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\(3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)^2=3[\left(x^4+2x^2+1\right)-x^2]-\left(x^2+x+1\right)^2\)\(=3[\left(x^2+1\right)^2-x^2]-\left(x^2+x+1\right)^2\)
\(=3\left(x^2-x+1\right)\left(x^2+x+1\right)-\left(x^2+x+1\right)^2\)
\(=\left(x^2+x+1\right)\left(2x^2-4x+2\right)=2\left(x-1\right)^2\left(x^2+x+1\right)\)
\(\left(x+1\right)^4+\left(x^2+x+1\right)^2\)
\(=\left(x+1\right)^4+x^2\cdot\left(x+1\right)^2+2x\left(x+1\right)+1\)
\(=\left(x+1\right)^2\cdot\left[\left(x+1\right)^2+x^2\right]+2x^2+2x+1\)
\(=\left(2x^2+2x+1\right)\left(x^2+2x+1+1\right)\)
\(=\left(2x^2+2x+1\right)\left(x^2+2x+2\right)\)
\(3\left(x^2+x+1\right)-\left(x^2+x+1\right)^2\)
\(=\left(x^2+x+1\right)\left(3-x^2-x-1\right)\)
\(=\left(x^2+x+1\right)\left(2-x^2-x\right)\)
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)
1. = (x-2)^2 - y^2 = (x - 2 - y)(x-2+y)
2. = (x-y-x-y)(x-y+x+y) = 2(-y)2x = -4xy
Ta có:
\(x^3-x^2-x-2=x^3-2x^2+x^2-2x+x-2\)
\(=x^2\left(x-2\right)+x\left(x-2\right)+x-2=\left(x-2\right)\left(x^2+x+1\right)\)
\(x^2\left(x-2\right)^2-\left(x-2\right)^2-x^2+1\)
\(=\left(x-2\right)^2\left(x^2-1\right)-\left(x^2-1\right)\)
\(=\left[\left(x-2\right)^2-1\right]\left(x^2-1\right)\)
Làm tiếp cho chắc nhé
\(=\left(x-2-1\right)\left(x-2+1\right)\left(x-1\right)\left(x+1\right)\)
\(=\left(x-3\right)\left(x-1\right)\left(x-1\right)\left(x+1\right)\)
\(=\left(x-3\right)\left(x-1\right)^2\left(x+1\right)\)