Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(2^2\cdot5-\dfrac{\left(1^{10}+8\right)}{3^2}\)
\(=4\cdot5-\dfrac{1+8}{3}\)
=20-3
=17
b: \(5^8:5^6+4\left(3^2-1\right)\)
\(=5^2+4\left(9-1\right)\)
=25+4*8
=25+32
=57
c: \(400-\left\{36-20:\left[3^3-\left(8-3\right)\right]\right\}\)
\(=400-36+20:\left[27-5\right]\)
\(=364+\dfrac{20}{22}\)
\(=364+\dfrac{10}{11}=\dfrac{4014}{11}\)
A) 2².5 - (1¹⁰ + 8) : 3²
= 4.5 - (1 + 8) : 9
= 20 - 9 : 9
= 20 - 1
= 19
B) 5⁸ : 5⁶ + 4.(3² - 1)
= 5² + 4.(9 - 1)
= 25 + 4.8
= 25 + 32
= 57
C) 400 - {36 - 20 : [3³ - (8 - 3)]}
= 400 - [36 - 20 : (27 - 5)]
= 400 -(36 - 20 : 22)
= 400 - (36 - 10/11)
= 400 - 386/11
= 4014/11
a) 9 234 : [3 . 3. (1 + 83)] = 9 234 : [3 . 3 . (1 + 512)]
= 9 234 : [3 . 3 . 513] = 9 234 : 4617 = 2
b) 76 - {2 . [2 . 52 - (31 - 2 . 3)]} + 3 . 25
= 76 - {2 . [2 . 25 - (31 - 6)]} + 75
= 76 - {2 . [50 - 25]} + 75 = 76 - {2 . 25} + 75 = 76 - 50 + 75 = 101
Theo đề ta có:
\(\left(5+\frac{1}{5}-\frac{2}{9}\right)-\left(2-\frac{1}{23}-2\frac{3}{5}+\frac{5}{6}\right)\)\(-\left(8-\frac{2}{3}-\frac{1}{18}\right)\)
= \(5+\)\(\frac{1}{5}-\frac{2}{9}\)-\(2+\frac{1}{23}+2+\frac{3}{5}+\frac{5}{6}-8+\frac{2}{3}-\frac{1}{18}\)
=\(\left(5+2-8\right)+\left(\frac{1}{5}+\frac{3}{5}\right)-\left(\frac{2}{9}-\frac{5}{6}-\frac{2}{3}+\frac{1}{18}\right)+\frac{1}{23}\)
= -1 +\(\frac{4}{5}\)\(-\frac{-11}{9}\)+\(\frac{1}{23}\)
= -1 +\(\frac{4}{5}+\frac{11}{9}+\frac{1}{23}\)
\(\left(5+\frac{1}{5}-\frac{2}{9}\right)-\left(2-\frac{1}{23}-2\frac{3}{5}+\frac{5}{6}\right)-\left(8-\frac{2}{3}-\frac{1}{18}\right)\)
= \(5+\frac{1}{5}-\frac{2}{9}-2+\frac{1}{23}+2+\frac{3}{5}-\frac{5}{6}-8+\frac{2}{3}+\frac{1}{18}\)
= \(\left(5-8\right)+\left(\frac{1}{5}+\frac{3}{5}\right)-\left(\frac{2}{9}-\frac{1}{18}-\frac{2}{3}\right)-\left(2-2\right)+\frac{1}{23}-\frac{5}{6}\)
= \(\left(-3\right)+\frac{4}{5}+\frac{1}{2}+\frac{1}{23}-\frac{5}{6}\)
= \(\left(\left(-3\right)+\frac{4}{5}+\frac{1}{2}-\frac{5}{6}\right)+\frac{1}{23}\)
= \(-\frac{38}{15}+\frac{1}{23}\)
= \(-\frac{859}{345}\)
a)\(5-\left(-\frac{5}{11}\right)^0+\left(\frac{1}{3}\right)^2:3=5-1+\frac{1}{9}\cdot\frac{1}{3}=4+\frac{1}{27}=\frac{108}{27}+\frac{1}{27}=\frac{109}{27}\)
b)\(2^3+3.\left(\frac{1}{2}\right)^0+\left[\left(-2\right)^3:\frac{1}{2}\right]=8+3.1+\left[\left(-8\right)\cdot2\right]=8+3-16=-5\)
a/ \(5-\left(-\frac{5}{11}\right)^0+\left(\frac{1}{3}\right)^2:3=5-1+\frac{1}{9}:3=5-1+\frac{1}{27}=4+\frac{1}{27}=\frac{109}{27}\)
b/ \(2^3+3.\left(\frac{1}{2}\right)^0+\left[\left(-2\right)^3:\frac{1}{2}\right]=8+3.1+\left[-8:\frac{1}{2}\right]=11+-16=-5\)
\(\frac{1}{2.5}\)\(+\)\(\frac{1}{5.8}\)\(+\frac{1}{8.11}\)\(+...+\frac{1}{152.155}\)
=\(\frac{1}{2}\) \(-\frac{1}{5}\) \(+\frac{1}{5}\) \(-\frac{1}{8}\) \(+...+\frac{1}{152}\) \(-\frac{1}{155}\)
=\(\frac{1}{2}\)\(-\frac{1}{155}\)
=\(\frac{153}{310}\)
a) \(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)....\left(\frac{1}{1000}-1\right)=-\frac{1}{2}.\left(-\frac{2}{3}\right).\left(-\frac{3}{4}\right)...\left(-\frac{999}{1000}\right)\)
\(=-\frac{1.2.3...999}{2.3.4...1000}=-\frac{1}{1000}\)
b)\(B=\frac{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{11}}:\frac{3}{4}=\frac{3\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}:\frac{3}{4}=\frac{3}{4}:\frac{3}{4}=1\)
d) \(D=1+\frac{1}{2}+\frac{1}{4}+..+\frac{1}{512}+\frac{1}{1024}\)
=> \(2D=2+1+\frac{1}{2}+...+\frac{1}{256}+\frac{1}{512}\)
=> \(2D-D=\left(2+1+\frac{1}{2}+...+\frac{1}{256}+\frac{1}{512}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}+\frac{1}{1024}\right)\)
=> \(D=2-\frac{1}{1024}=\frac{2047}{1024}\)
\(a.\frac{2\cdot\left(-13\right)\cdot9\cdot10}{\left(-3\right)\cdot4\cdot\left(-5\right)\cdot26}\)
\(=\frac{2\cdot\left(-13\right)\cdot3\cdot3\cdot2\cdot5}{\left(-3\right)\cdot2\cdot2\cdot\left(-5\right)\cdot13\cdot2}\)
\(=-\frac{3}{2}\)
b) \(\frac{2^3\cdot3^4}{2^2\cdot3^2\cdot5}=\frac{2\cdot3^2}{5}=\frac{2\cdot9}{5}=\frac{18}{5}\)
\(\frac{2^4\cdot5^2\cdot11^2\cdot7}{2^3\cdot5^3\cdot7^2\cdot11}=\frac{2\cdot1\cdot11\cdot1}{1\cdot5\cdot7\cdot1}=\frac{22}{35}\)
c) \(\frac{121\cdot75\cdot130\cdot169}{39\cdot60\cdot11\cdot198}=\frac{11\cdot11\cdot13\cdot10\cdot169}{13\cdot3\cdot6\cdot10\cdot11\cdot11\cdot6\cdot3}\)
\(=\frac{169}{3\cdot6\cdot6\cdot3}=\frac{169}{324}\)
d) \(\frac{1998\cdot1990+3978}{1992\cdot1991-3984}\)
bạn tick giúp mình nhé