thực hiện phép tính :
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(\frac{2^{10}.3^{31}+2^{40}.3^6}{2^{11}.3^{31}+2^{41}.3^6}\)
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)S1=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
b)S2=\(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)
c)S3=\(\frac{1}{10.9}+\frac{1}{18.13}+\frac{1}{26.17}+...+\frac{1}{802.405}\)
\(S1=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}\)
\(S1=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-....-\frac{1}{101}=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)
\(S2=\frac{5}{1.3}+\frac{5}{3.5}+....+\frac{5}{99.101}\)
\(S2=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-.....-\frac{1}{101}\right)=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{5}{2}\cdot\frac{100}{101}=\frac{250}{101}\)
Thực hiện phép tính :
\(\frac{2^{10}.3^{31}+2^{90}.3^6}{2^{11}.3^{31}+2^{41}.3^6}\)
\(\frac{2^{10}.3^{31}+2^{90}.3^6}{2^{11}.3^{31}+2^{41}.3^6}=\) \(\frac{2^{10}.3^6.\left(3^{25}+2^{30}\right)}{2^{10}.3^6\left(3^{25}+2^{30}\right)}\) \(=1\)
\(A=\frac{1}{1.3}\)\(+\frac{1}{3.5}\)\(+\frac{1}{5.7}\)\(+...+\frac{1}{99.101}\)
\(A=\frac{3^2}{10}+\frac{3^2}{40}\)\(+\frac{3^2}{88}\)\(+...+\frac{3^2}{340}\)
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{101}\right)\)
\(A=\frac{1}{2}.\frac{100}{101}\)
\(A=\frac{50}{101}\)
\(A=\frac{3^2}{10}+\frac{3^2}{40}+\frac{3^2}{88}+...+\frac{3^2}{340}\)
\(A=\frac{3^2}{2.5}+\frac{3^2}{5.8}+\frac{3^2}{8.11}+...+\frac{3^2}{17.20}\)
\(A=\frac{3^2}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\right)\)
\(A=3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\right)\)
\(A=3\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(A=3.\frac{9}{20}\)
\(A=\frac{27}{20}\)
k nhá bn!
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{5}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(2A=1-\frac{1}{101}\)
\(2A=\frac{100}{101}\)
\(\Rightarrow A=\frac{50}{101}\)
\(A=\frac{3^2}{10}+\frac{3^2}{40}+\frac{3^2}{88}+...+\frac{3^2}{340}\)
\(A=\frac{3^2}{2.5}+\frac{3^2}{5.8}+\frac{3^2}{8.11}+...+\frac{3^2}{17.20}\)
\(\Rightarrow A=3\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{17.20}\right)\)
\(A=3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\right)\)
\(A=3\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(A=3.\frac{9}{20}\)
\(A=\frac{27}{20}\)
Thực hiện phép tính:
A=\(\left(3\frac{1}{3}+2,5\right):\left(3\frac{1}{6}-4\frac{1}{5}\right)-\frac{11}{31}\)
B=\(\left(-6\right).10:\left[-0,25+\frac{1}{2}:\left(-2\right)\right]+1\frac{3}{4}\)
A \(=\frac{35}{6}.\left(\frac{-30}{31}\right)-\frac{11}{31}=\frac{-175}{31}-\frac{11}{31}=\frac{-186}{31}=-6\)
\(B=-60:\left(\frac{-1}{2}\right)+\frac{7}{4}=30+\frac{7}{4}=\frac{127}{4}\)
1) Thực hiện phép tính
a)\(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{91.94}+\frac{2}{94.97}\)
b)\(\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}\)PHẦN
\(\left(1\frac{3}{7}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)\)
2)Tìm x:
a)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x.\left(x+2\right)}=\frac{1005}{2011}\)
Bài nhìn vô muốn xỉu rồi ='((
1. a) \(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{91.94}+\frac{2}{94.97}\)
\(=\frac{2}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{91.94}+\frac{3}{94.97}\right)\)
\(=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}\right)\)
\(=\frac{2}{3}\left(1-\frac{1}{97}\right)=\frac{2}{3}.\frac{96}{97}=\frac{64}{97}\)
b) Bạn tự làm, làm nữa chắc xỉu =((( Khi nào rảnh mình sẽ làm, nếu bạn cần
2 )
a) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{1005}{2011}\)
\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{1005}{2011}\)
\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{x+2}\right)=\frac{1005}{2011}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{1005}{2011}:2=\frac{1005}{4022}\)
\(\Leftrightarrow\frac{1}{x+2}=1-\frac{1005}{4022}=\frac{3017}{4020+2}\)
\(\Rightarrow x=4020\)
Thực hiện phép tính: \(2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.100}\right)\)
\(2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.100}\right)\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.100}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
2*(1/1*3+1/3*5+.......+1/99*100)
=2*(2/1*3+2/3*5+.....+2/99*100)*1/2
=1/3-1/5+1/5-1/7+....+1/99-1/100
=1/3-1/100
=100/300-3/300
=97/300
2(1/1.3+1/3.5+1/5.7+...+1/99.100)
=2/1.3+2/3.5+2/5.7+...+2/99.100
=1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/100
=1-1/100
=99/100
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
1. tính
a) A=\(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
b) B=\(\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\)
c) C=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
d) D=\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{92.98}\)
1) a) A=\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{3}-\frac{1}{8}=\frac{5}{24}\)
c) C=\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(C=1-\frac{1}{101}\)
\(C=\frac{100}{101}\)
d) Sửa đề: thay \(\frac{3}{92.98}\)=\(\frac{3}{92.95}\)
\(D=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{92}-\frac{1}{95}\)
\(D=\frac{1}{2}-\frac{1}{95}\)
\(D=\frac{95-2}{190}=\frac{93}{190}\)
Các bài trên áp dụng theo tính chất: \(\frac{a}{b\left(b+a\right)}\frac{1}{b}-\frac{1}{b+a}\)