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
Tính nhanh \(\frac{7}{1.3}+\frac{7}{3.5}+\frac{7}{5.7}+...+\frac{7}{99.101}\)
\(\frac{7}{1.3}+\frac{7}{3.5}+\frac{7}{5.7}+....+\frac{7}{99.101}\)
\(=\frac{7}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}\right)\)
\(=\frac{7}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{7}{2}\left(1-\frac{1}{101}\right)=\frac{7}{2}.\frac{100}{101}=\frac{350}{101}\)
\(B=\frac{7}{1.3}+\frac{7}{3.5}+\frac{7}{5.7}+...+\frac{7}{99.101}\)
các bạn nhớ trình bày bài giải đầy đủ nhé mình tích cho.
B : 7/2 =2/1.3+2/3.5+...+2/99.101
B:7/2=1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101
B:7/2=1-1/101=100/101
B=100/101*7/2=700/202=350/101
B=7/2(2/1.3+2/3.5+ ...+2/99.101)
B=7/2(1-1/3+1/3-1/5+...+1/99-1/101)
B=7/2(1-1/101)=7/2.100/101=350/101
k nha bạn
B=1/1-1/3+1/3-1/5+1/5-1/7+....+1/99-1/101
B=1-1/101=100/101
vậy B=100/101
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\(\frac{97}{300}\)
=\(\frac{679}{600}\)
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
A =\(\frac{11}{1.3}\)+ \(\frac{47}{3.5}\)+ \(\frac{107}{5.7}\)+ \(\frac{191}{7.9}\)+...+ \(\frac{971}{17.19}\)
B = \(\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}\)- \(\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3-5^9.7^3.2^3}\)
C = 1.3+3.5+5.7+...+ (2n-1)(2n+1)
Giúp mình vs nhanh nhanh nha các bạn
tính tổng các phân số theo cách nhanh nhất
a) \(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)
b) \(\frac{1}{7}+\frac{1}{91}+\frac{1}{247}+\frac{1}{475}+\frac{1}{775}+\frac{1}{1147}\)
ghi rỏ cách làm nha mình tick cho nah
Tính hợp lí:
\(a)\frac{0,4-\frac{2}{9}+\frac{2}{11}}{0,6-\frac{3}{9}+\frac{3}{11}}+\frac{\frac{2}{3}+\frac{2}{7}-\frac{1}{14}}{-1-\frac{3}{7}+\frac{3}{28}}\)
\(b)1-\frac{2}{3.5}-\frac{2}{5.7}-\frac{2}{7.9}-.....-\frac{2}{97.99}\)
Giúp mình với! Mình cần gấp!
Ai nhanh mình tích cho!
a, Ta có:
\(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{0,6-\frac{3}{9}+\frac{3}{11}}+\frac{\frac{2}{3}+\frac{2}{7}-\frac{1}{14}}{-1-\frac{3}{7}+\frac{3}{28}}=\frac{2\left(0,2-\frac{1}{9}+\frac{1}{11}\right)}{3\left(0,2-\frac{1}{9}+\frac{1}{11}\right)}+\frac{2\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{28}\right)}{-3\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{28}\right)}=\frac{2}{3}+\frac{-2}{3}=0\)
k đúng cho mình nha. Thanks!!!
a, bày cho mình cách viết bằng phân số đi , mình trình bày cách làm cho. k đúng cho mình nha.
\(=\frac{2}{3}\times\frac{0,1-\frac{1}{9}+\frac{1}{11}}{0,1-\frac{1}{9}+\frac{1}{11}}+\frac{-\frac{2}{3}\times\left(-1-\frac{3}{7}+\frac{3}{28}\right)}{-1-\frac{3}{7}+\frac{3}{28}}\)
=\(\frac{2}{3}+\left(-\frac{3}{2}\right)\)
=\(-\frac{5}{6}\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{7}{5.7}+...+\frac{1}{99.101}\)
=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 \(\frac{1}{1.3}+\frac{1}{3.5}+.....+\frac{1}{99.101}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{1}-\frac{1}{101}\)
\(=\frac{100}{101}\)
Tính nhanh
a ) \(\frac{5}{7}.\frac{5}{11}+\frac{5}{7}.\frac{2}{11}-\frac{5}{7}.\frac{10}{11}\)
b) \(\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}.\frac{25}{4.6}.\frac{36}{5.7}\)
c) \(6\frac{4}{11}+\frac{8}{72}-\left(3\frac{8}{22}+\frac{1}{9}\right)\)