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a: Số cần tìm là 5,32:0,125=42,56
b: \(A=1+\dfrac{1}{2019}-1-\dfrac{1}{2018}+\dfrac{1}{2018}-\dfrac{1}{2019}=0\)
\(A=\frac{2020}{2019}-\frac{2019}{2018}+\frac{1}{2019\times2018}\)
\(=\frac{2020\times2018}{2019\times2018}-\frac{2019\times2019}{2019\times2018}+\frac{1}{2019\times2018}\)
\(=\frac{2020\times2018-2019\times2019+1}{2019\times2018}\)
\(=\frac{\left(2019+1\right)\times\left(2019-1\right)-2019\times2019+1}{2019\times2018}\)
\(=\frac{2019\times2019-2019+2019-1-2019\times2019+1}{2019\times2018}\)
\(=\frac{2019\times2019-1-\left(2019\times2019-1\right)}{2019\times2018}\)
\(=\frac{0}{2019\times2018}\)
\(=0\)
Vậy A = 0
ta có
A=2020*2018/2019*2018-2019*2019/2018*2019+1/2018*2019
=>A*(2018*2019)=2020*2018-2019*2019+1
=>A*(2018*2019)=(2019+1)*2018-(2018+1)*2019+1
=>A*(2018*2019)=(2019*2018+2018)-(2018*2019+2019)+1
=>A*(2018*2019)=2019*2018+2018-2018*2019-2019+1
=>A*(2018*2019)=2018-2019+1
=>A*(2018*2019)=2018+1-2019
=>A*(2018*2019)=0
=>A=0/(2018*2019)
=>A=0
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 = \(\dfrac{1009}{2020}\)
2020/2019 x 2019/2018 x 2018/2017 x....................3/2
= 2020/2
= 1010
Trả lời
A=(1-1/2)(1-1/4)(1-1/5).....(1-1/2018)(1-1/2019)
=1/2.3/4.4/5......2017/2018.2018/2019
=1/2.1/2019
=1/4038.
Nhưng theo mk nghĩ đề phải như thế này>
A=(1-1/3)(1-1/4)(1-1/5)........(1-2018)(1-2019)
=2/3.3/4.4/5......2017/2018.2018/2019
=2/2019.
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{2018}\right)\left(\frac{1}{2019}\right)\)
\(=\frac{1}{2}.\frac{3}{4}.\frac{4}{5}.....\frac{2017}{2018}.\frac{1}{2019}=\frac{1}{2}.\frac{3}{2018}.\frac{1}{2019}=\frac{1}{2716228}\)
Vậy\(A=\frac{1}{2716228}\)
\(A=\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2018\times2019}\)
\(A=\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2018}-\dfrac{1}{2019}\)
\(A=\left(\dfrac{2020}{2019}-\dfrac{1}{2019}\right)-\left(\dfrac{2019}{2018}-\dfrac{1}{2018}\right)\)
\(A=\left(\dfrac{2020-1}{2019}\right)-\left(\dfrac{2019-1}{2018}\right)\)
\(A=1-1\)
\(A=0.\)
\(A=\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2018\times2019}\)
\(A=\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2018}-\dfrac{1}{2019}\)
\(A=\left(\dfrac{2020}{2019}-\dfrac{1}{2019}\right)-\left(\dfrac{2019}{2018}-\dfrac{1}{2018}\right)\)
\(A=\dfrac{2019}{2019}-\dfrac{2018}{2018}\)
\(A=1-1\)
\(A=0\)
\(A=\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2018\times2019}\)
\(=\dfrac{2020\times2018}{2019\times2018}-\dfrac{2019\times2019}{2018\times2019}+\dfrac{1}{2018\times2019}\)
\(=\dfrac{2020\times2018-2019\times2019+1}{2019\times2018}\)
\(=\dfrac{\left(2019+1\right)\times2018-2019\times\left(2018+1\right)+1}{2019\times2018}\)
\(=\dfrac{2019\times2018+2018-2019\times2018-2019+1}{2019\times2018}\)
\(=\dfrac{2018-2019+1}{2019\times2018}\)
\(=\dfrac{\left(2018+1\right)-2019}{2019\times2018}=\dfrac{2019-2019}{2019\times2018}=0\)