Question

Va rog sa ma ajutati la 2 subpuncte la matematica:
125+{14·[40+(x+240):40]-34}·17=47147
si
S=7+7²+7³+···7(la puterea 2003)-7³(1+7+7²+...7(la puterea 2000)
Va rog sa imi spuneti si pasii

Answer 1

  
[tex]\displaystyle 1) \\ 125+\{14 \cdot [40+(x+240):40]-34\} \cdot 17=47147 \\ \{14 \cdot [40+(x+240):40]-34\} \cdot 17=47147-125 \\ \{14 \cdot [40+(x+240):40]-34\} \cdot 17=47022 \\ 14 \cdot [40+(x+240):40]-34= \frac{47022}{17} \\ 14 \cdot [40+(x+240):40]-34= 2766 \\ 14 \cdot [40+(x+240):40] = 2766+34 \\ 14 \cdot [40+(x+240):40] = 2800\\ 40+(x+240):40 = \frac{2800}{14} \\ 40+(x+240):40 = 200 \\ (x+240):40 =200-40 [/tex]
[tex](x+240):40 =160 \\ x+240 =160 \cdot 40 \\ x+240 = 6400 \\ x = 6400 - 240 = \boxed{6160} [/tex]


[tex]2) \\ S=7+7^2+7^3+ \hdots +7^{2003}- 7^3(1+7+7^2+ \hdots +7^{2000})= \\ =7+7^2+7^3+ \hdots +7^{2003}- (7^3+7^4+7^5+ \hdots +7^{2003})= \\ =7+7^2+7^3+ \hdots +7^{2003}- 7^3-7^4-7^5- \hdots -7^{2003}= \\ =7+7^2+(7^3 - 7^3)+(7^4 -7^4) \hdots +7^{2003} -7^{2003}= \\ =7+7^2 = 7+49 = \boxed{56} [/tex]