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3/5 A = 3/3.6 + 3/6.9 +..... + 3/96.99
= 1/3 - 1/6 + 1/6 - 1/9 + .... + 1/96 - 1/99 = 1/3 - 1/99 = 32/99
=> A = 160/297
k mk nha
a) Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(2A=1-\frac{1}{101}=\frac{100}{101}\)
\(A=\frac{100}{101}\div2=\frac{50}{101}\)
b) Đặt \(B=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+\frac{1}{12.15}\)
\(3B=\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+\frac{3}{12.15}\)
\(3B=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+\frac{1}{12}-\frac{1}{15}\)
\(3B=\frac{1}{3}-\frac{1}{15}=\frac{4}{15}\)
\(B=\frac{4}{15}\div3=\frac{4}{45}\)
Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(2A=1-\frac{1}{101}=\frac{100}{101}\)
\(A=\frac{100}{101}\div2=\frac{50}{101}\)
\(\frac{4}{3.6}+\frac{4}{6.9}+...+\frac{4}{27.30}=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{27}-\frac{1}{30}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{30}\right)=\frac{1}{2}.\frac{3}{10}=\frac{3}{20}\)
\(\frac{4}{3.6}+\frac{4}{6.9}+....+\frac{4}{27.30}\)
\(=\frac{4}{3}\left(\frac{3}{3.6}+\frac{3}{6.9}+...+\frac{3}{27.30}\right)\)
\(=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{27}-\frac{1}{30}\right)\)
\(=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{30}\right)\)
\(=\frac{4}{3}.\frac{3}{10}\)
\(=\frac{2}{5}\)
Ta có A=\(\frac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\)
=\(\frac{1.2+2.2.2+3.3.2+4.4.2+5.5.2}{3.4+3.2.4.2+3.3.4.3+3.4.4.4+3.5.4.5}\)
=\(\frac{2.\left(4+9+16+25\right)}{3.4.\left(4+9+16+25\right)}\)
=\(\frac{2}{3.4}=\frac{1}{3.2}=\frac{1}{6}\)
B=\(\frac{111111}{666666}=\frac{1}{6}\)
=>A=B
l-ike mình nha
\(A=\frac{1\cdot2+2\cdot4+3\cdot6+4\cdot8+5\cdot10}{3\cdot2\left(1\cdot2+2\cdot4+3\cdot6+4\cdot8+5\cdot10\right)}=\frac{1}{6}\)
\(B=\frac{111111}{666666}=\frac{1\cdot111111}{6\cdot111111}=\frac{1}{6}\)
Vì 1/6 =1/6 nên A=B
A = \(\frac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\)
A = \(\frac{1.2.\left(1+2+3+4+5\right)}{3.4.\left(1+2+3+4+5\right)}\)
A = \(\frac{2}{12}=\frac{222222}{1333332}\)
B = \(\frac{111111}{666665}=\frac{222222}{1333330}\)
Vì \(\frac{222222}{1333332}
\(=\frac{14\cdot101+15\cdot101+...+19\cdot101}{20\cdot101+21\cdot101+...+25\cdot101}=\frac{101\cdot\left(14+15+16+17+18+19\right)}{101\cdot\left(20+21+22+23+24+25\right)}\)
\(=\frac{14+15+16+17+18+19}{20+21+22+23+24+25}=\frac{\left(14+19\right)+\left(15+18\right)+\left(16+17\right)}{\left(20+25\right)+\left(21+24\right)+\left(22+23\right)}=\frac{33.3}{45.3}=\frac{33}{45}=\frac{11}{15}\)
\(\frac{\frac{1}{4}+\frac{1}{24}+\frac{1}{124}}{\frac{3}{4}+\frac{3}{24}+\frac{3}{124}}+\frac{\frac{2}{7}+\frac{2}{17}+\frac{2}{127}}{\frac{3}{7}+\frac{3}{17}+\frac{3}{127}}=\frac{\frac{1}{4}+\frac{1}{24}+\frac{1}{124}}{3\left(\frac{1}{4}+\frac{1}{24}+\frac{1}{124}\right)}+\frac{2\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{127}\right)}{3\left(\frac{1}{7}+\frac{1}{17}+127\right)}=\frac{1}{3}+\frac{2}{3}=\) \(1\)
\(\frac{2}{3}\div\frac{x}{5}+\frac{5}{7}=\frac{2}{7}\div\frac{3}{5}+\frac{10}{9}\)
\(\frac{2}{3}\div\frac{x}{5}+\frac{5}{7}=\frac{10}{21}+\frac{10}{9}\)
\(\frac{2}{3}\div\frac{x}{5}+\frac{5}{7}=\frac{300}{189}\)
\(\frac{2}{3}\div\frac{x}{5}=\frac{300}{189}-\frac{5}{7}\)
\(\frac{2}{3}\div\frac{x}{5}=\frac{55}{63}\)
\(\frac{x}{5}=\frac{2}{3}\div\frac{55}{63}\)
\(\frac{x}{5}=\frac{42}{55}\)
\(x=42\div\left(55\div5\right)=\frac{42}{11}\)