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x8 + x4 + 1. = ( x8+ 2x4 +1 ) - x4. = (x4 + 1)2 - x4. = ( x4 - x2 + 1)(x4+x2 +1). =( x4 - x2 + 1)(x4+2x2 -x2+1). = ( x4 - x2 + 1)[( x2+1)2-x2]. =( x4 - x2 + 1)(x2+1-x2)(x2+1+x2). =( x4 - x2 + 1).2x2.
\(x^{10}+x^8+x^6+x^4+x^2+1=x^8\left(x^2+1\right)+x^4\left(x^2+1\right)+\left(x^2+1\right)\)\(=\left(x^2+1\right)\left(x^8+x^4+1\right)=\left(x^2+1\right)\left(x^8-x^2+x^4+x^2+1\right)\)
\(=\left(x^2+1\right)[x^2\left(x-1\right)\left(x^3+1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\left(x^2-x+1\right)]\)
\(=\left(x^2+1\right)\left(x^2+x+1\right)\left(x^6-x^5+x^3-x+1\right)\)
a) \(9x^2+6x-8\)
\(=9x^2+12x-6x-8\)
\(=3x\left(3x+4\right)-2\left(3x+4\right)\)
\(=\left(3x+4\right)\left(3x-2\right)\)
b) \(x^2-7xy+10y^2\)
\(=x^2-2xy-5xy+10y^2\)
\(=x\left(x-2y\right)-5y\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x-5y\right)\)
c) \(x^8+x^7+1\)
\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)
x10+x5+1
= x10+x9+x8-x9-x8-x7+x7+x6+x5-x6-x5-x4+x5+x4+x3-x3-x2-x+x2+x+1
= x8(x2+x+1)-x7(x2+x+1)+x5(x2+x+1)-x4(x2+x+1)+x3(x2+x+1)-x(x2+x+1)+(x2+x+1)
= (x2+x+1)(x8-x7+x5-x4+x3-x+1)
x^4 + x^2 + 1
= x^4 + 2x^2 + 1 - x^2
= ( x^2 + 1)^2 - x^2
= ( x^2 - x + 1 )( x^2 + x + 1)
\(x5+x-1 = x5-x4+x3+x4-x3+x2-x2+x-1 = x3(x2-x+1)+x2(x2-x+1)-(x2-x+1) = (x2-x+1)(x3+x2-1) \)
hc tốt nha !!!!!!!!!
\(x^2+x-1\)=\(x^2+2x\frac{1}{2}+\frac{1}{4}-\frac{5}{4}\)=\(\left(x+\frac{1}{2}\right)^2-\left(\frac{\sqrt{5}}{2}\right)^2\)=\(\left(x+\frac{1}{2}-\frac{\sqrt{5}}{2}\right)\left(x+\frac{1}{2}+\frac{\sqrt{5}}{2}\right)\)
đề bài có sai ko vậy bạn
A = x( x^7 + 1 ) + 1