Đáp án:

`26)` `B<3/5`

`27)` `P=1` 

Giải thích các bước giải:

`26)` `B=1/3+1/{16}+1/{19}+1/{21}+1/{61}+1/{72}+1/{83}+1/{94}`

Ta có:

`1/{16}<1/{15};1/{19}<1/{15} ; 1/{21}<1/{15}`

`=>1/{16}+1/{19}+1/{21}<1/{15}+1/{15}+1/{15}=3/{15}=1/5`

$\\$

`1/{61}<1/{60}; 1/{72}<1/{60};1/{83}<1/{60}; 1/{94}<1/{60}`

`=>1/{61}+1/{72}+1/{83}+1/{94}<1/{60}+1/{60}+1/{60}+1/{60}=4/{60}=1/{15}`

$\\$

`B=1/3+(1/{16}+1/{19}+1/{21})+(1/{61}+1/{72}+1/{83}+1/{94})`

`< 1/3+1/5+1/{15}=5/{15}+3/{15}+1/{15}=9/{15}=3/5`

Vậy `B<3/5`

____________

`27)` Ta có:

`\qquad 1/{1.2}+1/{3.4}+...+1/{99.100}`

`=1/1-1/2+1/3-1/4+...+1/{99}-1/{100}`

`=(1+1/3+...+1/{99})-(1/2+1/4+...+1/{100})`

`=(1+1/2+1/3+1/4+...+1/{99}+1/{100})-(1/2+1/4+...+1/{100})-(1/2+1/4+...+1/{100})`

`=(1+1/2+1/3+1/4+...+1/{99}+1/{100})-2. (1/2+1/4+...+1/{100})`

`=(1+1/2+1/3+1/4+...+1/{50}+1/{51}+1/{52}+...+1/{100})-(1+1/2+...+1/{50})`

`=1/{51}+1/{52}+...+1/{100}`

$\\$

$P=\dfrac{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}{\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}}$

$=\dfrac{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}=1$

Vậy $P=1$

Đáp án:

`26. B < 3/5`

`27. P = 1`

Giải thích các bước giải:

`26.`

`B = 1/3 + 1/16 + 1/19 + 1/21 + 1/61 + 1/72 + 1/83 + 1/94`

`-> B = 1/3 + (1/16 + 1/19 + 1/21 ) + (1/61 + 1/72 + 1/83 + 1/94)`

$\bullet$ `1/16 + 1/19 + 1/21`

Vì `15 < 16`

`-> 1/16 < 1/15`

Tương tự ta có :

`1/19 < 1/15`

`1/21 < 1/15`

Cộng theo vế ta được :

`-> 1/16 + 1/19 + 1/21 < 1/15 + 1/15 + 1/15`

`-> 1/16 + 1/19 + 1/21 < 1/5` `(1)`

$\bullet$ `1/61 + 1/72 + 1/83 + 1/94`

Vì `60 < 61`

`-> 1/61 < 1/60`

Tương tự ta có :

`1/72 < 1/60`

`1/83< 1/60`

`1/94 < 1/60`

Cộng theo vế ta được :

`-> 1/61 + 1/72 + 1/83 + 1/94 < 1/60 + 1/60 + 1/60 + 1/60`

`-> 1/61 + 1/72 + 1/83 + 1/94 < 1/15` `(2)`

Lấy `(1) + (2)` vế với vế ta được :

`-> (1/16 + 1/19 + 1/21) + (1/61 + 1/72 + 1/83 + 1/94) < 1/5 + 1/15`

`-> 1/3 + (1/16 + 1/19 + 1/21 ) + (1/61 + 1/72 + 1/83 + 1/94) < 1/3 + 1/5 + 1/15`

`-> B  < 3/5`

Vậy `B < 3/5`

`27.`

$P = \dfrac{\dfrac{1}{51} + \dfrac{1}{52} + ... + \dfrac{1}{100} }{\dfrac{1}{1.2} + \dfrac{1}{3.4} + ... + \dfrac{1}{99.100} }$

Có : `1/(1.2) + 1/(3.4) + ... + 1/(99.100)`

`= (2-1)/(1.2) + (4-3)/(3.4) + ... + (100-99)/(99.100)`

`= 1 - 1/2 + 1/3 - 1/4 + ... + 1/99 - 1/100`

`= (1 +  1/3 + ... + 1/99) + (-1/2 - 1/4 - ... - 1/100)`

`= (1+1/3+...+1/99) - (1/2 + 1/4+...+1/100)`

`= (1+1/3+...+1/99) - (1/2 + 1/4+...+1/100) - (1/2 + 1/4 +... + 1/100) + (1/2 + 1/4+...+1/100)`

`= (1+1/3+...+1/99) - 2 (1/2 + 1/4 + 1/100) + (1/2 + 1/4 + ... + 1/100)`

`= (1+1/2+1/3+...+1/100) - (1 + 1/2 + ... + 1/50)`

`= 1+1/2+...+1/100 - 1 - 1/2-...-1/50`

`= (1-1) + (1/2 - 1/2) + ... + (1/50 - 1/50) + 1/51 + 1/52 + ... + 1/100`

`= 1/51 + 1/52 + ... + 1/100`

$→ P = \dfrac{\dfrac{1}{51} + \dfrac{1}{52} + ... + \dfrac{1}{100} }{\dfrac{1}{51} + \dfrac{1}{52} + ... + \dfrac{1}{100} }$

`-> P = 1`

Vậy `P=1`