Question

Ma poate ajuta cineva ?

Answer 1

a)7+3·[ 23+5(8+66÷6)]=

7+3·[23+5·(8+11)=

7+3·(23+5·19)=

7+3·(23+95)=

7+3·118=

7+354=

361

b)91+56÷2+33·(156-12·13)=

91+28+33·(156-156)=

91+28+33·0=

91+28+0=

119

c)[200-200÷(75-25)]÷4=

(200-200÷50)÷4=

(200-4)÷4=

196÷4=

49

d)1000·[105÷5+32÷(22-6·3)]=

1000·[21+32÷(22-18)]=

1000·(21+32÷4)=

1000·(21+8)=

1000·29=

29.000

e)  (1+2+3)^{4} +5·6·(7+8+9)=

6^{4}+5·6·24=

1296+720=

2016

f) 0^{1} + 1^{0} + 2^{3} - 3^{2} + 4^{5} - 349=

0+1+8-9+1024+625-349=

1300

g) la g nu stiu csf?....

h)(5^{0} + 5^{1} + 5^{2} + 5^{3} + 5^{4} ) ÷ (71·5+6)=

(1+5+25+5^{4} + 5^{4} ) ÷ (355+6)=

(31+5^{3} + 5^{4} ) ÷ 361=

31+5^{3} +5^{4} \ 361=

31+125+625\ 361=

781\361

Mai pe scurt \ inseamna deasupra si ^{2} la puterea numarului din paranteza.