x - 3h kolesnie velosipedi;
y - 2h kolesnie;
togda po usloviu mozhno sostavit' sistemu:
x+y=20;togda x=20-y;
3x+2y=55;ili 3(20-y)+2y=55;
60-3y+2y=55,togda y=5.
Otvet:5 2h kolesnih velosipedov.
Esli reshat' bez uravnenii:
1)dopustim vse velosipedi trehkolesnie,togda bi bilo 20*3=60 koles;
no ih 55,togda,2)esli uvelichivat' chislo dvuhkolesnih velosipedov na 1, budet umen'shatsia chislo trehkolesnih na 1,a chislo koles umenshitsia na 3-2=1.
3) no nam nado umenshit chislo koles s 60 do 55,znachit dvuhkolesnih velosipedov 60-55=5(no etot sposob reshenia bolshoi i ne ochen' udobnii dlia ponimania).
7 3/5-2 4/15= 5 (3/5-4/15)=5 (9-4/15)=
=5 1/3
5 1/3×9/32=16/3×9/32=3/2=1 1/2
2) (11/24+1/6)×1 3/5=1
11/24+1/6=11+4/24=15/24
15/24×1 3/5=15/24×8/5=3/3=1
3) (5 1/2+2 3/5)×5/9=4 1/2
5 1/2 +2 3/5=7 (1/2+3/5)=7 (5+6/10)=
=7 11/10=8 1/10
8 1/10×5/9=81/10×5/9=9/2 =4 1/2
4) (7-2 5/16)×4/25=3/4
7-2 5/16=6 16/16-2 5/16=4 11/16
4 11/16×4/25=75/16 ×4/25=3/4