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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 = \(-8\)

18 tháng 10 2018

\(C=\frac{2}{3}\sqrt{144}-\left(\frac{-3}{4}\right):\sqrt{\frac{225}{144}}\)

\(=\frac{2}{3}\cdot12+\frac{3}{4}:\frac{5}{4}\)

\(=8+\frac{3}{4}\cdot\frac{4}{5}\)

\(=8+\frac{3}{5}\)

\(=\frac{40}{5}+\frac{3}{4}=\frac{43}{5}\)

\(D=\frac{4^6\cdot25^5-2^{12}\cdot25^4}{2^{12}\cdot5^8-10^8\cdot64}=\frac{\left(2^2\right)^6\cdot\left(5^2\right)^5-2^{12}\cdot\left(5^2\right)^4}{2^{12}\cdot5^8-\left(2\cdot5\right)^8\cdot2^6}\)

\(=\frac{2^{12}\cdot5^{10}-2^{12}\cdot5^8}{2^{12}\cdot5^8-2^{14}\cdot5^8}=\frac{5^8\left(2^{12}\cdot5^2-2^{12}\right)}{5^8\left(2^{12}-2^{14}\right)}\)

\(=\frac{2^{12}\cdot5^2-2^{12}}{2^{12}-2^{14}}=\frac{2^{12}\left(5^2-1\right)}{2^{12}\left(1-2^2\right)}=\frac{24}{-3}=-8\)

31 tháng 7 2017

=\(-\frac{6}{5}\).\(\frac{-7}{6}\).\(\frac{-8}{7}\).\(\frac{-9}{8}\).\(\frac{-10}{9}\).\(\frac{-11}{10}\)

=\(\frac{7}{5}\).\(\frac{9}{7}\).\(\frac{11}{9}\)

=\(\frac{11}{5}\)

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 \(\frac{11}{5}\)

27 tháng 6 2019

Những câu từ D trở đi là các câu riêng biệt ak bạn

2 tháng 7 2019

\(A = {1\over2}-{3\over4}+{5\over6}-{7\over12}={6\over12}-{9\over12}+{10\over12}-{7\over12}\)\(={0\over12}=0\)

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Đọc tiếp

\(\left(\frac{-5}{12}+\frac{7}{4}-\frac{3}{8}\right)-\left[4\frac{1}{2}-7\frac{1}{3}\right]-\left(\frac{1}{4}-\frac{5}{2}\right)\)

\(\left[2\frac{1}{4}-5\frac{3}{2}\right]-\left(\frac{3}{10}-1\right)-5\frac{1}{2}+\left(\frac{1}{3}-\frac{5}{6}\right)\)

\(\frac{4}{7}-\left(3\frac{2}{5}-1\frac{1}{2}\right)-\frac{5}{21}+\left[3\frac{1}{2}-4\frac{2}{3}\right]\)

\(\frac{1}{8}-1\frac{3}{4}+\left(\frac{7}{8}-3\frac{7}{2}+\frac{3}{4}\right)-\left[\frac{7}{4}-\frac{5}{8}\right]\)

\(\left(\frac{3}{5}-2\frac{1}{10}+\frac{11}{20}\right)-\left[\frac{-3}{4}+1\frac{7}{2}\right]\)

\(\left[-2\frac{1}{5}-2\frac{2}{3}\right]-\left(\frac{1}{15}-5\frac{1}{2}\right)+\left[\frac{-1}{6}+\frac{1}{3}\right]\)

\(1\frac{1}{8}-\left(\frac{1}{15}-\frac{1}{2}+\frac{-1}{6}\right)+\left[\frac{5}{4}+\frac{3}{2}\right]\)

\(\frac{5}{6}-\left(1\frac{1}{3}-1\frac{1}{2}\right)+\left[\frac{5}{12}-\frac{3}{4}-\frac{1}{6}\right]\)

\(1\frac{1}{4}-\left(\frac{7}{12}-\frac{2}{3}-1\frac{3}{8}\right)+\left[\frac{5}{24}-2\frac{1}{2}\right]-\frac{1}{6}-\left[\frac{-3}{4}\right]\)

\(-2\frac{1}{5}+2\frac{3}{10}-\left(\frac{6}{20}-\left[\frac{2}{8}-1\frac{1}{2}\right]\right)+\left[\frac{7}{20}-1\frac{1}{4}\right]\)

\(-\left[1\frac{2}{3}-3\frac{1}{2}+\frac{1}{4}\right]+\left(\frac{2}{6}-\frac{5}{12}\right)-\left(\frac{1}{3}-\left[\frac{1}{4}-\frac{1}{3}\right]\right)\)

\(-\frac{4}{5}-\left(1\frac{1}{10}-\frac{7}{10}\right)+\left[\frac{3}{4}-1\frac{1}{5}\right]+1\frac{1}{2}\)

\(\frac{3}{21}-\frac{5}{14}+\left[1\frac{1}{3}-5\frac{1}{2}+\frac{5}{14}\right]-\left(\frac{1}{6}-\frac{3}{7}+\frac{1}{3}\right)\)

\(-1\frac{2}{5}+\left[1\frac{3}{10}-\frac{7}{20}-1\frac{1}{4}\right]-\left(\frac{1}{5}-\left[\frac{3}{4}-1\frac{1}{2}\right]\right)\)

\(2\frac{1}{3}-\left(\frac{1}{2}-2\frac{1}{6}+\frac{3}{4}\right)+\left[\frac{5}{12}-1\frac{1}{3}\right]-\frac{7}{8}+3\frac{1}{2}\)

\(2\frac{1}{4}-1\frac{3}{5}-\left(\frac{9}{20}-\frac{7}{10}\right)+\left[1\frac{3}{5}-2\frac{1}{2}\right]+\frac{3}{4}\)

\(\left[\frac{8}{3}-5\frac{1}{4}+\frac{1}{6}\right]-\frac{7}{4}+\frac{-5}{12}-\left(1-1\frac{1}{2}+\frac{1}{3}\right)\)

\(\left(\frac{1}{4}-\left[1\frac{1}{4}-\frac{7}{10}\right]+\frac{1}{2}\right)-2\frac{1}{5}-1\frac{3}{10}+\left[1-\frac{1}{2}\right]\)

TRÌNH BÀY GIÚP MÌNH NHA 

0
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.

\(\frac{9}{10}+\left(-\frac{7}{8}\right)-\left(-\frac{2}{5}\right)-\frac{4}{3}\)

\(=\frac{9}{10}-\frac{7}{8}+\frac{2}{5}-\frac{4}{3}=-\frac{109}{120}\)

14 tháng 9 2016

a) \(\left(-5\right)^2\).\(\left(-5\right)^3\) = -3125

10 tháng 10 2015

Phần c khó để tớ giải cho

25 tháng 6 2019

1.ta có :

\(\left(10^3+10^2+10+1\right)^2\) 

=\(\left(1111\right)^2\) 

=1234321

hc tốt