tìm số tự nhiên x biết:
1/3+1/6+1/10+.....+2/(x+1)=2019/2021
Tìm số tự nhiên x biết rằng 1/3+1/6+1/10+...+2/x.(x-1)=2019/2021
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{2021}\)
<=> \(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2019}{2021}\)
<=> \(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2019}{2021}\)
<=> \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{4042}\)
<=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2019}{2042}\)
<=> \(\frac{1}{x+1}=\frac{1}{2021}\)
<=> x + 1 = 2021
<=> x = 2020
Có phải là bình 6a3 học trường THCS Nguyễn Trãi đúng không
Bài 1 : Thực hiện phép tính
[(35−5):3] mũ 3+3
Bài 2 : Tìm số tự nhiên x biết
16 x +40 = 10.3 mũ2+ 5.( 1 + 2 +3)
Bài 3: Tính
S= 1 + 2- 3 – 4 + 5 + 6 -7 – 8 + 9 +10 -…+2018 -2019-2020+2021
Bài 1 : Thực hiện phép tính [(35−5):3] mũ3+3
Bài 2 : Tìm số tự nhiên x biết
16 x +40 = 10.3 mũ2+ 5.( 1 + 2 +3)
Bài 3: Tính
S= 1 + 2- 3 – 4 + 5 + 6 -7 – 8 + 9 +10 -…+2018 -2019-2020+2021
Giúp mình với mn!
Bài 2:
Ta có: \(16x+40=10\cdot3^2+5\left(1+2+3\right)\)
\(\Leftrightarrow16x+40=90+30\)
\(\Leftrightarrow16x=80\)
hay x=5
Bài 1 :
[( 35 - 5 ) : 3 ]3 + 3
= [30 : 3]3 + 3
= 103 + 3
= 1000 + 3
= 1003
Đây nha bạn!!!
Chúc bạn học tốt!!!
Tìm số tự nhiên x, biết rằng:
1/3+1/6+1/10+...+2/x.(x+1)=2018/2019
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x.\left(x+1\right)}=\frac{2018}{2019}\)
\(\Rightarrow\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x.\left(x+1\right)}=\frac{2018}{2019}\)
\(\Rightarrow2.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2018}{2019}\)
\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2018}{2019}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1009}{2019}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{4038}\)
\(\Rightarrow x+1=4038\)
\(\Rightarrow x=4037\)
Vậy \(x=4037\)
\(\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+...+\frac{2}{x.\left(x+1\right)}=\frac{2018}{2019}\)
\(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x.\left(x+1\right)}=\frac{2018}{2019}\)
\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}+\frac{1}{x+1}\right)=\frac{2018}{2019}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1009}{2019}\)
\(\frac{1}{x+1}=\frac{1}{4038}\)
\(x=4037\)
Cho 1 dố tự nhiên gồm các số tự nhiên liên tiếp nhau từ 1 đến 2021 được viết theo thứ tự 1 2 3 4 5 6 7 8 9. 10 11 12 13...... 2019 2020 2021 tính tổng các chữ số đó
Số số hạng của dãy đó là : ( 2021 - 1 ) : 1 + 1 = 2021 ( chữ số )
Tổng của các số hạng đó là : ( 2021 + 1 ) x 2021 : 2 = 2043231
Đáp số 2043231
Tìm x biết:
( 1/2 + 1/3 + ... + 1/2021 ).x = 2021/1 +2019/2 + ... + 2/2019 + 1/2020
Tìm x biết:
( 1/2 + 1/3 + ... + 1/2021 ).x = 2021/1 +2019/2 + ... + 2/2019 + 1/2020
Tìm số tự nhiên x, biết:
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2017}{2019}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x-1\right)}=\)\(\frac{2017}{2019}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x-1\right)}=\frac{2017}{2019}\)
\(2\left[\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right]=\frac{2017}{2019}\)
\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\)\(\frac{2017}{2019}\)
\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2017}{2019}:2\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2017}{4038}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{2017}{4038}\)
\(\frac{1}{x+1}=\frac{1}{2019}\)
x + 1 =2019
x = 2019-1 =2018
Vậy x = 2018
\(2\left(\frac{1}{3}.\frac{1}{2}+\frac{1}{6}.\frac{1}{2}+\frac{1}{10}.\frac{1}{2}+....+\frac{2}{x\left(x+1\right)}.\frac{1}{2}\right)=\frac{2017}{2019}\)
=>\(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)
=>\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\left(x+1\right)}\right)\)\(=\frac{2017}{2019}\)
=>\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
=> \(2[\frac{1}{2}+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+....+\left(\frac{1}{x}-\frac{1}{x}\right)-\frac{1}{x+1}]=\frac{2017}{2019}\)
=>\(2\left(\frac{1}{2}+0+0+....+0-\frac{1}{x-1}\right)=\frac{2017}{2019}\)
=>\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
=>\(\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2017}{4038}\)
=>\(\frac{1}{x+1}=\frac{1}{2019}\)
=> x+1=2019
=>x=2018
TÌm x biết
a) \(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+....+\dfrac{1}{x\left(x+1\right):2}=1\dfrac{2019}{2021}\)
\(\Leftrightarrow1+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{x\left(x+1\right)}=1+\dfrac{2019}{2021}\)
\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2019}{2021}\)
\(\Leftrightarrow1-\dfrac{2}{x+1}=\dfrac{2019}{2021}\)
\(\Leftrightarrow\dfrac{2}{x+1}=1-\dfrac{2019}{2021}\)
\(\Leftrightarrow\dfrac{2}{x+1}=\dfrac{2}{2021}\)
\(\Leftrightarrow x+1=2021\)
\(\Leftrightarrow x=2020\)