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\(A=\frac{81^4.3^{10}.27^5:3^{12}}{3^{18}:9^3.243^2}=\frac{3^{16}.3^{10}.3^{15}:3^{12}}{3^{18}:3^6.3^{10}}=\frac{3^{29}}{3^{22}}=3^7\)
\(B=\frac{2.55^2-9.55^{21}}{25^{10}}:\frac{5\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}=\frac{2.55^2-9.55^2.55^{19}}{25^{10}}:\frac{5\left(21.7^{14}-19.7^{14}\right)}{7.7^{15}+3.7^{15}}=\frac{55^2\left(55^{19}.9-2\right)}{25^{10}}:\frac{5.7^{14}.2}{7^{15}.10}=\frac{55^2\left(55^{19}.9-2\right)}{25^{10}}.\frac{7^{15}.10}{5.7^{14}.2}\)Chịu ==
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: \(=\dfrac{5^{42}\cdot7^{30}}{5^{40}\cdot7^{30}\cdot10}=\dfrac{5^2}{10}=\dfrac{25}{10}=\dfrac{5}{2}\)
b: \(=\dfrac{3^{18}\cdot5^{10}}{5^{10}\cdot3^{10}\cdot3^9}=\dfrac{1}{3}\)
A = \(\frac{1}{2}-\frac{3}{4}+\frac{5}{6}-\frac{7}{12}\)
A = \(\left(-\frac{1}{4}\right)+\frac{5}{6}-\frac{7}{12}\)
A = \(\frac{7}{12}-\frac{7}{12}\)
A = \(0\).
Mình làm câu A thôi nhé.
Chúc bạn học tốt!
=\(\frac{2^{30}3^{16}.5^{50}.7+2^{30}.3^{15}.3^2.11.5^{48}}{11.2^{25}.2^4.5^4.3^{18}.5^{18}.5^{30}-2^{30}.5^{30}.5^{22}.3^{18}}\)
=\(\frac{2^{30}.3^{16}.5^{50}.7+2^{30}.3^{17}.11.5^{48}}{11.2^{29}.5^{52}.3^{18}-2^{30}.5^{52}.3^{18}}\)
=\(\frac{2^{30}.3^{16}.5^{48}.11\left(5^2.7+3\right)}{2^{29}.5^{52}.3^{18}.11\left(1-2\right)}\)
=\(\frac{2.\left(5^2.7+3\right)}{5^4.3^2\left(1-2\right)}\)
=\(\frac{2.178}{5^4.3^2.\left(-1\right)}\)
=\(\frac{356}{-5625}\)
Hk tốt