\(A=\dfrac{6^{12}.3^2.5-7.9^7.2^{13}}{2.4^7.5-2^{14}.3^2}\)
K
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NN
1
4 tháng 4 2017
\(A=\dfrac{6^{12}\cdot3^3\cdot5-7\cdot9^7\cdot2^{13}}{2\cdot4^7\cdot5-2^{14}\cdot3^2}\)
\(=\dfrac{2^{12}\cdot3^{12}\cdot3^3\cdot5-7\cdot3^{14}\cdot2^{13}}{2\cdot2^{14}\cdot5-2^{14}\cdot3^2}\\ =\dfrac{2^{12}\cdot3^{15}\cdot5-7\cdot2^{13}\cdot3^{14}}{2^{15}\cdot5-2^{14}\cdot3^2}\)
\(=\dfrac{2^{12}\cdot3^{14}\left(3\cdot5-7\cdot2\right)}{2^{14}\left(5-3^2\right)}\)
\(=\dfrac{3^{14}\cdot1}{-4}=\dfrac{-3^{14}}{4}=\dfrac{-4782969}{4}=-1195742,25\)
H
20 tháng 8 2023
d)
\(\dfrac{3^9.3^{20}.2^8}{3^{24}.243.2^6}\\ =\dfrac{3^{29}.2^6.2^2}{3^{24}.3^5.2^6}\\ =\dfrac{3^{29}.2^6.4}{3^{29}.2^6}\\ =4\)
e)
\(\dfrac{2^{15}.5^3.2^6.3^4}{8.2^{18}.81.5}\\ =\dfrac{2^{21}.5^3.3^4}{2^3.2^{18}3^4.5}\\ =\dfrac{2^{21}.5.5^2.3^4}{2^{21}.3^4.5}\\ =5^2\\ =25\)
f)
\(=\dfrac{24\left(315+561+124\right)}{\dfrac{\left(1+99\right).50}{2}-500}\\ =\dfrac{24.1000}{2500-500}\\ =12\)
T
20 tháng 8 2023
\(a,\dfrac{-14.15}{21.\left(-10\right)}=\dfrac{-7.2.3.5}{7.3.\left(-2\right).5}=1\)
\(b,\dfrac{5.7-7.9}{7.2+6.7}=\dfrac{7\left(5-9\right)}{7\left(2+6\right)}=\dfrac{-4}{8}=-\dfrac{1}{2}\)
\(c,\dfrac{\left(-7\right).3+2.\left(-14\right)}{\left(-5\right).7-2.7}=\dfrac{-7.\left(3+4\right)}{7\left(-5-2\right)}\)
\(=\dfrac{\left(-7\right).7}{7.\left(-7\right)}=1\)
\(d,\dfrac{3^9.3^{20}.2^8}{3^{24}.243.2^6}=\dfrac{3^{29}.2^8}{3^{24}.3^5.2^6}=\dfrac{3^{29}.2^8}{3^{29}.2^6}=2^2=4\)
\(e,\dfrac{2^{15}.5^3.2^6.3^4}{8.2^{18}.81.5}=\dfrac{2^{21}.3^4.5^3}{2^{18}.2^3.3^4.5}=\dfrac{2^{21}.3^4.5^3}{2^{21}.3^4.5}=5^2=25\)
\(f,\dfrac{24.315+3.561.8+4.124.6}{1+3+5+...+97+99-500}\)
\(=\dfrac{24.315+24.561+24.124}{1+3+5+...+97+99-500}\)
\(=\dfrac{24\left(315+561+124\right)}{1+3+5+...+97+99-500}\)
\(=\dfrac{24.1000}{1+3+5+...+97+99-500}\) (1)
Đặt A = 1 + 3 + 5 + ... + 97 + 99
Số số hạng trong A là: (99 - 1) : 2 + 1 = 50 (số)
Tổng A bằng: (99 + 1) . 50 : 2 = 2500
Thay A = 2500 vào biểu thức (1), ta được:
\(\dfrac{24.1000}{2500-500}=\dfrac{24.1000}{2.1000}=12\)
LM
Lê Minh Vũ
VIP
14 tháng 6 2023
\(\left(-\dfrac{1}{2}\right)^2\div\dfrac{1}{4}-2\times\left(-\dfrac{1}{2}\right)^2\\= \dfrac{1}{4}\div\dfrac{1}{4}-2\times\dfrac{1}{4}\\ =1-\dfrac{1}{2}\\ =\dfrac{1}{2}\)
\(\left(-2\right)^3\times-\dfrac{1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right)\div\dfrac{5}{12}\)
= \(-6\times-\dfrac{1}{24}+\left(\dfrac{4}{3}-\dfrac{11}{6}\right)\div\dfrac{5}{12}\)
= \(\dfrac{1}{4}+-\dfrac{1}{2}\div\dfrac{5}{12}\)
= \(\dfrac{1}{4}+-\dfrac{6}{5}\)
= \(\dfrac{1}{4}-\dfrac{6}{5}\)
= \(-\dfrac{19}{20}\)
\(\left(6\dfrac{4}{9}+\dfrac{7}{11}\right)-\left(4\dfrac{4}{9}-2\dfrac{4}{11}\right)\\ =\dfrac{58}{9}+\dfrac{7}{11}-\dfrac{40}{9}+\dfrac{26}{11}\\ =\dfrac{58}{9}-\dfrac{40}{9}+\dfrac{7}{11}+\dfrac{26}{11}\\ =12+3\\ =15\)
TA
14 tháng 6 2023
\(a,\left(\dfrac{-1}{2}\right)^2:\dfrac{1}{4}-2\left(-\dfrac{1}{2}\right)^2\)
\(=\left(-\dfrac{1}{2}\right)^2\left(4-2\right)\)
\(=\dfrac{1}{4}.2=\dfrac{1}{2}\)
\(b,\left(-2\right)^3.\dfrac{-1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right):\dfrac{5}{12}\)
\(=\left(-8\right).\dfrac{-1}{24}+\left(-\dfrac{1}{2}\right).\dfrac{12}{5}\)
\(=\dfrac{1}{3}+\left(-\dfrac{1}{5}\right)=\dfrac{2}{15}\)
\(c,\left(6\dfrac{4}{9}+\dfrac{7}{11}\right)-\left(4\dfrac{4}{9}-2\dfrac{4}{11}\right)\)
\(=\dfrac{701}{99}-\dfrac{206}{99}=\dfrac{495}{99}=5\)
\(d,10\dfrac{1}{5}-5\dfrac{1}{2}.\dfrac{60}{11}+\dfrac{3}{15\%}\)
\(=\dfrac{51}{5}-30+20=\dfrac{1}{5}\)
\(e,\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\)
\(=\dfrac{5}{7}\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)=\dfrac{5}{7}.\left(-\dfrac{7}{11}\right)\)
\(=-\dfrac{5}{11}\)
\(f,\dfrac{-5}{7}.\dfrac{2}{11}+\left(-\dfrac{5}{7}\right).\dfrac{9}{11}+1\dfrac{5}{7}\)
\(=\left(-\dfrac{5}{7}\right)\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\)
\(=\left(-\dfrac{5}{7}\right)+\dfrac{12}{7}=1\)
C
7 tháng 7 2018
{ x2 - [ 62 - ( 82 - 9.7)3 - 7.5]3 - 5.3 }3 = 1
{ x2 + [ 36 - (64 - 63)3 - 35]3 - 15}3 = 1
[ x2 - ( 36 - 13 - 35 ) - 15 ]3 = 1
[ x2 - ( 36 - 1 - 35 ) - 15]3 = 1
[ x2 - ( 35 - 35 ) - 15]3 = 1
[ x2 - 0 - 15]3 = 1
( x2 - 15 )3 = 1
<=> ( x2 - 15)3 = 13
=> x2 - 15 = 1
<=> x2 = 16
=> x = 4
MV
3 tháng 8 2017
a,
\(\dfrac{4^2\cdot4^3}{2^{10}}=\dfrac{4^5}{2^{10}}=\dfrac{\left(2^2\right)^5}{2^{10}}=\dfrac{2^{10}}{2^{10}}=1\)
b,
\(\dfrac{\left(0,6\right)^5}{\left(0,2\right)^6}=\dfrac{\left(0,2\cdot3\right)^5}{\left(0,2\right)^5\cdot0,2}=\dfrac{\left(0,2\right)^5\cdot3^5}{\left(0,2\right)^5\cdot0,2}=\dfrac{243}{0,2}=\dfrac{243}{\dfrac{1}{5}}=243\cdot5=1215\)
c,
\(\dfrac{2^7\cdot9^3}{6^5\cdot8^2}=\dfrac{2^7\cdot\left(3^2\right)^3}{\left(2\cdot3\right)^5\cdot\left(2^3\right)^2}=\dfrac{2^6\cdot2\cdot3^6}{2^5\cdot3^5\cdot2^6}=\dfrac{3}{2^4}=\dfrac{3}{16}\)
d,
\(\dfrac{6^3+3\cdot6^2+3^3}{-13}=\dfrac{\left(2\cdot3\right)^3+3\cdot\left(2\cdot3\right)^2+3^3}{-13}=\dfrac{2^3\cdot3^3+3\cdot2^2\cdot3^2+3^3}{-13}=\dfrac{2^3\cdot3^3+2^2\cdot3^3+3^3}{-13}\dfrac{3^3\left(2^3+2^2+1\right)}{-13}=\dfrac{3^3\cdot13}{-13}=-3^3=-27\)
T
28 tháng 7 2018
a)\(\dfrac{6^2.6^3}{3^5}=\dfrac{2^2.3^2.2^3.3^3}{3^5}=2^5=32\)
b)\(\dfrac{25^2.4^2}{5^5\left(-2\right)^5}=\dfrac{5^2.5^2.2^2.2^2}{5^5.\left(-2\right)^5}=\dfrac{1}{-10}=-\dfrac{1}{10}\)
c)\(\dfrac{2^7.9^3}{8^2.3^6}=\dfrac{2^7.\left(3^2\right)^3}{\left(2^3\right)^2.3^6}=\dfrac{2^7.3^6}{2^6.3^6}=2\)
d)\(\dfrac{6^3+3.6^2+3^3}{-13}=\dfrac{2^3.3^3+3.2^2.3^2+3^3}{-13}=\dfrac{3^3\left(2^3+2^2+1\right)}{-13}\)
\(=\dfrac{3^3.13}{-13}=-3^3=-27\)
4 tháng 8 2017
a) \(\dfrac{4^2.4^3}{(2^2)^5}=\dfrac{4^2.4^3}{4^5}=\dfrac{4^3}{4^3}=1\)
b) = 1215
c) = \(\dfrac{3}{16}\)
d) = (-27)
1 tháng 10 2021
\(a,=\dfrac{3}{2}-\dfrac{5}{6}:\dfrac{1}{4}+\sqrt{\dfrac{1}{4}-\dfrac{1}{2}}=\dfrac{3}{2}-\dfrac{10}{3}+\sqrt{\dfrac{1}{2}}=-\dfrac{11}{6}+\dfrac{\sqrt{2}}{2}=\dfrac{-33+3\sqrt{2}}{6}\)
\(b,=-\dfrac{4}{3}\cdot\dfrac{9}{2}+\dfrac{13}{12}\cdot\left(-\dfrac{8}{13}\right)=6-\dfrac{2}{3}=\dfrac{16}{3}\\ c,=\dfrac{1}{4}-\left(-\dfrac{1}{6}:4-8\cdot\dfrac{1}{16}\right)=\dfrac{1}{4}-\left(-\dfrac{1}{24}-\dfrac{1}{2}\right)\\ =\dfrac{1}{4}-\dfrac{13}{24}=-\dfrac{7}{24}\\ d,=\dfrac{3^{11}\cdot5^{11}\cdot5^7\cdot3^4}{5^{18}\cdot3^{18}}=\dfrac{1}{3^3}=\dfrac{1}{27}\)
\(A=\dfrac{6^{12}.3^3.5-7.9^7.2^{13}}{2.4^7.5-2^{14}.3^2}\)
\(=\dfrac{2^{12}.3^{12}.3^3.5-7.3^{14}.2^{12}}{2.2^{14}.5-2^{14}.3^2}\)
\(=\dfrac{2^{12}.3^{15}.5-7.3^{14}.2^{13}}{2.3^{14}.5-2^{14}.3^2}\)
\(=\dfrac{2^{12}.3^{14}.\left(15-14\right)}{2^{14}.\left(2.5-3\right)}\)
\(=\dfrac{3^{14}.1}{2^2}\)
\(=\dfrac{314}{4}\)