Giải thích các bước giải:

Bài 4:

a.Xét $\Delta ABI,\Delta ACI$ có:

Chung $AI$

$AB=AC$

$IB=IC$

$\to \Delta ABI=\Delta ACI(c.c.c)$

$\to \widehat{AIB}=\widehat{AIC}$

Mà $\widehat{AIB}+\widehat{AIC}=180^o$

$\to \widehat{AIB}=\widehat{AIC}=90^o$

$\to AI\perp BC$

b.Xét $\Delta IAB,\Delta ICK$ có:

$IA=IK$

$\widehat{AIB}=\widehat{CIK}$

$IB=IC$

$\to \Delta AIB=\Delta KIC(c.g.c)$

$\to AB=CK$

c.Từ b $\to \widehat{IBA}=\widehat{ICF}$

$\to AB//CK$

Mà $IE\perp AB, IF\perp CK$

$\to EI\perp AB, IF\perp AB$

$\to I, E, F$ thẳng hàng

Bài 5:

Ta có:

$S=\dfrac34+\dfrac89+\dfrac{15}{16}+...+\dfrac{n^2-1}{n^2}$

$\to S=1-\dfrac14+1-\dfrac19+1-\dfrac1{16}+...+1-\dfrac1{n^2}$

$\to S=n-(\dfrac14+\dfrac19+\dfrac1{16}+...+\dfrac1{n^2})$

$\to S=n-(\dfrac1{2^2}+\dfrac1{3^2}+\dfrac1{4^2}+...+\dfrac1{n^2})$

Ta có: 

$\dfrac1{2.3}+\dfrac1{3.4}+\dfrac1{4.5}+...+\dfrac1{n(n+1)}<\dfrac1{2^2}+\dfrac1{3^2}+\dfrac1{4^2}+...+\dfrac1{n^2}<\dfrac1{1.2}+\dfrac1{2.3}+\dfrac1{3.4}+...+\dfrac1{(n-1)n}$

$\to \dfrac12-\dfrac13+\dfrac13-\dfrac14+...+\dfrac1n-\dfrac1{n+1}<\dfrac1{2^2}+\dfrac1{3^2}+\dfrac1{4^2}+...+\dfrac1{n^2}<1-\dfrac12+\dfrac12-\dfrac13+...+\dfrac1{n-1}-\dfrac1n$

$\to \dfrac12-\dfrac1{n+1}<\dfrac1{2^2}+\dfrac1{3^2}+\dfrac1{4^2}+...+\dfrac1{n^2}<1-\dfrac1n$

$\to 0<\dfrac1{2^2}+\dfrac1{3^2}+\dfrac1{4^2}+...+\dfrac1{n^2}<1$

$\to \dfrac1{2^2}+\dfrac1{3^2}+\dfrac1{4^2}+...+\dfrac1{n^2}\notin Z$

$\to n-(\dfrac1{2^2}+\dfrac1{3^2}+\dfrac1{4^2}+...+\dfrac1{n^2})\notin Z$

$\to S\notin Z$