2x^4-x^3+2x^2+3x-2

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

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

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

Bạn ơi , tìm x 

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

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

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

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

Ta có: \(2x^3+3x^2+2x+3=0\)

\(\Leftrightarrow x^2\left(2x+3\right)+\left(2x+3\right)=0\)

\(\Leftrightarrow2x+3=0\)

hay \(x=-\dfrac{3}{2}\)

mình biết nội quy rồi nên đưng đăng nội quy

ai chơi bang bang 2 kết bạn với mình

mình có nick có 54k vàng đang góp mua pika 

ai kết bạn mình cho

a) \(\left(2x+5\right)^2-8^2\)

\(=\left(2x+5-8\right)\left(2x+5+8\right)\)

\(=\left(2x-3\right)\left(2x+13\right)\)

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

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

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

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

a) xy+3x-7y-21

=x(y+3)-7(x+3)

=(x-7)(y+3)

b)2xy-15-6x-5y

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

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

c)2x^2y+2xy^2-2x-2y

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

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

x(x+3)-5x(x-5)-5(x+3)

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

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

Câu cuối mình bị nhầm dòng cuối phải là (x-5)(x+3+x-5)=(x-5)(2x-2)nha bạn

a) xy+3x-7y-21=(xy+3x)-(7y+21)= x(y+3)-7(y+3)=(y+3)(x-7)

b)2xy-15-6x+5y=(2xy-6x)+(5y-15)=2x(y-3)+5(y-3)=(y-3)(2x+5)

c)2x^2y+2xy^2-2x-2y=2xy(x+y)-2(x+y)=2(x+y)(xy-1)

d) x(x+3)-5x(x-5)-5(x+3)=[x(x+3)-5(x+3)]-5x(x-5)=(x+3)(x-5)-5x(x-5)=(x-5)(x+3-5x)=(x-5)(3-4x)

Bài 1: 

a: \(=6x^3-10x^2+6x\)

b: \(=-2x^3-10x^2-6x\)

Bài 4: 

a: =>3x+10-2x=0

=>x=-10

c: =>3x²-3x²+6x=36

=>6x=36

hay x=6

Bài 1:

\(a,=6x^3-10x^2+6x\\ b,=-2x^3-10x^2-6x\)

Bài 4:

\(a,\Leftrightarrow3x+10-2x=0\Leftrightarrow x=-10\\ b,\Leftrightarrow x\left(2x^2+9x-5\right)-\left(2x^3+9x^2+x+4,5\right)=3,5\\ \Leftrightarrow2x^3+9x^2-5x-2x^3-9x^2-x-4,5=3,5\\ \Leftrightarrow-6x=8\Leftrightarrow x=-\dfrac{4}{3}\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\)

Bài 1:

\(a,=7xy\left(2x-3y+4xy\right)\\ b,=x\left(x+y\right)-5\left(x+y\right)=\left(x-5\right)\left(x+y\right)\\ c,=\left(x-y\right)\left(10x+8\right)=2\left(5x+4\right)\left(x-y\right)\\ d,=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\\ =2x\left(4x+2\right)=4x\left(2x+1\right)\\ e,=5\left[\left(x-y\right)^2-4z^2\right]=5\left(x-y-2z\right)\left(x-y+2z\right)\\ f,=x^2+8x-x-8=\left(x+8\right)\left(x-1\right)\\ g,\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\\ =\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\\ h,=x^2+3x+x+3=\left(x+3\right)\left(x+1\right)\)

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

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

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

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

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

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

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

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

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

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