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24 tháng 3 2017

a/ \(x^5+x+1=\left(x^5+x^4+x^3\right)+\left(-x^4-x^3-x^2\right)+\left(x^2+x+1\right)\)

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

24 tháng 3 2017

c/ \(x\sqrt{x}-3x+4\sqrt{x}-2=\left(x\sqrt{x}-x\right)+\left(-2x+2\sqrt{x}\right)+\left(2\sqrt{x}-2\right)\)

\(=\left(\sqrt{x}-1\right)\left(x-2\sqrt{x}+2\right)\)

27 tháng 9 2021

a) \(=x^4-14x^2+40-72=x^4-14x^2-32=\left(x-4\right)\left(x+4\right)\left(x^2+2\right)\)

b) \(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1=\left(x^2+5x\right)^2+2\left(x^2+5x\right)+1=\left(x^2+5x+1\right)^2\)

c) \(=x^4+3x^3-3x^2+3x^3+9x^2-9x+x^2+3x-3-5=x^4+6x^3+7x^2-6x-8=\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x+4\right)\)

a: Ta có: \(\left(x^2-4\right)\left(x^2-10\right)-72\)

\(=x^4-14x^2-32\)

\(=\left(x^2-16\right)\left(x^2+2\right)\)

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

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

\(=\left(x^2+5x+6\right)\left(x^2+5x+4\right)+1\)

\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+24+1\)

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

1 tháng 4 2017

mình tưởng lopw 9 mới học căn

2 tháng 4 2018

em chịu bác lớp 7 học rui

14 tháng 2 2020

\(x\sqrt{x}-3x+4\sqrt{x}-2=x\sqrt{x}-x-2x+2\sqrt{x}+2\sqrt{x}-2\)

\(=x\left(\sqrt{x}-1\right)-2\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)\)

\(=\left(\sqrt{x}-1\right)\left(x-2\sqrt{x}+2\right)\)

16 tháng 11 2021

\(1,\\ a,=6x^4-15x^3-12x^2\\ b,=x^2+2x+1+x^2+x-3-4x=2x^2-x-2\\ c,=2x^2-3xy+4y^2\\ 2,\\ a,=7x\left(x+2y\right)\\ b,=3\left(x+4\right)-x\left(x+4\right)=\left(3-x\right)\left(x+4\right)\\ c,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\ d,=x^2-5x+3x-15=\left(x-5\right)\left(x+3\right)\\ 3,\\ a,\Leftrightarrow3x\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

16 tháng 11 2021

Câu 1

a)\(3x^2\left(2x^2-5x-4\right)=6x^4-15x^3-12x^2\)

b)\(\left(x+1\right)^2+\left(x-2\right)\left(x+3\right)-4x=x^2+2x+1+x^2+3x-2x-6-4x=2x^2-x-5\)

 

Câu b đề sai nha bạn.

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a) \(x^3+9x^2+27x+27=\left(x+3\right)^3\)

b) \(3\sqrt{3x^3}+18x^2+12\sqrt{3x}+8=\left(\sqrt{3x}+2\right)^3\)

c) \(\dfrac{1}{4}-x^2=\left(\dfrac{1}{2}-x\right)\left(\dfrac{1}{2}+x\right)\)

18 tháng 7 2021

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

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

c) \(x^3-4x^2+12x-27=x^3-3x^2-x^2+3x+9x-27=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)=\left(x-3\right)\left(x^2-x+9\right)\)

d) \(a^6-a^4+2a^3+2a^2=a^2\left(a^4-a^2+2a+2\right)=a^2\left[a^2\left(a-1\right)\left(a+1\right)+2\left(a+1\right)\right]=a^2\left(a+1\right)\left(a^3-a^2+2\right)=a^2\left(a+1\right)\left[a^3+a^2-2a^2+2\right]=a^2\left(a+1\right)\left[a^2\left(a+1\right)-2\left(a-1\right)\left(a+1\right)\right]=a^2\left(a+1\right)^2\left(a^2-2a+2\right)\)

a) Ta có: \(x^2-2x-4y^2-4y\)

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

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

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

b) Ta có: \(x^3+2x^2+2x+1\)

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

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

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

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: \(x^2-6xy+9y^2+4x-12y\)

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

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