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a) \(x^4+324=\left(x^2-6x+18\right)\left(x^2+6x+18\right)\)
c) \(x^{13}+x^5+1=\left(x^2+x+1\right)\left(x^{11}-x^{10}+x^8-x^7+x^5-x^4+x^3-x+1\right)\)
d) \(x^{11}+x+1=\left(x^2+x+1\right)\left(x^9-x^8+x^6-x^5+x^3-x^2+1\right)\)
e) \(x^8+3x^4+4=\left(x^4+x^2+2\right)\left(x^4-x^2+2\right)\)
\(x^4+4\)
\(=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-4x^2\)
\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
a, \(x^5+x^4+1\)
\(\Leftrightarrow x^5+x^4-x^2+\frac{1}{4}-\frac{1}{4}+x^2\)
\(\Leftrightarrow x^5+\left(x^2-\frac{1}{2}\right)^2-\frac{1}{4}+x^2\)
\(\Leftrightarrow x^2\left(x^3+1\right)+\left(x^2-\frac{1}{2}\right)^2-\frac{1}{4}\)
\(\Leftrightarrow x^2\left(x+1\right)\left(x^2-x+1\right)+\left(x^2-\frac{1}{2}+\frac{1}{2}\right)\left(x^2-\frac{1}{2}-\frac{1}{2}\right)\)
ta có :x^5 +x^4 +1=x^5-x^2 +x^4 -x +x^2 +x +1=x^2(x^3-1) +x(x^3 -1)+x^2 +x +1=x^2(x-1)(x^2+x+1)+x(x-1)(x^2 +x+1) +x^2 +x +1=(x^2 +x +1)(x^3 -x^2 +x^2 -x +1)=(x^2 +x+1)(x^3-x+1)
Ta có : x5 + x + 1
= x5 + x4 + x3 + x2 + x + 1 - x4 - x3 - x2
= (x5 + x4 + x3) + (x2 + x + 1) - (x4 + x3 + x2)
= x3(x2 + x + 1) + (x2 + x + 1) - x2(x2 + x + 1)
= (x2 + x + 1)(x3 - x2 + 1) .
Ta có : x5 + x + 1
= x5 + x4 + x3 + x2 + x + 1 - x4 - x3 - x2
= (x5 + x4 + x3) + (x2 + x + 1) - (x4 + x3 + x2)
= x3(x2 + x + 1) + (x2 + x + 1) - x2(x2 + x + 1)
= (x2 + x + 1)(x3 - x2 + 1) .
a, \(x^8+x^7+1=x^8-x^2+x^7-x+x^2+x+1=x^2\left(x^6-1\right)+x\left(x^6-1\right)+\left(x^2+x+1\right)\)
\(=x^2\left(x^3-1\right)\left(x^3+1\right)+x\left(x^3-1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\)
\(=x^2\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)+x\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left[x^2\left(x-1\right)\left(x^3+1\right)+x\left(x-1\right)\left(x^3+1\right)+1\right]\)
\(=\left(x^2+x+1\right)\left[\left(x^3-x^2\right)\left(x^3+1\right)+\left(x^2-x\right)\left(x^3+1\right)+1\right]\)
\(=\left(x^2+x+1\right)\left(x^6+x^3-x^5-x^2+x^5+x^2-x^4-x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)
b, \(x^8+x^4+1=x^8+2x^4+1-x^4=\left(x^4+1\right)^2-x^4=\left(x^4-x^2+1\right)\left(x^4+x^2+1\right)\)
\(=\left(x^4-x^2+1\right)\left(x^4+2x^2+1-x^2\right)=\left(x^4-x^2+1\right)\left[\left(x^2+1\right)-x^2\right]=\left(x^4-x^2+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)\)
c, \(x^5+x+1=x^5-x^2+x^2+x+1=x^2\left(x^3-1\right)+\left(x^2+x+1\right)=x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
khó quá mk nản chí rùi huhu!!
3463465655775676876897756232544545465657578768
x^8 + 4 = x^8 + 4x^4 + 4 - 4 x^4
= ( x^ 4 + 2 )^2 - (2x^2)^2
= ( x^4 + 2x^2 + 2 )( x^4 - 2x^2 + 2)
+) \(A=x^8+x+1=\left(x^8-x^2\right)+\left(x^2+x+1\right)=x^2\left(x^6-1\right)+\left(x^2+x+1\right)\)
Ta có : \(x^6-1=\left(x^3+1\right)\left(x^3-1\right)=\left(x^3+1\right)\left(x-1\right)\left(x^2+x+1\right)\)
Thay vào A được : \(A=x^2\left(x^3+1\right)\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left[x^2\left(x^3+1\right)\left(x-1\right)+1\right]=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)
Câu dưới tương tự...
x8 + x + 1 = x8 + x4 - x4 + x2 - x2 + x + 1
= ( x8 + x4 + 1 ) - ( x4 + x2 + 1 ) + ( x2 + x + 1 ) ( 1 )
trong đó :
x4 + x2 + 1 = x2 + 2x2 - x2 + 1
= ( x4 + 2x2 + 1 ) - x2
= ( x2 + 1 )2 - x2
= ( x2 - x + 1 )( x2 + x + 1 ) ( 2 )
x8 + x4 + 1 = x8 + 2x4 - x4 + 1
= ( x8 + 2x4 + 1 ) - x4
= ( x4 + 1 ) - ( x2 )2
= ( x4 - x2 + 1 )( x4 + x2 + 1 )
= ( x4 - x2 + 1 )( x2 - x + 1 )( x2 + x + 1 )
Thế ( 2 ) , ( 3 ) vào ( 1 ) ta được :
x8 + x + 1 = ( x4 - x2 + 1 )( x2 - x + 1 )( x2 + x + 1 ) - ( x2 - x + 1 )( x2 + x + 1 ) + ( x2 + x + 1 )
= ( x2 + x + 1 )[ ( x4 - x2 + 1 )( x2 - x + 1 ) - ( x2 - x + 1 ) + 1 )
= ( x2 + x + 1 )( x6 - x5 + x3 - x2 + 1 )