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\(x^4+2019x^2+2018x+2019\)
\(=x^4+x^2+1+2018\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+1\right)+2018\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2019\right)\)
\(x^7+x^5+x^4+x^3+x^2+1\)
\(=x^7+x^6-x^6-x^5+2x^5+2x^4-x^4-x^3+2x^3+2x^2-x^2-x+x+1\)
\(=\left(x^7+x^6\right)-\left(x^6+x^5\right)+\left(2x^5+2x^4\right)-\left(x^4+x^3\right)+\left(2x^3+2x^2\right)-\left(x^2+x\right)+\left(x+1\right)\)
\(=x^6.\left(x+1\right)-x^5.\left(x+1\right)+2x^4\left(x+1\right)-x^3\left(x+1\right)+2x^2\left(x+1\right)-x\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^6-x^5+2x^4-x^3+2x^2-x+1\right)\)
\(3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)^2=3[\left(x^4+2x^2+1\right)-x^2]-\left(x^2+x+1\right)^2\)\(=3[\left(x^2+1\right)^2-x^2]-\left(x^2+x+1\right)^2\)
\(=3\left(x^2-x+1\right)\left(x^2+x+1\right)-\left(x^2+x+1\right)^2\)
\(=\left(x^2+x+1\right)\left(2x^2-4x+2\right)=2\left(x-1\right)^2\left(x^2+x+1\right)\)
a) x2 - 4 + (x - 2)2 = 0
=> (x - 2)(x + 2) + (x2 - 4x + 4) = 0
x2 + 2x - 2x - 4 + x2 - 4x + 4 = 0
2x2 - 4x = 0
2x(x - 2) = 0
=> 2x = 0 => x = 0
x - 2 = 0 x = 2
Ta có \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\text{[}\left(x+2\right)\left(x+5\right)\text{]}.\text{[}\left(x+3\right)\left(x+4\right)\text{] -24}\)
\(=\left(x^2+7x+10\right)\times\left(x^2+7x+12\right)-24\)
Đặt \(x^2+7x+10=y\)
\(x^2+7x+12=y+2\)
\(\Rightarrow y\left(y+2\right)-24=y^2+2y-24\)
\(=y^2-4y+6y-24\)
\(=y\left(y-4\right)+6\left(y-4\right)\)
\(=\left(y-4\right)\left(y+6\right)\left(1\right)\)
Thay\(y=x^2+7x+10\) vào (1) ta được
\(\left(x^2+7x+10-4\right)\left(x^2+7x+10+6\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x^2+x+6x+6\right)\left(x^2+7x+16\right)\)
\(=\text{[}x\left(x+1\right)+6\left(x+1\right)\text{]}\left(x^2+7x+6\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+6\right)\)
Vậy \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left(x+1\right)\left(x+6\right)\left(x^2+7x+6\right)\)