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MK bổ sung thêm nhé: abc=1
\(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ac+c+1}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{a\left(bc+b+1\right)}+\frac{abc}{ab\left(ac+c+1\right)}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{abc+ab+a}+\frac{abc}{abc.a+abc+ab}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}+\frac{1}{ab+a+1}\) ( thay abc = 1 )
\(=\frac{ab+a+1}{ab+a+1}=1\)
P=2.(5^2-1).(5^2+1).(5^4+1).(5^8+1).(5^16+1)
=2.(5^4-1).(5^4+1).(5^8+1).(5^16+1)
= 2.(5^8-1).(5^8+1).(5^16+1)
= 2.(5^16-1).(5^16+1)
= 2.(5^32-1)
1)P= 12(5^2+1)(5^4+1)(5^8+1)(5^16+1)
=> 2P = 24(5^2+1)(5^4+1)(5^8+1)(5^16+1)
=(5^2-1)(5^2+1)(5^4+1)(5^8+1)(5^16+1)
=(5^4-1)(5^4+1)(5^8+1)(5^16+1)
=(5^8-1)(5^8+1)(5^16+1)
=(5^16-1)(5^16+1)
=5^32-1
=> P = (5^32-1)/2
= 1/x+1 + 1/x+1 - 1/x+2 + 1/x+2 - 1/x+3 + .... + 1/x+2016 - 1/x+2017
= 2/x+1 - 1/x+2017
k mk nha
\(A=\left(10a^2\right)^3-1=1000a^6-1\)