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 \(x=-\dfrac{3}{2}\)

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

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

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

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

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

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a: Ta có: \(2-x=2\left(x-2\right)^3\)

\(\Leftrightarrow2\left(x-2\right)^3+x-2=0\)

\(\Leftrightarrow\left(x-2\right)\left[2\left(x-2\right)^2+1\right]=0\)

\(\Leftrightarrow x-2=0\)

hay x=2

c: Ta có: \(\left(x-1.5\right)^6+2\left(1.5-x\right)^3=0\)

\(\Leftrightarrow\left(x-1.5\right)^6-2\left(x-1.5\right)^3=0\)

\(\Leftrightarrow\left(x-1.5\right)^3\cdot\left[\left(x-1.5\right)^3-2\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1.5\\x=\sqrt[3]{2}+1.5\end{matrix}\right.\)

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

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

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

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

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

bạn sử dụng phương pháp hệ số bất định nha để tìm đáp số 

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

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:

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

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

a, 3x(x^2-4)=0

+3x=0=>x=0

+x^2-4=0

=>x^2=4

=>x=+-2

c,x(x+2)-3x-6=0

x(x+2)-3(x+2)=0

(x+2)(x-3)=0

TH1 :x+2=0

x=-2

TH2 : x-3=0

x=3

câu b bạn chờ mình chúc nha

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