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a) \(\frac{1}{3}-\frac{-1}{6}=\frac{1}{3}+\frac{1}{6}=\frac{1}{2}\)
b) \(2\frac{1}{3}+4\frac{1}{5}=\frac{7}{3}+\frac{21}{5}=\frac{98}{15}\)
c) \(\frac{4}{9}-\frac{13}{3}-\frac{4}{9}-\frac{10}{3}=\left(\frac{4}{9}-\frac{4}{9}\right)-\left(\frac{13}{3}+\frac{10}{3}\right)\)
\(=0-\frac{23}{3}=\frac{-23}{3}\)
d) \(4-\left(2-\frac{5}{2}\right)+0,5=4-2+\frac{5}{2}+\frac{1}{2}=2+3=5\)
b)=(2/3 +2/7 - 2/28)/(-3/3 -3/7 + 3/28)
=[2(1/3+1/7-1/28)]/[(-3)(1/3+1/7-1/28)]
=2/-3
=-2/3
Bài 1:
a, \(\frac{1}{-16}-\frac{3}{45}=\frac{-1}{16}-\frac{1}{15}\)
\(=\frac{-15}{240}-\frac{16}{240}\)
\(=\frac{-31}{240}\)
b, \(=\frac{-10}{12}-\frac{-12}{12}\)
\(=\frac{2}{12}=\frac{1}{6}\)
c, \(=\frac{-30}{6}-\frac{1}{6}\)
\(=\frac{-31}{6}\)
Bài 2:
a, \(x=-\frac{1}{2}-\frac{3}{4}\)
\(x=-\frac{1}{4}\)
b, \(\frac{1}{2}+x=-\frac{11}{2}\)
\(x=-\frac{11}{2}-\frac{1}{2}\)
\(x=-6\)
Bạn nhớ k đúng và chọn câu trả lời này nhé!!!! Mình giải đúng và chính xác hết ^_^
a) \(B=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{302\cdot305}\)
\(B=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{302\cdot305}\right)\)
\(B=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{302}-\frac{1}{305}\right)\)
\(B=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{305}\right)=\frac{1}{3}\cdot\frac{303}{610}=\frac{101}{610}\)
b) \(C=\frac{6}{1\cdot4}+\frac{6}{4\cdot7}+....+\frac{6}{202\cdot205}\)
\(C=2\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{202\cdot205}\right)=2\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{202}-\frac{1}{205}\right)\)
\(=2\left(1-\frac{1}{205}\right)=2\cdot\frac{204}{205}=\frac{408}{205}\)
c) \(D=\frac{5^2}{1\cdot6}+\frac{5^2}{6\cdot11}+...+\frac{5^2}{266\cdot271}\)
\(D=5\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+...+\frac{5}{266\cdot271}\right)\)
\(D=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{266}-\frac{1}{271}\right)=5\left(1-\frac{1}{271}\right)=5\cdot\frac{270}{271}=\frac{1350}{271}\)
d) \(E=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{5}{16}\cdot...\cdot\frac{9999}{10000}=\frac{3\cdot8\cdot15\cdot...\cdot9999}{4\cdot9\cdot16\cdot...\cdot10000}=\frac{3}{10000}\)
e) \(F=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{50^2}\right)\)
\(F=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{2500}\right)\)
\(F=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{2499}{2500}=\frac{3\cdot8\cdot15\cdot...\cdot2499}{4\cdot9\cdot16\cdot...\cdot2500}=\frac{3}{2500}\)
a. \(B=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{302.305}\)
\(\Rightarrow3B=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{302.305}\)
\(\Rightarrow3B=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{302}-\frac{1}{305}\)
\(\Rightarrow3B=\frac{1}{2}-\frac{1}{305}\)
\(\Rightarrow3B=\frac{303}{610}\)
\(\Rightarrow B=\frac{101}{610}\)
b. \(C=\frac{6}{1.4}+\frac{6}{4.7}+...+\frac{6}{202.205}\)
\(\Rightarrow\frac{1}{2}C=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{202.205}\)
\(\Rightarrow\frac{1}{2}C=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{202}-\frac{1}{205}\)
\(\Rightarrow\frac{1}{2}C=1-\frac{1}{205}\)
\(\Rightarrow\frac{1}{2}C=\frac{204}{205}\)
\(\Rightarrow C=\frac{408}{205}\)
c. \(D=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{266.271}\)
\(\Rightarrow\frac{1}{5}D=\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{266.271}\)
\(\Rightarrow\frac{1}{5}D=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{266}-\frac{1}{271}\)
\(\Rightarrow\frac{1}{5}D=1-\frac{1}{271}\)
\(\Rightarrow\frac{1}{5}D=\frac{270}{271}\)
\(\Rightarrow D=\frac{1350}{271}\)