Va rog mult..!!
Rezolvati,in multimea numerelor naturale,ecuatiile:22+5x-7-3(x-2)=x+23
2·x+3+2²·x_3²=42+1
11+12+13+14+15x+16x=3²
7·{x-2·[20-3·(11·3-27)]}+x=2109
11·{x-2·[200+10·(24+24/4)]}=1210
3x+2-x-5<x+21
12²+12x-11²-x-21≤7x+10²+3


Urgent!!!!

1

Răspunsuri

2014-11-09T22:27:25+02:00
22+5x-7-3(x-2)=x+23
22+5x-7-3x+6=x+23
2x+21=x+23
2x-x=23-21
x=2

2·x+3+2²·x-3²=42+1
2x+3+4x-9=43
6x-6=43
6x=43+6
x=49/6
11+12+13+14+15x+16x=3²
50+31x=9
31x=9-50
x=-41/31

7·{x-2·[20-3·(11·3-27)]}+x=2109
7·{x-2·[20-3·6]}+x=2109
7[x-2·2]+x=2109
7x-28+x=2109
8x=2109+28
x=2137/8=267,125

11·{x-2·[200+10·(24+24/4)]}=1210
11{x-2[200+10·(96+24)/4)]}=1210
11{x-2[200+10·30]}=1210
11{x-2·500}=1210
11x-11000=1210
11x=1210+11000
x=12210/11=1110

3x+2-x-5<x+21
2x-3<x+21
2x-x<21+3
x<24
x∈(-α,24)

12²+12x-11²-x-21≤7x+10²+3
144+12x-121-x-21
≤7x+100+3
11x-8≤7x+103
11x-7x≤103+8
4x≤111
x≤111/4
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