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\(\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{45.50}\)
\(=\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{45}-\frac{1}{50}\)
\(=\frac{1}{5}-\frac{1}{50}=\frac{10}{50}-\frac{1}{50}=\frac{9}{50}\)
\(=\frac{10-5}{5.10}+\frac{15-10}{10.15}+...+\frac{50-45}{45.50}\)
\(=\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{45}-\frac{1}{50}\)
\(=\frac{1}{5}-\frac{1}{50}=\frac{9}{50}\)
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 \(\frac{222222}{1333332}<\frac{222222}{1333330}\)
=> A < B
\(=\frac{14\cdot101+15\cdot101+...+19\cdot101}{20\cdot101+21\cdot101+...+25\cdot101}=\frac{101\cdot\left(14+15+16+17+18+19\right)}{101\cdot\left(20+21+22+23+24+25\right)}\)
\(=\frac{14+15+16+17+18+19}{20+21+22+23+24+25}=\frac{\left(14+19\right)+\left(15+18\right)+\left(16+17\right)}{\left(20+25\right)+\left(21+24\right)+\left(22+23\right)}=\frac{33.3}{45.3}=\frac{33}{45}=\frac{11}{15}\)
A=1999/2000
B=199/200
C=511/512
hok tốt
Đáp án
mình lười trình bày cách làm lém, để đáp án thui nha
A = \(\frac{1999}{2000}\)
B = \(\frac{199}{200}\)
C = \(\frac{511}{512}\)
a)\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}......\frac{99}{100}\)
\(=\frac{1.2.3.4.....99}{2.3.4.5.6.....100}\)
\(=\frac{1}{100}\)
b) Tương tự như câu a
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=\frac{1}{1}-\frac{1}{8}\)
\(=\frac{7}{8}\)'
\(a.1\frac{1}{3}+2\frac{1}{2}\)
\(=\frac{4}{3}+\frac{5}{2}\)
\(=\frac{8}{6}+\frac{15}{6}\)
\(=\frac{23}{6}\)
\(b.3\frac{2}{5}-1\frac{1}{7}\)
\(=\frac{17}{5}-\frac{8}{7}\)
\(=\frac{119}{35}-\frac{40}{35}\)
\(=\frac{79}{35}\)
\(c.3\frac{1}{2}.1\frac{1}{7}\)
\(=\frac{7}{2}.\frac{8}{7}\)
\(=4\)
\(d.4\frac{1}{6}:2\frac{1}{3}\)
\(=\frac{25}{6}:\frac{7}{3}\)
\(=\frac{25}{6}.\frac{3}{7}\)
\(=\frac{25}{14}\)
\(A=\frac{1}{5}x\left(\frac{1}{5}+\frac{1}{10}+\frac{1}{10}+\frac{1}{995.1000}\right)\)
\(A=\frac{1}{5}x\left(\frac{1}{5}-\frac{1}{1000}\right)\)
\(A=\frac{1}{5}x\frac{199}{1000}\)
\(A=\frac{199}{5000}\)
Nếu muốn thì thử lại :
\(=\frac{1}{5}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+..+\frac{1}{995}-\frac{1}{1000}\right)...\)
\(=\frac{1}{5}\left(1-\frac{1}{1000}\right)=\frac{1}{5}\cdot\frac{995}{1000}\)
tự tính nốt nha