Câu a, b, c, e bạn áp dụng quy tắc đấu ngoặc để mở ngoặc rồi tính
Câu d, g bạn tính như bình thường

\(a.83-\left(-756-17\right)+\left(50-756\right)=83+756+17+50-756=\left(83+17\right)+50+\left(756-756\right)=100+50=150\)

\(b.-105+\left(-34-95\right)-166=-105+\left(-34\right)-95-166=\left(-105-95\right)-\left(34+166\right)=\left(-200\right)-200=-400\)

\(c,-\left(39+228-407\right)+\left(118-161\right)=-39-228+407+118-161=\left(-39-161\right)-\left(228-118\right)+407=\left(-200\right)-110=-310\)

\(d,5^{10}:5^8+60:12+\left(-10\right)=5^2+5+\left(-10\right)=5\left(5+1-2\right)=5.4=20\)

\(e,-342-\left(161-342\right)-39=-342-161+342-39=\left(-342+342\right)-\left(161+39\right)=-200\)\(g,7^5:7^3+\left(-187-149\right)-213=7^2+\left(-187\right)-149-213=\left(49-149\right)-\left(187+213\right)=\left(-100\right)-400=-500\)

a)\(\text{ -19+84+19+16 }\) 

     \(=\left(-19+19\right)+\left(84+16\right)\)

\(=0+100=100\)

\(\left(437-25\right)-\left(175+437\right)\)

\(=437-437-\left(25+175\right)=-200\)

\(c\))\(2^3.\left(-7\right).125\)

\(=\left(8.125\right).\left(-7\right)\)

\(=1000.\left(-7\right)=-7000\)

a). -19+84+19+16= 19+ (-84) +19+16=  19+ ( -84 + 16) = 19 + (-68 )= - 49
c). 2³. (-7) . 125 = 8. (-7) . 125 = (125.8). (-7) = 1000. (-7 ) = -7000 
d). (-17) . 39 + 17.( -161) = 17. (-39 ) + 17. (-161) = 17. \([\left(-39\right)+\left(-161\right)]\) = 17. (-200)= -3400
nhớ k cho mik nha <3
 

\(\left(17-3x\right)^3+36=161\)

=> \(\left(17-3x\right)^3=161-36=125\)

=> \(\left(17-3x\right)^3=5^3\)

=> \(17-3x=5\)

=> \(3x=12\)

=> \(x=4\)

We have: 1 . 3 + 2 = 5

               5 . 3 + 2 = 17

               17 . 3 + 2 = 53

               53 . 3 + 2 = 161

=> The next number in the sequence is: 161 . 3 + 2 = 485

\(17\times17+17\times44+39\times17=17\left(17+44+39\right)\)

\(=17\times100=1700\)

17x17+17x44+39x17

= 17x(17+44+39)

= 17x(100)

=1700

17 x 17 + 17 x 44 + 39 x 17

= 17 x ( 17 + 44 + 39)

= 17 x 100

=1700

\(\dfrac{x-241}{17}+\dfrac{x-220}{19}+\dfrac{x-195}{21}+\dfrac{x-161}{23}=10\)

<=>\(\dfrac{x-241}{17}-1+\dfrac{x-220}{19}-2+\dfrac{x-195}{21}-3\dfrac{x-161}{23}-4=0\)

<=>\(\dfrac{x-258}{17}+\dfrac{x-258}{19}+\dfrac{x-158}{21}+\dfrac{x-158}{23}=0\)

<=>\(\left(x-258\right)\left(\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{21}+\dfrac{1}{23}\right)=0\)

vì 1/17+1/19+1/21+1/23 khcas 0

=>x-258=0<=>x=258

vậy...............

chcus bạn học tốt ^^

bằng 0

ai bảo là 0

cho mình câu trả lời chi tiết đi\

a) 7.x - x = 5²¹ : 5¹⁹ + 3.2² - 7⁰

6x = 25 + 12 - 1

6x = 36

x = 6

b) 7x - 2x = 6¹⁷ : 6¹⁵ + 44 : 11

5x = 36 + 4

5x = 40

x = 8

c) 5x + x = 39 - 3¹¹ : 3⁹

6x = 39 - 9

6x = 30

x = 5

d) [(6x - 39) : 7]. 4 = 12

(6x - 39) : 7 = 12 : 4

(6x - 39) : 7 = 3

6x - 39 = 3 . 7

6x - 39 = 21

6x = 21 + 39

6x = 60

x = 10

x =