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\(=\frac{4}{2x4}+\frac{4}{4x6}+\frac{4}{6x8}+...+\frac{4}{18x20}\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{18}-\frac{1}{20}\right)\)
\(=2x\left(\frac{1}{2}-\frac{1}{20}\right)\\ =2x\frac{9}{20}\\ =\frac{9}{10}\)
\(S=\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{20.22}\)
\(2S=\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{20.22}\)
\(2S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{20}-\frac{1}{22}\)
\(2S=1-\frac{1}{22}=\frac{21}{22}\)
\(S=\frac{21}{22}:2=\frac{21}{44}\)
Ta nhận thấy mẫu số của các phân số có qui luật 1x3; 2x4; 3x5; 4x6...... => mẫu số của phân số thứ 98 là 98x100
\(\Rightarrow A=\frac{4}{3}x\frac{9}{8}x\frac{16}{15}x\frac{25}{24}x\frac{36}{35}x...x\frac{9801}{9800}\)
\(A=\frac{2x2x3x3x4x4x5x5x6x6x...x99x99}{1x2x3x3x4x4x5x5x...x96x96x97x97x98x98x99x100}=\frac{2x99}{100}=\frac{99}{50}=1\frac{49}{50}\)
\(B=\frac{2}{8}+\frac{2}{24}+\frac{2}{48}+...+\frac{2}{18\cdot20}\)
\(B=\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{18\cdot20}\)
\(B=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{18}-\frac{1}{20}\)
\(B=\frac{1}{2}-\frac{1}{20}\)
\(B=\frac{9}{20}\)
=))
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(A=\frac{1}{2}+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+...+\left(\frac{1}{9}-\frac{1}{9}\right)-\frac{1}{10}\)
\(A=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
Đặt biểu thức trên là A ta có:
A = \(\frac{1}{3}\)+ \(\frac{1}{6}\)+ \(\frac{1}{12}\)+ \(\frac{1}{24}\)+ \(\frac{1}{48}\)+ \(\frac{1}{96}\)
A x 3 = \(1\)+ \(\frac{1}{2}\)+ \(\frac{1}{4}\)+ \(\frac{1}{8}\)+ \(\frac{1}{16}\)+ \(\frac{1}{32}\)
A x 3 = \(1\)+ \(1\)- \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{8}\)+ \(\frac{1}{8}\)- \(\frac{1}{16}\)+ \(\frac{1}{16}\)- \(\frac{1}{32}\)
A x 3 = 2 - \(\frac{1}{32}\)= \(\frac{63}{32}\)
A = \(\frac{63}{32}\): 3 = \(\frac{63}{96}\)
\(1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}.....1\frac{1}{99}\)
\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}......\frac{100}{99}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.....\frac{10.10}{9.11}\)
\(=\frac{\left(2.3.4...10\right)\left(2.3.4....10\right)}{\left(1.2.3...9\right)\left(3.4.5....11\right)}\)
\(=\frac{10.2}{11}=\frac{20}{11}\)
\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}.\frac{25}{24}...\frac{100}{99}\)
\(1\frac{1}{3}\times1\frac{1}{8}\times1\frac{1}{15}\times1\frac{1}{24}\times...\times1\frac{1}{99}\)
\(=\frac{4}{3}\times\frac{9}{8}\times\frac{16}{15}\times\frac{25}{24}\times...\times\frac{100}{99}\)
\(=\frac{2\times2}{1\times3}\times\frac{3\times3}{2\times4}\times\frac{4\times4}{3\times5}\times\frac{5\times5}{4\times6}\times...\times\frac{10\times10}{9\times11}\)
\(=\frac{2}{1}\times\frac{10}{11}\)
\(=\frac{20}{11}\)
\(\dfrac{1}{8}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{120}+\dfrac{1}{168}+\dfrac{1}{224}+\dfrac{1}{288}+\dfrac{1}{360}\)
\(=\dfrac{1}{2\text{x}4}+\dfrac{1}{4\text{x}6}+...+\dfrac{1}{18\text{x}20}\)
\(=\dfrac{1}{2}\text{x}\left(\dfrac{2}{2\text{x}4}+\dfrac{2}{4\text{x}6}+...+\dfrac{2}{18\text{x}20}\right)\)
\(=\dfrac{1}{2}\text{x}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{18}-\dfrac{1}{20}\right)\)
\(=\dfrac{1}{2}\text{x}\left(\dfrac{1}{2}-\dfrac{1}{20}\right)=\dfrac{1}{2}\text{x}\dfrac{9}{20}=\dfrac{9}{40}\)
81+241+481+1201+1681+2241+2881+3601
=12x4+14x6+...+118x20=2x41+4x61+...+18x201
=12x(22x4+24x6+...+218x20)=21x(2x42+4x62+...+18x202)
=12x(12−14+14−16+...+118−120)=21x(21−41+41−61+...+181−201)
=12x(12−120)=12x920=940=21x(21−201)=21x209=409