Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\frac{9}{8}-\frac{8}{9}+\frac{3}{25}+\frac{1}{4}-\frac{5}{16}+\frac{19}{25}-\frac{1}{9}+\frac{2}{25}-\frac{1}{81}\)
\(A=\left(\frac{9}{8}+\frac{1}{4}-\frac{5}{16}\right)-\left(\frac{8}{9}+\frac{1}{9}-\frac{1}{81}\right)+\left(\frac{3}{25}+\frac{19}{25}+\frac{2}{25}\right)\)
\(A=\frac{17}{16}-\frac{80}{81}+\frac{24}{25}\)
\(A=\frac{33529}{32400}\)
Bài 1:
A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + (\(\frac45\) - \(\frac{3}{17}\) + \(\frac13\)) - \(\frac17\) + (- \(\frac{14}{30}\))
A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + \(\frac45\) - \(\frac{3}{17}\) + \(\frac13\) - \(\frac17\) - \(\frac{14}{30}\)
A = (\(\frac15\) + \(\frac45\)) + (\(\frac{3}{17}\) - \(\frac{3}{17}\)) - (\(\frac43-\frac13\)) - \(\frac{30}{210}\) - \(\frac{98}{210}\)
A = 1 + 0 - 1 - (\(\frac{30}{210}+\frac{98}{210}\))
A = 1 - 1 - \(\frac{228}{210}\)
A = 0 - \(\frac{128}{210}\)
A = - \(\frac{64}{105}\)
Bài 2:
B= (\(\frac58\) - \(\frac{4}{12}\) + \(\frac32\)) - (\(\frac58\) + \(\frac{9}{13}\)) - (\(\frac{-3}{2}\)) + \(\frac{7}{-15}\)
B = \(\frac58\) - \(\frac{4}{12}\) + \(\frac32\) - \(\frac58\) - \(\frac{9}{13}\) + \(\frac32\) - \(\frac{7}{15}\)
B = (\(\frac58\) - \(\frac58\)) + (\(\frac32\) + \(\frac32\)) - (\(\frac13\) + \(\frac{9}{13}\) + \(\frac{7}{15}\))
B = 0 + 3 - (\(\frac{65}{195}\) + \(\frac{135}{195}\) + \(\frac{91}{195}\))
B = 3 - (\(\frac{200}{195}\) + \(\frac{91}{195}\))
B = 3 - \(\frac{97}{65}\)
B = \(\frac{195}{65}\) - \(\frac{97}{65}\)
B = \(\frac{98}{65}\)
a: \(\frac{4^{30}\cdot3^{43}}{2^{53}\cdot27^{15}}\)
\(=\frac{\left(2^2\right)^{30}\cdot3^{43}}{2^{53}\cdot\left(3^3\right)^{15}}\)
\(=\frac{2^{60}}{2^{53}}\cdot\frac{3^{43}}{3^{45}}=\frac{2^7}{3^2}=\frac{128}{9}\)
b: \(\left(-27\right)^{10}\cdot16^{25}\)
\(=-3^{30}\cdot\left(2^4\right)^{25}=-3^{30}\cdot2^{100}\)
\(6^{30}\cdot\left(-32\right)^{15}\)
\(=2^{30}\cdot3^{30}\cdot\left(-1\right)\cdot\left(2^5\right)^{15}=-2^{105}\cdot3^{30}\)
\(\frac{\left(-27\right)^{10}\cdot16^{25}}{6^{30}\cdot\left(-32\right)^{15}}\)
\(=\frac{-2^{100}\cdot3^{30}}{-2^{105}\cdot3^{30}}=\frac{1}{2^5}=\frac{1}{32}\)
c: \(\left(-\frac{2}{17}\right)^3\cdot\left(-\frac{17}{35}\right)^3\cdot\left(\frac{35}{3}\right)^3\)
\(=\left(-\frac{2}{17}\cdot\frac{-17}{35}\cdot\frac{35}{3}\right)^3\)
\(=\left(\frac23\right)^3=\frac{8}{27}\)
e: \(2^2+4^2+\cdots+24^2\)
\(=2^2\left(1^2+2^2+\cdots+12^2\right)\)
\(=4\cdot650=2600\)
\(a)=\frac{7}{25}+\frac{4}{13}-\frac{5}{2}+\frac{18}{25}-\frac{17}{13}\)
\(=1-1-\frac{5}{2}\)
\(=-\frac{5}{2}\)
29-x/21 + 27-x/23 + 25-x/25 + 23-x/27 + 21-x/29 = -5
1 + 29-x/21 + 1 + 27-x/23 + 1 + 25-x/25 + 1 + 23-x/27 + 1 + 21-x/29 = 0
50-x/21 + 50-x/23 + 50-x/25 + 50-x/27 + 50-x/29 = 0
(50-x) (1/21 + 1/23 + 1/25 + 1/27 + 1/29) = 0
Vì: 1/21 + 1/23 + 1/25 + 1/27 + 1/2 > 0
=> 50 - x = 0
x = 50
Vậy x = 50
\(\frac{-1}{3}+\frac{0,2-0,3+\frac{5}{11}}{-0,3+\frac{9}{16}-\frac{15}{12}}\)
\(=\frac{-1}{3}+\frac{\frac{2}{10}-\frac{3}{10}+\frac{5}{11}}{\frac{-3}{10}+\frac{9}{16}-\frac{15}{12}}\)
\(=\frac{-1}{3}+\frac{\frac{39}{110}}{\frac{-79}{80}}\)
\(=\frac{-1}{3}-\frac{312}{869}\)
\(=\frac{-1805}{2607}\)
\(A=\frac{81^4.3^{10}.27^5:3^{12}}{3^{18}:9^3.243^2}=\frac{3^{16}.3^{10}.3^{15}:3^{12}}{3^{18}:3^6.3^{10}}=\frac{3^{29}}{3^{22}}=3^7\)
\(B=\frac{2.55^2-9.55^{21}}{25^{10}}:\frac{5\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}=\frac{2.55^2-9.55^2.55^{19}}{25^{10}}:\frac{5\left(21.7^{14}-19.7^{14}\right)}{7.7^{15}+3.7^{15}}=\frac{55^2\left(55^{19}.9-2\right)}{25^{10}}:\frac{5.7^{14}.2}{7^{15}.10}=\frac{55^2\left(55^{19}.9-2\right)}{25^{10}}.\frac{7^{15}.10}{5.7^{14}.2}\)Chịu ==