a) 15.40-180:6
b) 12.6+24:(-8)
c) 2021+33:\([9+2.\left(3.11-21\right)]\)
d)186:\([\)51-7+2.(81:32)\(]\)
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21 tháng 7 2023
a: \(=\dfrac{2^9\cdot5^9\cdot3^{40}}{2^{12}\cdot5^{10}\cdot3^{20}}=\dfrac{3^{20}}{5\cdot2^3}\)
b: \(=\dfrac{-3^8\cdot2^{10}\cdot5^6}{2^9\cdot\left(-1\right)\cdot3^6\cdot5^7}=\dfrac{-2}{5}\cdot3^2=-\dfrac{18}{5}\)
c: \(=\dfrac{3^{186}\cdot5^{100}}{5^{100}\cdot3^{187}}=\dfrac{1}{3}\)
7 tháng 1
a: \(A=\dfrac{9^4}{3^2}=\dfrac{\left(3^2\right)^4}{3^2}=\dfrac{3^8}{3^2}=3^6\)=729
b: \(B=81\left(\dfrac{5}{3}\right)^4=81\cdot\dfrac{5^4}{3^4}=\dfrac{81}{3^4}\cdot5^4=5^4=625\)
c: \(C=\left(\dfrac{4}{7}\right)^{-4}\cdot\left(\dfrac{2}{7}\right)^3\)
\(=\left(\dfrac{7}{4}\right)^4\cdot\left(\dfrac{2}{7}\right)^3\)
\(=\dfrac{7^4}{4^4}\cdot\dfrac{2^3}{7^3}\)
\(=\dfrac{2^3}{4^4}\cdot7\)
\(=\dfrac{2^3}{2^8}\cdot7=\dfrac{7}{2^5}=\dfrac{7}{32}\)
d: \(D=7^{-6}\cdot\left(\dfrac{2}{3}\right)^0\left(\dfrac{7}{5}\right)^6\)
\(=7^{-6}\left(\dfrac{7}{5}\right)^6\)
\(=\dfrac{1}{7^6}\cdot\dfrac{7^6}{5^6}=\dfrac{1}{5^6}=\dfrac{1}{15625}\)
e: \(E=8^3:\left(\dfrac{2}{3}\right)^5\cdot\left(\dfrac{1}{3}\right)^2\)
\(=2^6:\dfrac{2^5}{3^5}\cdot\dfrac{1}{3^2}\)
\(=2^6\cdot\dfrac{3^5}{2^5}\cdot\dfrac{1}{3^2}\)
\(=\dfrac{2^6}{2^5}\cdot\dfrac{3^5}{3^2}=3^3\cdot2=54\)
f: \(F=\left(\dfrac{7}{9}\right)^{-2}\cdot\left(\dfrac{1}{\sqrt{3}}\right)^8\)
\(=\left(\dfrac{9}{7}\right)^2\cdot\left(\dfrac{1}{3}\right)^4\)
\(=\dfrac{9^2}{7^2}\cdot\dfrac{1}{3^4}=\dfrac{9^2}{3^4}\cdot\dfrac{1}{7^2}=\dfrac{81}{81}\cdot\dfrac{1}{49}=\dfrac{1}{49}\)
g: \(G=\left(-\dfrac{4}{5}\right)^{-2}\cdot\left(\dfrac{2}{5}\right)^2\cdot\left(\sqrt{2}\right)^3\)
\(=\left(-\dfrac{5}{4}\right)^2\cdot\left(\dfrac{2}{5}\right)^2\cdot2\sqrt{2}\)
\(=\dfrac{25}{16}\cdot\dfrac{4}{25}\cdot2\sqrt{2}=\dfrac{4}{16}\cdot2\sqrt{2}=\dfrac{8\sqrt{2}}{16}=\dfrac{\sqrt{2}}{2}\)
VQ
9 tháng 5 2022
98775 - 32 x 85
=98775 -2720
=96055
67500 - 24 x 236
= 67500 -5664
=61836
568 + 101598 : 287
= 568 +354
=922
6875 + 980 -180
=7855 -180
=7675
VQ
9 tháng 5 2022
\(\dfrac{2}{5}+\dfrac{3}{10}-\dfrac{1}{2}\)
\(=\dfrac{7}{10}-\dfrac{1}{2}\)
= \(\dfrac{1}{5}\)
\(\dfrac{8}{11}+\dfrac{8}{33}x\dfrac{3}{4}\)
\(=\dfrac{8}{11}+\dfrac{2}{11}\)
\(=\dfrac{10}{11}\)
\(\dfrac{7}{9}x\dfrac{3}{14}:\dfrac{5}{8}\)
\(=\dfrac{1}{6}:\dfrac{5}{8}\)
\(=\dfrac{1}{6}x\dfrac{8}{5}\)
\(=\dfrac{8}{30}\)
\(=\dfrac{4}{15}\)
\(\dfrac{5}{12}-\dfrac{7}{32}:\dfrac{21}{16}\)
\(=\dfrac{5}{12}-\dfrac{7}{32}x\dfrac{16}{21}\)
\(=\dfrac{5}{12}-\dfrac{1}{6}\)
\(=\dfrac{5}{12}-\dfrac{2}{12}\)
\(=\dfrac{3}{12}=\dfrac{1}{4}\)
S
10 tháng 3 2019
\(a,=\left(-\frac{3}{7}+\frac{3}{7}\right)+\frac{5}{13}=0+\frac{5}{13}=\frac{5}{13}\)
\(b,=-\frac{5}{21}+\left(-\frac{2}{21}\right)+\frac{8}{24}=-\frac{1}{3}+\frac{1}{3}=0\)
\(c,=\left[-\frac{6}{11}+\frac{\left(-5\right)}{11}\right]+2=-1+2=1\)
\(d,=\frac{-1+\left(-15\right)}{32}+\frac{1}{2}=-\frac{1}{2}+\frac{1}{2}=0\)
10 tháng 3 2019
a) \(\frac{-3}{7}+\frac{5}{13}+\frac{3}{7}\)
\(=\left(\frac{-3}{7}+\frac{3}{7}\right)+\frac{5}{13}\)
\(=0+\frac{5}{13}=\frac{5}{13}\)
b) \(\frac{-5}{21}+\frac{-2}{21}+\frac{8}{24}\)
\(=\frac{-7}{21}+\frac{1}{3}\)
\(=\frac{-1}{3}+\frac{1}{3}=0\)
c) \(\frac{-5}{11}+\left(\frac{-6}{11}+2\right)\)
\(=\frac{-5}{11}+\frac{-6}{11}+2\)
\(=\left(-1\right)+2=1\)
d) \(\left(\frac{-1}{32}+\frac{1}{2}\right)+\frac{-15}{32}\)
\(=\frac{-1}{32}+\frac{1}{2}+\frac{-15}{32}\)
\(=\left(\frac{-1}{32}+\frac{-15}{32}\right)+\frac{1}{2}\)
\(=\frac{-16}{32}+\frac{1}{2}\)
\(=\frac{-1}{2}+\frac{1}{2}=0\)
PH
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TT
5
VN
Võ Ngọc Phương
VIP
5 tháng 9 2023
a) Vì tổng A có 25 số hạng nên A = \(\dfrac{\left(1+25\right).25}{2}=325\)
b) Số số hạng là:
\(\left(50-2\right):2+1=25\) \(\left(số\right)\)
Tổng là:
\(\left(2+50\right).25:2=650\)
c) Số số hạng là:
\(\left(51-3\right):2+1=25\) \(\left(số\right)\)
Tổng là:
\(\left(3+51\right).25:2=675\)
d) Số số hạng là:
\(\left(81-1\right):4+1=21\) \(\left(số\right)\)
Tổng là:
\(\left(1+81\right).21:2=861\)
\(#Wendy.Dang\)
a. 600 - 30 =570
b. 72 + (-3) = 69
c. 2021 + 33:(9 +2x12)
=2021 + 33: 33
=2021+1
=2022
d. 186: (44+2x9)
=186 : 62
=3