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10 tháng 8 2015

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

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

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

2 tháng 11 2018

\(x^8+x+1\)

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

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

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

6 tháng 9 2021

x+ 2500

= (x4)2 + 2.x4.50 + (50)2 - 2.x4.50

= (x4 + 50)2 - (10x2)2

= (x4 + 50 - 10x2)(x4 + 50 + 10x2)

10 tháng 10 2021

e) \(8\left(x+3y\right)-16x\left(x+3y\right)=\left(x+3y\right)\left(8-16x\right)=8\left(x+3y\right)\left(1-2x\right)\)

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

g) \(3\left(x-y\right)-5x\left(y-x\right)=3\left(x-y\right)+5x\left(x-y\right)=\left(3+5x\right)\left(x-y\right)\)

10 tháng 10 2021

mn ơiiiiiiiiiiiii

a) \(x^3-16x=x\left(x^2-4\right)=x\left(x-2\right)\left(x+2\right)\)

b) \(3x^2+3y^2-6xy-12=3\left(x^2-2xy+y^2-4\right)=3\left(x-y-2\right)\left(x-y+2\right)\)

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

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

5 tháng 8 2021

\(\left(x-1,5\right)^6+2.\left(1,5-x\right)^2=0\\ \Leftrightarrow\left(x-1,5\right)^6+2.\left(x-1,5\right)^2=0\\ \Leftrightarrow\left(x-1,5\right)^2.\left[\left(x-1,5\right)^4+2\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-1,5\right)^2=0\\\left(x-1,5\right)^4+2\ge0\forall x\in R\end{matrix}\right.\\ \Leftrightarrow x=1,5\)

Vậy x=1,5

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

\(\Leftrightarrow x-1.5=0\)

hay x=1,5

NV
7 tháng 4 2021

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

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

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

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

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

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

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

26 tháng 12 2021

\(-8x^3+1=1^3-\left(2x\right)^3=\left(1-2x\right)\left(1+2x+4x^2\right)\)

2 tháng 9 2021

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

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

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

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