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xm+2+xm = xm(x^2 + 1)
xx+1-xx-1 = x^x . x - x^x - \(\frac{x^x}{x^x}\)=x^x ( x - 1 - 1/x^x) ( cái này mik ko bik đúng hay sai)
a^3 - 1 = a^3 - 1^3 = ( a-1)(a^2 + a + 1)
a^5 - b^5 = \(\left(\sqrt{a^5}\right)^2-\left(\sqrt{b^5}\right)^2=\left(\sqrt{a^5}+\sqrt{b^5}\right)\left(\sqrt{a^5}-\sqrt{b^5}\right)\)
\(x^7+x^5+x^4+x^3+x^2+1\)
\(=x^7+x^6-x^6-x^5+2x^5+2x^4-x^4-x^3+2x^3+2x^2-x^2-x+x+1\)
\(=\left(x^7+x^6\right)-\left(x^6+x^5\right)+\left(2x^5+2x^4\right)-\left(x^4+x^3\right)+\left(2x^3+2x^2\right)-\left(x^2+x\right)+\left(x+1\right)\)
\(=x^6.\left(x+1\right)-x^5.\left(x+1\right)+2x^4\left(x+1\right)-x^3\left(x+1\right)+2x^2\left(x+1\right)-x\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^6-x^5+2x^4-x^3+2x^2-x+1\right)\)
1/ đề sai vd: 2+3=5 là số nguyên tố
2/ \(4x^2-a^2+y^2-16b^2+4xy+8ab\)
\(=\left[\left(2x\right)^2+2.2xy+y^2\right]-\left[a^2+2.4ab-\left(4b\right)^2\right]\)
\(=\left(2x+y\right)^2-\left(a-4b\right)^2\)
\(=\left(2x+y+a-4b\right)\left(2x+y-a+4b\right)\)
3/
\(M=\left(x-1\right)\left(x+5\right)\left(x^2+4x+5\right)\)
\(=\left(x^2+5x-x-5\right)\left(x^2+4x+5\right)\)
\(=\left(x^2+4x-5\right)\left(x^2+4x+5\right)\)
\(=\left(x^2+4x\right)^2-5^2\)
\(=\left(x^2+4x\right)^2-25\)
Vì \(\left(x^2+4x\right)^2\ge0\)
\(\Rightarrow\left(x^2+4x\right)^2-25\ge-25\)
\(\Rightarrow M\ge-25\)
Dấu "=" xảy ra khi x = 0 hoặc x = -4
Vậy Mmin = -25 khi x = 0 hoặc x = -4
\(3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)^2=3[\left(x^4+2x^2+1\right)-x^2]-\left(x^2+x+1\right)^2\)\(=3[\left(x^2+1\right)^2-x^2]-\left(x^2+x+1\right)^2\)
\(=3\left(x^2-x+1\right)\left(x^2+x+1\right)-\left(x^2+x+1\right)^2\)
\(=\left(x^2+x+1\right)\left(2x^2-4x+2\right)=2\left(x-1\right)^2\left(x^2+x+1\right)\)