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Ta có:
\(\left(2^3+1\right)\left(3^3+1\right)...\left(100^3+1\right)\)
\(=\left(2+1\right)\left(4-2+1\right)\left(3+1\right)\left(9-3+1\right)...\left(100+1\right)\left(100^2-100+1\right)\)
\(=3.3.4.7...101.9901\)
\(=\left(3.4.5...101\right)\left(3.7.13...9901\right)\)
\(\left(2^3-1\right)\left(3^3-1\right)...\left(100^3-1\right)\)
\(=\left(2-1\right)\left(4+2+1\right)\left(3-1\right)\left(9+3+1\right)...\left(100-1\right)\left(100^2+100+1\right)\)
\(=1.7.2.13.3.21...99.10101\)
\(=\left(1.2.3...99\right)\left(7.13.21.10101\right)\)
=> \(\frac{\left(2^3+1\right)\left(3^3+1\right)...\left(100^3+1\right)}{\left(2^3-1\right)\left(3^3-1\right)...\left(100^3-1\right)}\)
\(=\frac{\left(3.4.5...101\right)\left(3.7.13...9901\right)}{\left(1.2.3...99\right)\left(7.13.21.10101\right)}=\frac{\left(100.101\right).3}{\left(1.2\right).10101}=\frac{30300}{20202}\)
\(=3.3.4.7...101.9901\)
\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{63}+1\right).\)
\(=\frac{\left(3+1\right)\left(3-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{63}+1\right)}{2}\)
\(=\frac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{63}+1\right)}{2}\)
\(=\frac{\left(3^{64}-1\right)\left(3^{63}+1\right)}{2}\left(\text{bn xem lại chỗ }3^{63}\text{ nhé!! ko thì ko lm đc tiếp đâu}\right)\)
\(A=x^3-8-\left(x^3+3x^2+3x+1\right)+3\left(x^2-1\right)\)
\(=x^3-8-x^3-3x^2-3x-1+3x^2-3\)
\(=\left(x^3-x^3\right)+\left(-3x^2+3x^2\right)-3x-8-3\)
\(=-3x-11\)