tính tổng
A=7/ 10.11+7/11.12 + 7/12.13 +...+7/ 69.70
a)= 20 + 21 + 22 +...+22010
b)= 1 + 3 + 32+...+ 3100
tính;
A= 7/10.11 +7/11.12 =7/12.13 + ... + 7/69.70
A=7.(1/10.11+1/11.12+...+1/69.70)
A=7.(1/10-1/11+1/11-1/12+...+1/69-1/70)
A=7.(1/10-1/70)
A=7. 3/35
A= 3/5
chúc bạn học tốt nha
\(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(\Rightarrow A=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(\Rightarrow A=7\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(\Rightarrow A=7.\frac{6}{70}=\frac{6}{10}=\frac{3}{5}\)
\(A=\frac{7}{10\cdot11}+\frac{7}{11\cdot12}+\frac{7}{12\cdot13}+...+\frac{7}{69\cdot70}\)
\(\Rightarrow7A=\frac{1}{10\cdot11}+\frac{1}{11\cdot12}+\frac{1}{12\cdot13}+...+\frac{1}{69\cdot70}\)
\(\Rightarrow7A=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\)
\(\Rightarrow7A=\frac{1}{10}-\frac{1}{70}\)
\(\Rightarrow7A=\frac{7}{70}-\frac{1}{70}\)
\(\Rightarrow7A=\frac{3}{35}\)
\(\Rightarrow A=\frac{3}{35}:7\)
\(\Rightarrow A=\frac{3}{245}\)
Tính tổng: A= 7/10.11+7/11.12+7/12.13+...+7/69.70
\(A=\frac{1}{7}.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{68}-\frac{1}{70}\right)\)
\(A=\frac{1}{7}.\left(\frac{1}{10}-\frac{1}{70}\right)=\frac{1}{7}.\frac{3}{35}=\frac{3}{245}\)
A=\(\frac{7}{10.11}\)+\(\frac{7}{11.12}\)+\(\frac{7}{12.13}\)+...+\(\frac{7}{69.70}\)
A=\(\frac{7}{10}\)-\(\frac{7}{11}\)+\(\frac{7}{11}\)-\(\frac{7}{12}\)+\(\frac{7}{12}\)-\(\frac{7}{13}\)+...+\(\frac{7}{69}\)-\(\frac{7}{70}\)
A=\(\frac{7}{10}-\frac{7}{70}\)
A=\(\frac{7}{10}-\frac{1}{10}\)
Ạ=\(\frac{6}{10}=\frac{3}{5}\).
\(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+..+\frac{7}{69.70}\)
=\(7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+..+\frac{1}{69}-\frac{1}{70}\right)\)
=\(7.\left(\frac{1}{10}-\frac{1}{70}\right)=7.\frac{3}{35}=\frac{21}{35}\)
tính tổng
A=20+21+22+2 2021
B=1+3+32+......+3100
a: \(2A=2^1+2^2+...+2^{2022}\)
\(\Leftrightarrow A=2^{2022}-1\)
\(A=1+2+2^2+...+2^{2021}\)
\(2A=2+2^2+2^3+...+2^{2020}\)
\(2A-A=\left(2+2^2+2^3+...+2^{2020}\right)-\left(1+2+2^2+...+2^{2021}\right)\)
\(A=2^{2020}-1\)
a) Tính:
\(\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}.......+\frac{7}{69.70}\)
= 7.( \(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-....-\frac{1}{70}\))
= 7.( \(\frac{1}{10}-\frac{1}{70}\))
= 7.(\(\frac{7}{70}-\frac{1}{70}\))
= 7.\(\frac{6}{70}\)
= \(\frac{3}{5}\)
Tính các tổng sau :
A = 7/10.11+7/11.12+7/12.13+...+7/69.70
A = ?
\(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(\Rightarrow A=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+....+\frac{1}{69.70}\right)\)
\(\Rightarrow A=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(\Rightarrow A=7\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(\Rightarrow A=7.\frac{3}{35}\)
\(\Rightarrow A=\frac{3}{5}\)
A=7/10.11+7/11.12+7/12.13+...+7/69.70
B=1/25.27+1/27.29+1/29.31+...+1/73.75
Bài 6: Tính các tổng sau:
A= \(\dfrac{7}{10.11}\)+\(\dfrac{7}{11.12}\)+\(\dfrac{7}{12.13}\)+...+\(\dfrac{7}{69.70}\)
B= \(\dfrac{1}{25.27}\)+\(\dfrac{1}{27.29}\)+\(\dfrac{1}{29.31}\)+...+\(\dfrac{1}{73.75}\)
\(A=7\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)
\(=7\left(\dfrac{1}{10}-\dfrac{1}{70}\right)=\dfrac{7.60}{700}=\dfrac{420}{700}=\dfrac{3}{5}\)
\(B=\dfrac{1}{2}\left(\dfrac{1}{25}-\dfrac{1}{27}+\dfrac{1}{27}-\dfrac{1}{29}+...+\dfrac{1}{73}-\dfrac{1}{75}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{25}-\dfrac{1}{75}\right)=\dfrac{1}{75}\)
Tính hộ mik với : H=7/10.11 + 7/11.12 + 7/12.13 + .... + 7/69.70
H = \(\frac{7}{\text{10.11}}+\frac{7}{\text{11.12}}+\frac{7}{\text{12.13}}+...+\frac{7}{\text{69.70}}\)
H = 7 . \(\left(\frac{1}{\text{10.11}}+\frac{1}{\text{11.12}}+\frac{1}{\text{12.13}}+...+\frac{1}{\text{69.70}}\right)\)
H = 7 . \(\left(\frac{1}{\text{10}}-\frac{1}{11}+\frac{1}{\text{11}}-\frac{1}{12}+\frac{1}{\text{12}}-\frac{1}{13}+...+\frac{1}{\text{69}}-\frac{1}{70}\right)\)
H = 7 . \(\left(\frac{1}{10}-\frac{1}{70}\right)\)
H = 7 . \(\frac{3}{35}\)
H = \(\frac{3}{5}\)
1) A = 7/ 10.11 + 7/ 11.12 + 7/ 12.13 + .... + 7/69.70
2 ) B = 1/ 25.27 + 1/ 27.29 + 1/29.31+ ......+ 1/ 73.75
1) A=7/10.11+7/11.12+7/12.13+...+7/69.70
A=7.(1/10.11+1/11.12+1/12.13+...+1/69.70)
A= 7.(1/10-1/11+1/11-1/12+1/12-1/13+...+1/69-1/70)
A= 7.(1/10-1/70)
A=7.3/35=3/5
2)B=1/25.27+1/27.29+1/29.31+...+1/73.1/75
B=1/25-1/27+1/27-1/29+1/29-1/31+...+1/73-1/75
B=1/25-1/75=2/75
A = 7/ 10.11 + 7/ 11.12 + 7/ 12.13 + .... + 7/69.70
(1/7).A=1/10.11+1/11.12+...+1/69.70
=1/10-1/11+1/11-1/12+...+1/69-1/70
=1/10-1/70=3/35
=>A=7.(3/35)
=3/5
2 ) B = 1/ 25.27 + 1/ 27.29 + 1/29.31+ ......+ 1/ 73.75
=>(1/2).B=2/25.27+...+2.73.75
=1/25-1/27+...+1/73-1/75
=1/25-1/75
=2/75
=>B=4/75
A=7.(1/10.11+1/11.12+1/12.13+...+1/69.70)
A= 7.(1/10-1/11+1/11-1/12+1/12-1/13+...+1/69-1/70)
A= 7.(1/10-1/70)
A=7.3/35=3/5
2)B=1/25.27+1/27.29+1/29.31+...+1/73.1/75
B=1/25-1/27+1/27-1/29+1/29-1/31+...+1/73-1/75
B=1/25-1/75=2/75
\(\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
vãi nồi :
S=7/10.11+7/11.12+...+7/69.70
=>S= 7.(1/10.11+1/11.12+...+1/69.70)
=> S= 7. ( 1/10-1/11+...+1/69-1/70)
=> S= 7. ( 1/10 - 1/70)
=> S = 7. 3/35
=> S = 3/5
We are one_Nguyễn Ngọc Sáng
\(\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(=7.\frac{3}{35}\)
\(=\frac{3}{5}\)
\(\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(=7\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
\(=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(=7\left(\frac{1}{10}-\frac{1}{70}\right)=7\left(\frac{7}{70}-\frac{1}{70}\right)=7.\frac{6}{70}=\frac{6}{10}=\frac{3}{5}\)