Yêu cầu đề bài là gì vậy bạn?

tính f(2013)

Đặt S = ( 1/1.2 + 1/3.4 + 1/5.6 + ... + 1/2017.2018 )

Đặt A = ( 1/1.2 + 1/3.4  + ... + 1/2017.2018)

= 1 - 1/2 + 1/3 - 1/4  + ... + 1/2017  - 1/2018

= ( 1 + 1/3 + ... + 1/2017 ) - ( 1/2 + 1/4 + ... + 1/2018 )

= ( 1 + 1/2 + ... + 1/2018 ) - 2 ( 1/2 + 1/4 + ... + 1/2018) )

= ( 1 + 1/2 + ... + 1/2018 ) - ( 1 + 1/2 + ... + 1/1009 )

= 1/1010 + 1/1011 + ... + 1/2018

=> A - ( 1/1010 + 1/1011 + ... + 1/2017 ) = 1/2018

=> S = 1/2018

Vậy S = 1/2018

thanks bạn nhiều

\(\frac{33-33:\frac{11}{6}}{\frac{6}{11}}\)

\(=\frac{33-33\cdot\frac{6}{11}}{\frac{6}{11}}\)

\(=\frac{33-33\cdot1}{1}\)

\(=33-33=0\)

\(\frac{33-33:\frac{11}{6}}{\frac{6}{11}}\)

\(=\frac{33-33\cdot\frac{6}{11}}{\frac{6}{11}}\)

\(=\frac{33-18}{\frac{6}{11}}\)

\(=\frac{15}{\frac{6}{11}}\)

\(=\frac{55}{2}\)

\(a+b+c = 1 ; 1/a + 1/b + 1/c = 1 \)

\(=> (a+b+c)(1/a +1/b+1/c) = 1\)

\(<=> a/b + b/a + a/c + c/a + b/c + c/b + 3 - 1 = 0\)

\(<=> (a^2+b^2)/ab + (a^2+c^2)/ac + (b^2+c^2)/bc + 2 =0\)

\(<=> (a^2 + b^2).c + (a^2+c^2).b + (b^2+c^2).a + 2abc = 0\)

\(<=> a^2c + b^2c + a^2b + c^2b + ab^2 + ac^2 + 2abc =0 \)

\(<=> a^2c + ac^2 + abc + a^2b+ ab^2 + abc + b^2c + bc^2 =0\)

\(<=> ac(a+b+c) + ab(a+b+c) + bc(b+c) =0 \)

\(<=> a(b+c)(a+b+c) + bc(b+c) =0 \)

\(<=> (b+c)(a^2 + ab + ac + bc ) = 0 \)

\(<=> (b+c)[a(a+b) + c(a+b)] =0\)

\(<=> (b+c)(a+b)(a+c) =0 \)

<=> 1 trong 3 số \(b+c;a+b ; a+c = 0\)

\(a+b=0 => a= -b => a + b + c = 1 <=> c = 1 ; a = b = 0\)

Thay vào S ta được : \(\Rightarrow S=0^{2019}+0^{2019}+1^{2019}=1\)