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\(6+\frac{4}{12}\)\(-\frac{2}{15}\)
\(C1:6+\left(\frac{1}{3}-\frac{2}{15}\right)\)
\(\frac{90}{15}+\left(\frac{5}{15}-\frac{2}{15}\right)\)\(=\frac{93}{15}\)\(=\frac{31}{5}\)
\(C2:6+\frac{4}{12}-\frac{2}{15}\)
\(6+\frac{1}{3}+\frac{2}{15}\)
\(\frac{90}{15}+\frac{5}{15}-\frac{2}{15}\)
\(b\)Tương tự như câu a em tự giải coi đó như bài tập nha
Phá ngoặc trc nó là dấu cộng thì giữa nguyên dấu
trc nó là dấu trừ thì đổi dấu các số hạng trong ngoặc
\(\)
\(6+\left(\frac{4}{12}-\frac{2}{15}\right)=6+\left(\frac{5}{15}-\frac{2}{15}\right)=\frac{30}{5}+\frac{1}{5}=\frac{31}{5}\)
\(1+\left(\frac{5}{6}+\frac{7}{3}+\frac{2}{12}\right)=1+\left(\frac{10+28+2}{12}\right)=1+\frac{40}{12}=\frac{3}{3}+\frac{10}{3}=\frac{13}{3}\)
\(A=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}=>2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...-\frac{1}{101}\)
A= 1/2(1/3 - 1/101)
A= 49/303
a, 3/5 - 1/15 x 10/13
= 3/5 - 2/39
= 117/195 - 10/195
= 107/195
b, (3/5 - 3/20) x 4/5
= (12/20 - 3/20) x 4/5
= 9/20 x 4/5
= 9/25
Chúc bạn học tốt
\(A=\frac{1}{2}:0,25-\frac{1}{4}:0,25+\frac{1}{8}:0,125-\frac{1}{10}:0,1\)
\(A=2-1+1+1\)
\(A=1+1+1\)
\(A=1.3\)
\(A=3\)
Vậy \(A=3\)
\(a=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{45}\)
\(a=\frac{1}{1.3}+\frac{1}{2.3}+\frac{1}{2.5}+\frac{1}{3.5}+...+\frac{1}{5.9}\)
\(a=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}\right)\)
\(a=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(a=2\left(\frac{1}{2}-\frac{1}{10}\right)\)
=> \(a=2.\frac{2}{5}\)
=> \(a=\frac{4}{5}\)
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{45}\)
\(\Rightarrow\frac{1}{2}A=\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}\right)\cdot\frac{1}{2}\)
\(=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
\(=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
\(\Rightarrow A=\frac{2}{5}:\frac{1}{2}=\frac{4}{5}\)