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Toán lớp 7        bye mk đi hc đây hihi

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Ta có : \(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}\)

= \(\frac{1}{3}+\)( \(\frac{1}{31}+\frac{1}{35}+\frac{1}{37}\)) \(+\)( \(\frac{1}{47}+\frac{1}{53}+\frac{1}{61}\) ) \(< \)\(\frac{1}{3}+\)( \(\frac{1}{30}+\frac{1}{30}+\frac{1}{30}\)) \(+\)( \(\frac{1}{45}+\frac{1}{45}+\frac{1}{45}\)) = \(\frac{1}{2}\)

Vậy \(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}< \frac{1}{2}\)

1/5+1/14+1/27+1/43+1/61+1/89+1/111=0,368...( khi đem tử chia cho mẫu)

vi 1:2=0,5 ne 0,5>0,368...

CMR: 0,5>0,368..

nen 1/2 lon hon

ai giúp mình cho

\(\frac{1}{5}+\frac{1}{14}+\frac{1}{27}+\frac{1}{43}+\frac{1}{61}+\frac{1}{89}+\frac{1}{111}=0,368,..\) khi đem tu chia cho mau

1:2=0,5 CMR=0,5>0,368..

ta có A=1/5+1/14+1/27+1/43+1/61+1/89+1/111

=1/5+(1/14+1/27+1/43)+(1/61+1/89+1/111)<1/5 +(1/12+1/12+1/12)+(1/60+1/60+1/60)=1/5+1/4+1/20=1/2

ta suy ra A<1/2(đpcm)

Đặt \(A=\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{64^2}\)

Đặt \(B=\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{64^2}\)

Ta có: \(\frac{1}{5^2}< \frac{1}{4.5}\)

           \(\frac{1}{6^2}< \frac{1}{5.6}\)

            ....................

          \(\frac{1}{64^2}< \frac{1}{63.64}\)

\(\Rightarrow B< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{63.64}\)

\(\Rightarrow B< \frac{1}{4}-\frac{1}{64}< \frac{1}{4}\)

\(\Rightarrow B< \frac{1}{4}\)

\(\Rightarrow A< \frac{1}{4^2}+\frac{1}{4}\)

\(\Rightarrow A< \frac{5}{16}\)

Ta có S =\(\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{64^2}\)

= \(\frac{1}{4.4}+\frac{1}{5.5}+\frac{1}{6.6}+...+\frac{1}{64.64}\)

< \(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{63.64}\)

= \(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{63}-\frac{1}{64}\)

= \(\frac{1}{3}-\frac{1}{64}\)

= \(\frac{61}{192}\)> \(\frac{60}{192}=\frac{5}{16}\)

S <  \(\frac{61}{192}>\frac{5}{16}\)

=> sai đề