\(B=\frac{7}{1.3}+\frac{7}{3.5}+\frac{7}{5.7}+...+\frac{7}{99.101}\)
các bạn nhớ trình bày bài giải đầy đủ nhé mình tích cho.
Tính tổng:
\(\frac{7}{1.3}+\frac{7}{3.5}+\frac{7}{5.7}+...+\frac{7}{99.101}\)
Mình chỉ cần mấy bạn giải giúp khúc nhân cái tổng đó với 2, làm chi tiết khúc đó lên nhé! Nhưng phải đúng. Tick cho ( 3 tick)
7/1.3 + 7/3.5 + 7/5.7 + ... + 7/99.101
= 7.(1/1.3 + 1/3.5 + 1/5.7 + ... + 1/99.101)
= 7/2 . 2 . (1/1.3 + 1/3.5 + 1/5.7 + ... + 1/99.101)
= 7/2 . (2/1.3 + 2/3.5 + 2/5.7 + ... + 2/99.101)
= 7/2 . (1 - 1/3 + 1/3 - 1/5 + ... + 1/99 - 1/101)
= 7/2 . (1 - 1/101)
= 7/2 . 100/101
= 350/101
\(\frac{7}{1.3}+\frac{7}{3.5}+...+\frac{7}{99.101}\)
\(=7\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\right)\)
\(=\)\(\frac{7}{2}.2.\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\right)\)
\(=\)\(\frac{7}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
=\(\frac{7}{2}x\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{99}-\frac{1}{100}\right)\)
=\(\frac{7}{2}x\left(\frac{1}{3}-\frac{1}{100}\right)\)
=\(\frac{7}{2}\)x\(\frac{97}{300}\)
=\(\frac{679}{600}\)
Tính nhanh \(\frac{7}{1.3}+\frac{7}{3.5}+\frac{7}{5.7}+...+\frac{7}{99.101}\)
\(\frac{7}{1.3}+\frac{7}{3.5}+\frac{7}{5.7}+....+\frac{7}{99.101}\)
\(=\frac{7}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}\right)\)
\(=\frac{7}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{7}{2}\left(1-\frac{1}{101}\right)=\frac{7}{2}.\frac{100}{101}=\frac{350}{101}\)
7/1.3+7/3.5+7/5.7+.....+7/99.101 ai làm đầy đủ mình tick cho
\(\frac{7}{1.3}+\frac{7}{3.5}+\frac{7}{5.7}+...+\frac{7}{99.101}\)
\(=\frac{7}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(=\frac{7}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{7}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{7}{2}.\frac{100}{101}\)
\(=\frac{350}{101}\)
k mk nha
7/1.3+7/3.5+7/5.7+...+7/99.101
=7(1/1.3+1/3.5+1/5.7+...+1/99.101)
=7(1/1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101)
=7(1-1/101)
=7.100/101
=700/101
Đầy đủ ko bỏ bước nào lun!!
K CHO MK NHA!!!
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{7}{5.7}+...+\frac{1}{99.101}\)
=1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101
=1-1/101
=100/101
k cho mình nha
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{101}\right)=\frac{1}{2}.\frac{100}{101}=\frac{50}{101}\)
TA CÓ \(\frac{1}{1.3}+\frac{1}{3.5}+.....+\frac{1}{99.101}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{1}-\frac{1}{101}\)
\(=\frac{100}{101}\)
giá trị x >0 thỏa mãn:
\(-\frac{7}{3}< \left|\frac{2}{7}-x\right|-\frac{5}{2}< -\frac{7}{4}\)
Trình Bày bài giải đầy đủ nhé các bạn
Bài 1: Tính tổng sau :
A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
B =\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
C =\(\frac{3^2}{10}+\frac{3^2}{40}+\frac{3^2}{88}+...+\frac{3^2}{340}\)
D =\(\frac{7}{1.3}+\frac{7}{3.5}+\frac{7}{5.7}+...+\frac{7}{99.101}\)
E =\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\)
G =\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{99}\right)\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}=\frac{99}{100}\)
\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(B=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{101-99}{99.101}\)
\(B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(B=1-\frac{1}{101}=\frac{100}{101}\)
\(C=\frac{3^2}{10}+\frac{3^2}{40}+\frac{3^2}{88}+...+\frac{3^2}{340}\)
\(C=3\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\right)\)
\(C=3\left(\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+...+\frac{20-17}{17.20}\right)\)
\(C=3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right)\)
\(C=3\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{27}{20}\)
\(D=\frac{7}{1.3}+\frac{7}{3.5}+\frac{7}{5.7}+...+\frac{7}{99.101}\)
\(D=\frac{7}{2}B=\frac{7}{2}.\frac{100}{101}=\frac{350}{101}\)
\(E=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\)
\(3E=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\)
\(3E-E=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\right)\)
\(2E=1-\frac{1}{3^8}\)
\(E=\frac{3^8-1}{2.3^8}\)
\(G=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{99}\right)\)
\(G=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{98}{99}=\frac{1}{99}\)
TÍNH NHANH:
A=\(\frac{7}{1.3}+\frac{7}{3.5}+\frac{7}{5.7}+....+\frac{7}{99.100}\)
ai làm được mình tích cho
\(A=\frac{7}{1.3}+\frac{7}{3.5}+.............+\frac{7}{99.101}\)
\(=\frac{7}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+........+\frac{2}{99.101}\right)\)
\(=\frac{7}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.......+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{7}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{7}{2}.\frac{100}{101}\)
\(=\frac{350}{101}\)
Chứng minh: \(\frac{202}{1.3}\)+\(\frac{202}{3.5}\)+\(\frac{202}{5.7}\)+...+ \(\frac{202}{99.101}\)là số chính phương
Mong các bạn có lời giải thật đầy đủ ạ! Rất mong!
\(\frac{202}{1.3}+\frac{202}{3.5}+\frac{202}{5.7}+...+\frac{202}{99.101}\)
\(=202\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)
\(=202.\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=101\left(1-\frac{1}{101}\right)\)
\(=101.\frac{100}{101}\)
\(=100=10^2\Rightarrowđpcm\)
Tính tổng hợp lí:
a) \(\frac{3}{4}+\frac{3}{28}+\frac{3}{70}+\frac{3}{130}+\frac{3}{208}+\frac{3}{304}+\frac{3}{418}+\frac{3}{550}\)
b) \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{\left(2n+1\right).\left(2n+3\right)}\)
C) \(\frac{7+\frac{7}{13}-\frac{7}{48}+\frac{7}{95}}{15+\frac{15}{13}-\frac{15}{48}+\frac{15}{95}}-\frac{7070707}{15151515}\)
BÀI HƠI KHÓ GIẢI ĐẦY ĐỦ+ CHI TIẾT VÀ DỄ HIỂU CHO MÌNH NHA
a) \(\frac{3}{4}+\frac{3}{28}+\frac{3}{70}+\frac{3}{130}+\frac{3}{208}+\frac{3}{304}+\frac{3}{418}+\frac{3}{550}\)
= \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}+\frac{3}{16.19}+\frac{3}{19.22}+\frac{3}{22.25}\)
= \(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}+\frac{1}{19}-\frac{1}{22}+\frac{1}{22}-\frac{1}{25}\)
= \(\frac{1}{1}-\frac{1}{25}\)
= \(\frac{24}{25}\)
b) \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2n+1\right).\left(2n+3\right)}\)
= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n+1}-\frac{1}{2n+3}\)
= \(\frac{1}{1}-\frac{1}{2n+3}\)
= \(\frac{2n+2}{2n+3}\)
c) \(\frac{7+\frac{7}{13}-\frac{7}{48}+\frac{7}{95}}{15+\frac{15}{13}-\frac{15}{48}+\frac{15}{95}}-\frac{7070707}{15151515}\)
= \(\frac{7\left(1+\frac{1}{13}-\frac{1}{48}+\frac{1}{95}\right)}{15\left(1+\frac{1}{13}-\frac{1}{48}+\frac{1}{95}\right)}-\frac{7.1010101}{15.1010101}\)
= \(\frac{7}{15}-\frac{7}{15}\)
= 0
a) 24/25
b) (2n+2)/(2n+3)
c) 0
sai thì thôi nhé