Giải thích các bước giải:

Bài 4:

a.Xét $\Delta AMB,\Delta AMC$ có:

Chung $AM$

$MB=CM$

$AB=AC$

$\to \Delta AMB=\Delta AMC(c.c.c)$

b.Xét $\Delta MAC, \Delta MDB$ có:

$MA=MD$

$\widehat{AMC}=\widehat{BMD}$
$MC=MB$
$\to \Delta AMC=\Delta DMB(c.g.c)$

$\to \widehat{MAC}=\widehat{MDB}$

$\to AC//DB$

c.Xét $\Delta MAI,\Delta MDK$ có:

$MA=MK$

$\widehat{MAI}=\widehat{MDB}$ vì $AC//BD$

$AI=DK$

$\to \Delta MAI=\Delta MDK(c.g.c)$

$\to \widehat{AMI}=\widehat{DMK}$

$\to I, M, K$ thẳng hàng

Bài 5:

Ta có:

$A=1^2+2^2+3^2+...+8^2+9^2$

$\to A=1\cdot (2-1)+2\cdot (3-1)+3\cdot (4-1)+...+8\cdot (9-1)+9\cdot (10-1)$

$\to A=1\cdot 2-1+2\cdot 3-2+3\cdot 4-3+...+8\cdot 9-8+9\cdot 10-9$

$\to A=(1\cdot 2+2\cdot 3+...+8\cdot 9+9\cdot 10)-(1+2+3+...+8+9)$

Đặt:

$B=1\cdot 2+2\cdot 3+...+8\cdot 9+9\cdot 10$

$\to 3B=1\cdot 2\cdot 3+2\cdot 3\cdot 3+...+8\cdot 9\cdot 3+9\cdot 10\cdot 3$

$\to 3B=1\cdot 2\cdot (3-0)+2\cdot 3\cdot (4-1)+...+8\cdot 9\cdot (10-7)+9\cdot 10\cdot (11-8)$

$\to 3B=-0\cdot 1\cdot 2+1\cdot 2\cdot 3-1\cdot 2\cdot 3+2\cdot 3\cdot 5+...-7\cdot 8\cdot9+8\cdot 9\cdot 10-8\cdot 9\cdot 10+9\cdot 10\cdot 11$

$\to 3B=9\cdot 10\cdot 11$

$\to B=\dfrac{9\cdot 10\cdot 11}3$

$\to A=\dfrac{9\cdot 10\cdot 11}3-(1+2+3+...+8+9)$

$\to A=\dfrac{9\cdot 10\cdot 11}3-\dfrac{9\cdot (9+1)}2$

$\to A=285$