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a ) \(5\frac{3}{4}:3+2\frac{1}{4}.\frac{1}{3}-\frac{3}{8}=\frac{23}{4}:\frac{3}{1}+\frac{9}{4}.\frac{1}{3}=\frac{23}{12}+\frac{3}{4}=\frac{8}{3}\)
b ) \(\frac{3}{5}:\frac{5}{6}:\frac{6}{7}:\frac{7}{8}+\frac{7}{8}+\frac{2}{5}+\frac{23}{35}=\frac{3.6.7.8}{5.5.6.7}+\frac{7}{8}+\frac{2}{5}+\frac{23}{35}=\frac{24}{25}+\frac{7}{8}+\frac{2}{5}+\frac{23}{35}=\frac{4049}{1400}\)
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A=1/3.5+1/5.7+1/7.9+...+1/99.101
2A= 2/3.5+2/5.7+2/7.9+...+2/99.101
2A= 1/3-1/5+1/5-1/7-1/7+1/7-1/9+...+1/99-1/101
2A=1/3-1/101=98/303
A=(98/303)/2=49/303
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\(A=1/3.5+1/5.7+1/7.9+…+1/99.101\)
A.2=2/3.5+2/5.7+2/7.9+…+2/99.101
A.2=1/3-1/5+1/5-1/7+1/7-1/9+...+1/99-1/101
Vậy
A.2=1/3-1/101=98/303
A=98/303/2=49/303
Đúng không
A = 1/15 + 1/35 + 1/63 + 1/99 + ... + 1/9999
= 1/3x5 + 1/5x7 + 1/7x9 + 1/9x11 + ... + 1/99x101
A x 2 = 2/3x5 + 2/5x7 + 2/7x9 + 2/9x11 + ... + 2/99x101
= 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 + ... + 1/99 - 1/101
= 1/3 - 1/101 = 98/303
Vậy A = 98/303 : 2 = 49/303
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T = \(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{624}{625}\)
T = \(\frac{1.3.2.4.3.5.....24.26}{2.2.3.3.4.4.....25.25}\)
T = \(\frac{\left(1.2.3.....24\right)\left(3.4.5.....26\right)}{\left(2.3.4.....25\right)\left(2.3.4.....25\right)}\)
T = \(\frac{26}{25.2}\)
T = \(\frac{13}{25}\)
\(\frac{8}{35}+\frac{8}{63}+\frac{8}{99}+...+\frac{8}{575}\)
= \(4x\left(\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+...+\frac{2}{23x25}\right)\)
= \(4x\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{23}-\frac{1}{25}\right)\)
=\(4x\left(\frac{1}{3}-\frac{1}{25}\right)\)
= \(4x\frac{22}{75}\)
=\(\frac{88}{75}\)
=\(4.\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{23.25}\right)\)
=\(4.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{23}-\frac{1}{25}\right)\)
=\(4.\left(\frac{1}{5}-\frac{1}{25}\right)\)
=\(4.\frac{22}{75}\)
=\(\frac{88}{75}\)