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\(\frac{2019}{1\times2}+\frac{2019}{2\times3}+\frac{2019}{3\times4}+...+\frac{2019}{2018\times2019}\)
\(=2019\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{2018\times2019}\right)\)
\(=2019\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(=2019\left(1-\frac{1}{2019}\right)\)
\(=2019\left(\frac{2019}{2019}-\frac{1}{2019}\right)\)
\(=2019\times\frac{2018}{2019}\)\(=\frac{2019\times2018}{2019}=2018\)
a/ \(A=2018\cdot2018\)
\(=\left(2019-1\right)\cdot2018=2019\cdot2018-2018\)
\(B=2017\cdot2019\)
\(=\left(2018-1\right)\cdot2019=2018\cdot2019-2019\)
\(\Rightarrow A>B\)
b/
\(A=2018\cdot2019\)
\(=\left(2017+1\right)\cdot2019=2017\cdot2019+2019\)
\(B=2017\cdot2020\)
\(=2017\cdot\left(2019+1\right)=2017\cdot2019+2017\)
\(\Rightarrow A>B\)
a) \(2021^{2020}-2021^{2019}=2021^{2019}.\left(2021-1\right)=2021^{2019}.2020\)
b) Ta có :\(7x-140=3.7^2\)
\(\implies\) \(7x-140=3.49\)
\(\implies\) \(7x-140=147\)
\(\implies\) \(7x=287\)
\(\implies\) \(x=41\)
1. \(\dfrac{2019}{2020}-\left(\dfrac{2019}{2020}-\dfrac{2020}{2021}\right)\)
\(=\dfrac{2019}{2020}-\dfrac{2019}{2020}+\dfrac{2020}{2021}\)
\(=0+\dfrac{2020}{2021}=\dfrac{2020}{2021}\)
Giải:
1) \(\dfrac{2019}{2020}-\left(\dfrac{2019}{2020}-\dfrac{2020}{2021}\right)\)
\(=\dfrac{2019}{2020}-\dfrac{2019}{2020}+\dfrac{2020}{2021}\)
\(=\left(\dfrac{2019}{2020}-\dfrac{2019}{2020}\right)+\dfrac{2020}{2021}\)
\(=0+\dfrac{2020}{2021}\)
\(=\dfrac{2020}{2021}\)
2) \(\dfrac{2}{9}+\dfrac{7}{9}:\left(\dfrac{42}{5}-\dfrac{7}{5}\right)\)
\(=\dfrac{2}{9}+\dfrac{7}{9}:7\)
\(=\dfrac{2}{9}+\dfrac{1}{9}\)
\(=\dfrac{1}{3}\)
3) \(\dfrac{3}{4}+\dfrac{x}{4}=\dfrac{5}{8}\)
\(\dfrac{x}{4}=\dfrac{5}{8}-\dfrac{3}{4}\)
\(\dfrac{x}{4}=\dfrac{-1}{8}\)
\(\Rightarrow x=\dfrac{4.-1}{8}=\dfrac{-1}{2}\)
4) \(\left|3x+1\right|-\dfrac{1}{4}=\dfrac{-1}{4}\)
\(\left|3x-1\right|=\dfrac{-1}{4}+\dfrac{1}{4}\)
\(\left|3x-1\right|=0\)
\(3x-1=0\)
\(3x=0+1\)
\(3x=1\)
\(x=1:3\)
\(x=\dfrac{1}{3}\)
Chúc bạn học tốt!
a: =58(57+150-125)=58x82=4756
b: \(=9\cdot5-4\cdot7+83=45-28+83=100\)
c: =(2019-2019)+(-247-53)=-300
d: \(=13\cdot70-50\cdot\left[10:2+8\right]=910-50\cdot13=910-650=260\)
\(a,=58.\left(57+150-125\right)\\ =58.82=4756\\ b,=9.5-4.7+83.1\\ =45-28+83=100\)
a) 58.57+58.150-58.125
=58.(57+150-125)
=58. 82
= 4756
b)32.5-22.7+83.20190
=9.5-4.7+83
=45-28+83
=100
c)2019+(-247)+(-53)-2019
=(2019-2019)+[ (-247)+(-53)]
0+(-300)
= -300
d)13.70-50 [(19-32):2+23]
=13.70-50[10:2+8]
=13.70-50.13
=13.(70-50)
=13.20
=260
2.
a)x-36:18=12-5
x-36:18=6
x-36=6.18
x-36=108
x=108+36
x= 144
b)92-(17+x)=72
17+x=92-72
17+x=20
x=20-17
x=3
c)720:[41-(2x+5)]=40
41-(2x+5)=720:40
41-(2x+5)= 18
2x+5=41-18
2x+5=23
2x=23-5
2x=18
x=18:2
x=9
d) (x+2)3 -23=41
(x+2)3 =41+23
(x+2)3 =64
=> (x+2)3 =43
=>x+2=4
=>x=4-2
=>x=2
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(A=1-\frac{1}{2020}\)
\(A=\frac{2019}{2020}\)
\(B=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2017.2019}\)
\(2B=\frac{2}{1.3}+\frac{2}{3.5}=\frac{2}{5.7}+...+\frac{2}{2017.2019}\)
\(2B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}=\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\)
\(2B=1-\frac{1}{2019}\)
\(2B=\frac{2018}{2019}\)
\(B=\frac{2018}{2019}:2=\frac{1009}{2019}\)