Matematică
13death
4

I. Rezolvati ecuatiile.a) 2·x + 4·x + 5·x+...+102·x=2652 b) 4·(x+3) +2·(2·x-2) = 5·x+14 II . Rezolvati folosind regula factorului comun.  1. (2+4+6+8+20) : (1+2+3+4+5) 2. (5+10+15+...+40) : (1+2+3+...+8) 3. (4+8+12+...+60) : (1+2+3+...+15)

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(1) Răspunsuri
morarusimonaorl

I) a)[latex]2x + 4x + 5x+...+102x=2652[/latex] [latex]x(2+4+5+...+102)=2652[/latex] [latex]2+4+5+...102= \frac{(2+102)*51}{2} = \frac{104*51}{2} =52*51=2652[/latex] [latex]x*2652=2652[/latex] [latex]x= \frac{2652}{2652} [/latex] [latex]x=1[/latex] b) [latex]4*(x+3) +2*(2x-2) = 5x+14 [/latex] [latex]4x+12 +4x-4 = 5x+14 [/latex] [latex]8x+8 = 5x+14 [/latex] [latex]8x-5x=14-8[/latex] [latex]3x=6[/latex] [latex]x= \frac{6}{3} [/latex] sau [latex]x=6:3[/latex] [latex]x=2[/latex] II) 1. [latex](2+4+6+8+10) : (1+2+3+4+5)=2(1+2+3+4+5):[/latex][latex](1+2+3+4+5)=2[/latex] 2. [latex](5+10+15+...+40) : (1+2+3+...+8)=5(1+2+3+...+8): [/latex][latex](1+2+3+...+8)=5[/latex] 3.[latex](4+8+12+...+60) : (1+2+3+...+15)=4(1+2+3+...+15):[/latex][latex](1+2+3+...+15)=4[/latex]

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