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17 tháng 10 2018
C = \(\frac{2}{3}\sqrt{144}-\left(-\frac{3}{4}\right)\div\sqrt{\frac{225}{144}}\)
C = \(\frac{2}{3}.12+\frac{3}{4}\div\frac{5}{4}\)
C = \(8+\frac{3}{5}\)
C = \(8\frac{3}{5}\)
D = \(\frac{4^6.25^5-2^{12}.25^4}{2^{12}.5^8-10^8.64}\)
D = \(\frac{\left(2^2\right)^6.\left(5^2\right)^5-2^{12}.\left(5^2\right)^4}{2^{12}.5^8-\left(2.5\right)^8.2^6}\)
D = \(\frac{2^{12}.5^{10}-2^{12}.5^8}{2^{12}.5^8-2^8.5^8.2^6}\)
D = \(\frac{2^{12}.5^8.\left(25-1\right)}{2^{12}.5^8.\left(1-2^2\right)}\)
D = \(\frac{24}{-3}\)
D = \(-8\)
T
18 tháng 10 2018
\(C=\frac{2}{3}\sqrt{144}-\left(\frac{-3}{4}\right):\sqrt{\frac{225}{144}}\)
\(=\frac{2}{3}\cdot12+\frac{3}{4}:\frac{5}{4}\)
\(=8+\frac{3}{4}\cdot\frac{4}{5}\)
\(=8+\frac{3}{5}\)
\(=\frac{40}{5}+\frac{3}{4}=\frac{43}{5}\)
\(D=\frac{4^6\cdot25^5-2^{12}\cdot25^4}{2^{12}\cdot5^8-10^8\cdot64}=\frac{\left(2^2\right)^6\cdot\left(5^2\right)^5-2^{12}\cdot\left(5^2\right)^4}{2^{12}\cdot5^8-\left(2\cdot5\right)^8\cdot2^6}\)
\(=\frac{2^{12}\cdot5^{10}-2^{12}\cdot5^8}{2^{12}\cdot5^8-2^{14}\cdot5^8}=\frac{5^8\left(2^{12}\cdot5^2-2^{12}\right)}{5^8\left(2^{12}-2^{14}\right)}\)
\(=\frac{2^{12}\cdot5^2-2^{12}}{2^{12}-2^{14}}=\frac{2^{12}\left(5^2-1\right)}{2^{12}\left(1-2^2\right)}=\frac{24}{-3}=-8\)
BK
2 tháng 7 2019
\(A = {1\over2}-{3\over4}+{5\over6}-{7\over12}={6\over12}-{9\over12}+{10\over12}-{7\over12}\)\(={0\over12}=0\)
HH
30 tháng 7 2018
Giải:
1) \(7^8.\left(-\dfrac{1}{7}\right)^8\)
\(=7^8.\left(\dfrac{1}{7}\right)^8\)
\(=7^8.\dfrac{1^8}{7^8}\)
\(=1\)
2) \(\left(\dfrac{4}{3}\right)^{10}.\left(-\dfrac{3}{4}\right)^{10}\)
\(=\left(\dfrac{4}{3}\right)^{10}.\left(\dfrac{3}{4}\right)^{10}\)
\(=\dfrac{4^{10}}{3^{10}}.\dfrac{3^{10}}{4^{10}}\)
\(=1\)
3) \(\left(-\dfrac{7}{2}\right)^{2006}.\left(-\dfrac{2}{7}\right)^{2006}\)
\(=\left(\dfrac{7}{2}\right)^{2006}.\left(\dfrac{2}{7}\right)^{2006}\)
\(=1\)
4) \(\left(-\dfrac{5}{13}\right)^{2007}.\left(\dfrac{13}{5}\right)^{2006}\)
\(=\left(\dfrac{5}{13}\right)^{2007}.\left(\dfrac{13}{5}\right)^{2006}\)
\(=\dfrac{5^{2007}.13^{2006}}{13^{2007}.5^{2006}}\)
\(=\dfrac{5}{13}\)
Vậy ...
HN
31 tháng 7 2017
=\(-\frac{6}{5}\).\(\frac{-7}{6}\).\(\frac{-8}{7}\).\(\frac{-9}{8}\).\(\frac{-10}{9}\).\(\frac{-11}{10}\)
=\(\frac{7}{5}\).\(\frac{9}{7}\).\(\frac{11}{9}\)
=\(\frac{11}{5}\)
VT
31 tháng 7 2017
\(=\frac{-6}{5}\times\frac{-7}{6}\times\frac{-8}{7}\times\frac{-9}{8}\times\frac{-10}{9}\times\frac{-11}{10}\)
\(=\frac{\left(-6\right).\left(-7\right).\left(-8\right).\left(-9\right).\left(-10\right).\left(-11\right)}{5.6.7.8.9.10}\)
\(=\frac{6\times7\times8\times9\times10\times11}{5\times6\times7\times8\times9\times10}\)
Triệt tiêu các thừa số bằng nhau ở tử và mẫu, ta có kết quả là \(\frac{11}{5}\)
23 tháng 6 2017
1.
\(\frac{3}{10}-\left[\left(-\frac{5}{6}\right)-\left(-\frac{2}{3}\right)\right]\)
\(=\frac{3}{10}-\left(-\frac{13}{30}\right)=\frac{11}{15}\)
2.
\(\frac{9}{10}+\left(-\frac{7}{8}\right)-\left(-\frac{2}{5}\right)-\frac{4}{3}\)
\(=\frac{9}{10}-\frac{7}{8}+\frac{2}{5}-\frac{4}{3}=-\frac{109}{120}\)