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a/ \(3+2^{x-1}=24-\left[4^2-\left(2^2-1\right)\right]\\3+2^{x+1}=24-\left[16-\left(4-1\right)\right]\)
\(3+2^{x+1}=24-\left(16-3\right)\\ 3+2^{x-1}=24-13\\ 3+2^{x-1}=11\\ 2^{x+1}=11-3\\ 2^{x-1}=8\)
\(2^{x-1}=2^3\\ \Rightarrow x-1=3\\x=3+1\\ x=4\)
\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=205550\)
\(\left(x.100\right)+\left(1+2+3+....+100\right)=205550\)
Ta tính tổng \(1+2+3+...+100\\ \) trước
Số các số hạng: \(\left[\left(100-1\right):1+1\right]=100\)
Tổng :\(\left[\left(100+1\right).100:2\right]=5050\)
Thay số vào ta có được:
\(\left(x.100\right)+5050=205550\\ \\ x.100=205550-5050\\ \\x.100=20500\\ \\x=20500:100\\ \\\Rightarrow x=2005\)
a)
4 . 25 – 12 . 25 + 170 : 10
= (4 . 25) – (12 . 25) + (170 : 10)
= 100 - 300 + 17
= -183
b)
(7 + 33 + 32) . 4 – 3
= (7 + 27 + 9) .4 – 3
= 43 . 4 – 3
= (43 . 4) – 3
= 45
c)
12 : {400 : [500 – (125 + 25 . 7)}
= 12 : {400 : [500 – (125 + 175)}
= 12 : (400: 200)
= 12 : 2
= 6
d)
168 + {[2.(24 + 32) - 2560] : 72}.
= 168 + [2 . (16 + 9) – 1] : 49
= 168 + 49: 49
= 168 + 1
= 167
a)
4 . 25 – 12 . 25 + 170 : 10
= (4 . 25) – (12 . 25) + (170 : 10)
= 100 - 300 + 17
= -183
b)
(7 + 33 + 32) . 4 – 3
= (7 + 27 + 9) .4 – 3
= 43 . 4 – 3
= (43 . 4) – 3
= 45
c)
12 : {400 : [500 – (125 + 25 . 7)}
= 12 : {400 : [500 – (125 + 175)}
= 12 : (400: 200)
= 12 : 2
= 6
d)
168 + {[2.(24 + 32) - 2560] : 72}.
= 168 + [2 . (16 + 9) – 1] : 49
= 168 + 49: 49
= 168 + 1
= 167
(1 + x) + (2 + x) + (3 + x) + ... + (10 + x) = 75
=> (1 + 2 + 3 + ... + 10) + (x + x + x + ... + x) = 75
=> 55 + 10x = 75
=> 10x = 20
=> x = 2
vậy_
x : [(1800 + 600) : 30] = 560 : (315 - 35)
=> x : [2400 : 30] = 560 : 280
=> x : 80 = 2
=> x = 160
vậy_
a)\(\dfrac{2}{3}-\dfrac{3}{5}:\left(-1\dfrac{1}{5}\right)+\left(\dfrac{-2}{3}\right)\cdot\dfrac{3}{8}\)
\(=\dfrac{2}{3}-\dfrac{3}{5}\cdot\dfrac{-5}{6}+\left(\dfrac{-1}{4}\right)=\dfrac{5}{12}+\dfrac{1}{2}=\dfrac{11}{12}\)
b)\(17\dfrac{11}{9}-\left(6\dfrac{3}{13}+7\dfrac{11}{19}\right)+\left(10\dfrac{3}{13}-5\dfrac{1}{4}\right)=\dfrac{164}{9}-\left(\dfrac{81}{13}+\dfrac{144}{19}\right)+\left(\dfrac{133}{13}-\dfrac{21}{4}\right)=\dfrac{164}{9}-\dfrac{3411}{247}+\dfrac{259}{52}=\dfrac{6425}{684}\)
c)\(\left(\dfrac{-3}{2}\right)^2-\left[-2\dfrac{1}{3}-\left(\dfrac{3}{4}+\dfrac{1}{3}\right):2\dfrac{3}{5}\right]\cdot\left(\dfrac{-3}{4}\right)=\dfrac{9}{4}-\left[\dfrac{-7}{3}-\dfrac{13}{12}\cdot\dfrac{5}{13}\right]\cdot\left(\dfrac{-3}{4}\right)=\dfrac{9}{4}-\left(\dfrac{-11}{4}\right)\cdot\left(\dfrac{-3}{4}\right)=\dfrac{3}{16}\)
d)\(\dfrac{21}{33}:\dfrac{11}{5}-\dfrac{13}{33}:\dfrac{11}{5}+\dfrac{25}{33}:\dfrac{11}{5}+\dfrac{6}{11}=\dfrac{5}{11}\cdot\left(\dfrac{21}{33}-\dfrac{13}{33}+\dfrac{25}{33}\right)+\dfrac{6}{11}=\dfrac{5}{11}\cdot1+\dfrac{6}{11}=1\)
\(a)\dfrac{2}{3}-\dfrac{3}{5}:\left(-1\dfrac{1}{5}\right)+\left(\dfrac{-2}{3}\right).\dfrac{3}{8}\)
\(=\dfrac{2}{3}-\dfrac{3}{5}:\left(\dfrac{-6}{5}\right)+\left(\dfrac{-2}{3}\right).\dfrac{3}{8}\)
\(=\dfrac{2}{3}-\dfrac{-1}{2}+\left(\dfrac{-2}{3}\right).\dfrac{3}{8}\)
\(=\dfrac{2}{3}-\dfrac{-1}{2}+\dfrac{-1}{4}\)
\(=\dfrac{7}{6}+\dfrac{-1}{4}\)
\(=\dfrac{11}{12}\)
\(a,-7-\left[\left(-19\right)+\left(21\right)\right].\left(-3\right)-\left[\left(32\right)+\left(-7\right)\right]\)
\(=-7-\left[21-19\right].\left(-3\right)-\left[32-7\right]\)
\(=-7-2.\left(-3\right)-25\)
\(=-7+6-25=-26\)
\(b,\left(-2\right)^2.3-\left(1^{10}+8\right):\left(-3\right)^2\)
\(=4.3-9:9\)
\(=12-1=11\)
Bài 1:
a: \(A=\dfrac{\left(85+\dfrac{7}{30}-83-\dfrac{5}{18}\right):\dfrac{8}{3}}{\dfrac{1}{25}}\)
\(=\left(2+\dfrac{7}{30}-\dfrac{5}{18}\right)\cdot\dfrac{3}{8}\cdot25\)
\(=\dfrac{180+21-25}{90}\cdot\dfrac{75}{8}\)
\(=\dfrac{176}{90}\cdot\dfrac{75}{8}=\dfrac{55}{3}\)
=>12,5% của A là 55/8x1/8=55/64
b: \(B=\dfrac{\left(6+\dfrac{3}{5}-3-\dfrac{3}{14}\right)\cdot\dfrac{36}{5}}{19.75:2.5}\)
\(=\dfrac{\left(3+\dfrac{27}{70}\right)\cdot\dfrac{36}{5}}{\dfrac{79}{10}}\)
\(=\dfrac{\dfrac{210+27}{70}\cdot\dfrac{36}{5}}{\dfrac{79}{10}}\)
\(=\dfrac{4266}{175}\cdot\dfrac{10}{79}=\dfrac{108}{35}\)
=>5% là 108/35x1/20=27/175
a) \(\left(-7\right)-\left[\left(-19\right)+\left(-21\right)\right].\left(-3\right)-\left[\left(+32\right)+\left(-7\right)\right]\)
\(=\left(-7\right)-\left(-40\right).\left(-3\right)-25\)
\(=\left(-7\right)-120-25\)
\(=-152\)
b) \(\left(-2\right)^3.3-\left(1^{10}+8\right):\left(-3\right)^2\)
\(=\left(-8\right).3-\left(1+8\right):9\)
\(=\left(-24\right)-9:9\)
\(=\left(-24\right)-1\)
\(=-25\)
Bài giải
a, \(\left(-7\right)-\left[\left(-19\right)+\left(-21\right)\right]\cdot\left(-3\right)-\left[\left(+32\right)+\left(-7\right)\right]\)
\(=\left(-7\right)-\left(-40\right)\cdot\left(-3\right)-25\)
\(=-7-120-25\)
\(=-127-25\)
\(=-152\)
b, \(\left(-2\right)^3\cdot3-\left(1^{10}+8\right)\text{ : }\left(-3\right)^2\)
\(=-8\cdot3-\left(1+8\right)\text{ : }9\)
\(=-24-9\text{ : }9\)
\(=-24-1\)
\(=-25\)