Question

Answer 1

Hình 1:

Xét ΔABC có MN//BC

nên \(\dfrac{AM}{MB}=\dfrac{AN}{NC}\)

=>\(\dfrac{x}{3}=\dfrac{3}{4}\)

=>\(x=\dfrac{3}{4}\cdot3=2,25\)

Hình 2:

EH+HD=ED

=>ED=0,5+1,5=2

Xét ΔEFD có GH//FD

nên \(\dfrac{EG}{EF}=\dfrac{EH}{ED}\)

=>\(\dfrac{0.75}{EF}=\dfrac{0.5}{2}=\dfrac{1}{4}\)

=>EF=0,75*4=3

=>y=3

Hình 3:

Xét ΔMNI có KT//NI

nên \(\dfrac{MK}{MN}=\dfrac{MT}{MI}\)

=>\(\dfrac{6}{MN}=\dfrac{3.5}{10}\)

=>\(MN=6\cdot\dfrac{10}{3.5}=\dfrac{60}{3.5}=\dfrac{120}{7}\)

=>\(z=\dfrac{120}{7}\)