calculati in doua moduri:

1. a)4999x1995+1994x4999-3988x4999

b)2011x3211-3211+2010x1789

c)129x130+220x129-350x99

d)3022x102-3022+101x1978-5000x99

2. daca a,b,c sunt numere naturale astfel incat a+b+c=32si2xa+b+2xc=57, calculati (a+c)x(3xa+2xb+3xc)

1
ajutatima si pe mn va rog
trebuie sa plec la scoala deci ajutatima urgent va rog
2) a+b+c=32 aduni aceste relatii 2xa+b+2xc=57 si obtii 3xa+2xb+3xc=32+57 adica 3xa+2xb+3xc=89; apoi din relatia a2-a o scazi pe prima si ai 2xa+b+2xc-a-b-c=57-32 adica a+c=25
(a+c)x(3xa+2xb+3xc)=25 x 89=2225 bafta la scoala!
NU E CORECT

Răspunsuri

2014-10-03T11:42:03+03:00
1.  a)4999x1995+1994x4999-3988x4999=4999x(1995+1994-3988)=4999x1=4999  sau
 4999x1995+1994x4999-3988x4999=4999x(1995+1994)- 3988x4999= 4999x3989-  3988x4999=4999x(3989-3988)=4999
b)2011x3211-3211+2010x1789=3211x(2011-1)+2010x1789=3211x2010 +2010x1789=
=2010x(3211+1789)=2010x5000=10.050.000  sau
2011x3211-3211+2010x1789=
=(2010+1)x3211-3211+2010x1789=2010x3211+3211-3211+2010x1789=
=2010x(3211+1789)=2010x5000=10.050.000

c)129x130+220x129-350x99 =129x(130+220) -350x99 = 129x350-350x99 =
=350x(129-99)=350x30=10500


d)3022x102-3022+101x1978-5000x9
=3022x(102-1)+101x1978-5000x9=
=3022x101+101x1978-5000x9=101x(3022+1978)-5000x9=101x5000-5000x9=
=5000x(101-9)=5000x92=460.000


















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