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Ok em, để olm.vn giúp em nhá:
A = \(\dfrac{1}{2}\):3 + \(\dfrac{1}{3}\):4 + \(\dfrac{1}{4}\):5+...+\(\dfrac{1}{2018}\):2019 + \(\dfrac{1}{2019}\): 2020
A=\(\dfrac{1}{2}\times\dfrac{1}{3}+\dfrac{1}{3}\times\dfrac{1}{4}+\dfrac{1}{4}\times\dfrac{1}{5}+..+\dfrac{1}{2018}\times\dfrac{1}{2019}+\dfrac{1}{2019}\times\dfrac{1}{2020}\)
A = \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\)+....+ \(\dfrac{1}{2018}\) - \(\dfrac{1}{2019}\)+ \(\dfrac{1}{2019}\) - \(\dfrac{1}{2020}\)
A = \(\dfrac{1}{2}\) - \(\dfrac{1}{2020}\)
A = \(\dfrac{1009}{2020}\)
Ta có:\(\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times..\times\frac{2018}{2019}\times\frac{2019}{2020}\)\(=\frac{1}{2020}\)
Vậy biểu thức \(\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times..\times\frac{2018}{2019}\times\frac{2019}{2020}\)\(=\frac{1}{2020}\)
1/2 x 2/3 x 3/4 x ... x 2018/2019 x 2019/2020
= 1 x 2 x 3 x ... x 2018 x 2019 / 2 x 3 x 4 x ... x 2019 x 2020
Khử loại đi ta còn lại phân số 1/2020
Hok tốt ^^
\(A=\left(2020\times2019+2019\times2018\right)\times\left(1+\dfrac{1}{2}:1\dfrac{1}{2}-1\dfrac{1}{3}\right)\)
\(A=\left[2019\times\left(2020+2018\right)\right]\times\left(1+\dfrac{1}{2}:\dfrac{3}{2}-\dfrac{4}{3}\right)\)
\(A=4038\times2019\times\left(1+\dfrac{1}{3}-\dfrac{4}{3}\right)\)
\(A=4038\times2019\times0\)
\(A=0\)
\(A=\dfrac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2020\right)}{1\times2020+2\times2019+3\times2018+...+2020\times1}\)
Ta có: \(1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2020\right)\)
\(=\left(1+1+1+...+1\right)+\left(2+2+2+...+2\right)+\left(3+3+3+...+3\right)+...+\left(2019+2019\right)+2020\)
Trong đó có: 2020 số 1, 2019 số 2, 2018 số 3,..., 2 số 2019, 1 số 2020
Vậy: \(\left(1+1+...+1\right)+\left(2+2+...+2\right)+\left(3+3+...+3\right)+...+2020\)
\(=1\times2020+2\times2019+3\times2018+...+2020\times1\)
\(\Rightarrow A=\dfrac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2020\right)}{1\times2020+2\times2019+3\times2018+...+2020\times1}\)
\(A=\dfrac{1\times2020+2\times2019+3\times2018+...+2020\times1}{1\times2020+2\times2019+3\times2018+...+2020\times1}=1\)
1) 2*17*9+18*540+29*18
= 18*17+18*540+29*18
= 18*(17+540+29)
= 18*586
= 10548
2) 5*{26-[3*(5+2*5)+15]/15}
= 5*{26-[3*(5+10)+15]/15}
= 5*{26-[3*15+15]/15}
= 5*{26-[45+15]/15}
= 5*{26-60/15}
= 5*{26-4}
=5*22
=110
3) (2018*2019+2019*2020)*(45*120-15*360)*(1+5+9+13+17+...+2015+2019)
= (2018*2019+2019*2020)*(15*3*120-15*120*3)*(1+5+9+13+17+...+2015+2019)
= (2018*2019+2019*2020)*0*(1+5+9+13+17+...+2015+2019)
= 0
S = 2020 + 2019 - 2018 - 2017 + 2016 + 2015 - 2014 - 2013 + ... + 4 + 3 - 2 - 1
= ( 2020 + 2019 - 2018 - 2017 ) + ( 2016 + 2015 - 2014 - 2013 ) + ... + ( 4 + 3 - 2 - 1 ) (có tất cả 2020 : 4 = 505 nhóm)
= 4 + 4 + ... + 4
= 4. 505 = 2020
Vậy S = 2020.
Cácbạn ghi rõ lời giải giúp mình nhé.
Thanks các bạn!
ta có 1/2*2/3*...*2019/2020
=1*2*3*...*2019/2*3*4*..*2020
=1/2020 (rút gọn các số giống nhau)