\(\frac{1}{16}\)<\(\frac{1}{3\cdot4}\)tương tự=>\(\frac{1}{4}+\)\(\frac{1}{16}\)+.......+\(\frac{1}{196}< \frac{1}{3\cdot4}+......+\frac{1}{8\cdot9}=\frac{1}{3}\)--\(\frac{1}{9}\)+\(\frac{1}{4}\)=\(\frac{7}{18}< \frac{1}{2}\)

Vậy.................

em moi hoc lop 5 thoi

\(\frac{7}{3}\)

8/5:(8/5.5/4) / 16/25-1/25 + 1:4/7 / (50/9-9/4).36/17) + 0,6.0,5:2/5

= 8/5:8/5:5/4 / 3/5 + 7/4 / 50/9.36/17-9/4.36/17 + 0,3.5/2

= 4/5:3/5 + 7/4 / 200/17-81/17 + 3/10.5/2

= 4/5.5/3 + 7/4 / 119/17 + 3/4

= 4/3 + 7/4 : 7 + 3/4

= 4/3 + 4 + 3/4

= 16/12 + 48/12 + 9/12

= 73/12

☆★☆★☆

73/12

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1/302 < 1/301; 1/303<1/301; ...; 1/400<1/301

=> A < 1/2 + 1/301+1/301+...+1/301=1/2 + 100/301< 1/2+100/300=1/2+1/3=5/6<1

=> A<1 => đpcm

\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)

\(=\frac{1\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}{4.\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3.\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4.\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)

\(=\frac{1}{4}+\frac{3}{4}=1\)

\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)

\(=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)

\(=\frac{1}{4}+\frac{3}{4}\)

=1

ảo                         

\(\frac{1}{5}+\frac{4}{10}+...+\frac{81}{45}=\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+...+\frac{9}{5}=\frac{1+2+3+...+9}{5}=\frac{45}{5}=9\)

1/5 + 4/10 + 9/15 + 16/20 + 25/25 + 36/30 + 49/35 + 64/40 + 81/45

=1/5 + 2/5 + 3/5 + 4/5 + 5/5 + 6/5 + 7/5 + 8/5 + 9/5

=45/5 = 9