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\(a,A=\dfrac{1}{7.9}\) và \(B=\dfrac{1}{7}-\dfrac{1}{9}\)
\(\Leftrightarrow2A=\dfrac{2}{7.9}=\dfrac{1}{7}-\dfrac{1}{9}\)
Vì \(2A=B\left(=\dfrac{1}{7}-\dfrac{1}{9}\right)\)
\(\Leftrightarrow A< B\)
\(b,A=\dfrac{15}{301}\) và \(B=\dfrac{25}{499}\)
Ta thấy: \(\dfrac{15}{301}< \dfrac{15}{300}\) và \(\dfrac{25}{499}>\dfrac{25}{500}\)
\(\dfrac{15}{300}=\dfrac{1}{20}\) và \(\dfrac{25}{500}=\dfrac{1}{20}\)
Vì \(\dfrac{1}{20}=\dfrac{1}{20}\Leftrightarrow\dfrac{15}{300}=\dfrac{25}{500}\)
\(\Leftrightarrow\dfrac{15}{301}< \dfrac{1}{20}< \dfrac{25}{499}\)
\(\Leftrightarrow\dfrac{15}{301}< \dfrac{25}{499}\)
\(\Leftrightarrow A< B\)
\(c,A=\dfrac{5.6}{9.25}\) và \(B=\dfrac{18.4-18}{8.9+7.9}\)
\(A=\dfrac{5.6}{9.25}=\dfrac{1.2}{3.5}=\dfrac{2}{15}\)
\(B=\dfrac{18.4-18}{8.9+7.9}=\dfrac{18\left(4-1\right)}{9\left(8+7\right)}=\dfrac{18.3}{9.15}=\dfrac{2.1}{1.5}=\dfrac{2}{5}\)
Vì \(\dfrac{2}{15}< \dfrac{2}{5}\)
\(\Leftrightarrow A< B\)
Chúc bạn học tốt!
Là \(A< B\) mà. Cách trên khó hiểu chút. Theo cách này đi.
\(B=\dfrac{1}{7}-\dfrac{1}{9}\)
\(\Leftrightarrow B=\dfrac{2}{7.9}\)
Vì \(\dfrac{1}{7.9}< \dfrac{2}{7.9}\)
\(\Leftrightarrow A< B\)
1: \(B=\dfrac{1}{7}-\dfrac{1}{9}=\dfrac{2}{63}>\dfrac{1}{7\cdot9}=A\)
2: \(A=\dfrac{15}{301}< \dfrac{15}{300}=\dfrac{1}{20}=\dfrac{25}{500}< \dfrac{25}{499}\)
A = \(\frac{5}{1.2}\) + \(\frac{5}{2.3}\) +........+\(\frac{5}{99.100}\)
A = 5.(\(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) +......+\(\frac{1}{99.100}\) )
A = 5. ( \(\frac{1}{1}\) - \(\frac{1}{2}\) +\(\frac{1}{2}-\frac{1}{3}\) +......+\(\frac{1}{99}-\frac{1}{100}\) )
A= 5. (\(1-\frac{1}{100}\))
A= 5.\(\frac{99}{100}\)
A= \(\frac{99}{20}\)
B = \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+............+ \(\frac{1}{9.10}\)
= \(\frac{1}{2}\)- \(\frac{1}{3}\)+\(\frac{1}{3}\)- \(\frac{1}{4}\)+ ...................+\(\frac{1}{9}\)- \(\frac{1}{10}\)
= \(\frac{1}{2}\) - \(\frac{1}{10}\)
= \(\frac{2}{5}\)
\(A=\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+...+\frac{5}{99.100}\)
\(A=5\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(A=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=5\left(1-\frac{1}{100}\right)\)
\(A=5.\frac{99}{100}\)
\(A=\frac{99}{20}\)
\(B=\frac{1}{1.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(B=\frac{1}{2}-\frac{1}{10}\)
\(B=\frac{2}{5}\)
\(C=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\)
\(C=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(C=\frac{1}{3}-\frac{1}{15}\)
\(C=\frac{4}{15}\)
\(A=\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+...+\frac{5}{99.100}\)
\(A=5\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(A=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=5\left(1-\frac{1}{100}\right)\)
\(A=5.\frac{99}{100}\)
\(A=\frac{99}{20}\)
\(B=\frac{1}{1.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(B=\frac{1}{2}-\frac{1}{10}\)
\(B=\frac{2}{5}\)
\(C=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\)
\(C=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(C=\frac{1}{3}-\frac{1}{15}\)
\(C=\frac{4}{15}\)
\(A=\frac{1}{7\cdot9}\)
\(B=\frac{1}{7}-\frac{1}{9}\)
\(2A=\frac{1}{7}-\frac{1}{9}\)
\(\Rightarrow A< B\)
\(\text{Câu 1 :}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{12.13}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{12}-\frac{1}{13}\)
\(=\frac{1}{1}-\frac{1}{13}\)
\(=\frac{12}{13}\)
\(\text{Câu 2 :}\)
\(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)
\(=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\frac{100}{101}\)
\(=\frac{250}{101}\)
\(\Rightarrow A=\frac{5.2.3}{3.3.5.5}=\frac{1.2.1}{1.3.1.5}=\frac{2}{15}\)
\(B=\frac{3.3.2.2-3.3.2}{2.2.2.3.3+7.3.3}=\frac{1.1.1.1-1.1.2}{2.1.1.1.1+7.1.1}=\frac{-1}{9}\)
=> A > B
Ai thấy đúng k nha
\(A=\frac{5.6}{9.25}=\frac{1.2}{3.5}=\frac{2}{15}.\)
\(B=\frac{18.4-18}{8.9+7.9}=\frac{18\left(4-1\right)}{9\left(8+7\right)}=\frac{18.3}{9.15}=\frac{2.1}{15}=\frac{2}{15}.\)
\(\Rightarrow A=B.\)