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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 = \(\frac{98}{65}\)
\(\frac{9^{15}.\left(-6\right)^{30}}{27^{21}.8^{11}}=\frac{\left(3^2\right)^{15}.\left(-6\right)^{30}}{\left(3^3\right)^{21}.8^{11}}=\frac{3^{30}.\left(-6\right)^{30}}{3^{63}.8^{11}}\)
Sau đó bạn tách ra rồi rút gọn là xong
Chúc bạn học tốt!
a)\(\frac{3^6.45^4-15^{13}.5^{-9}}{27^4.25^3+45^6}=\frac{3^6.\left(3^2.5\right)^4-\left(3.5\right)^{13}.5^{-9}}{\left(3^3\right)^4.\left(5^2\right)^3+\left(3^2.5\right)^6}=\frac{3^6.3^8.5^4-3^{13}.5^{13}.5^{-9}}{3^{12}.5^6+3^{12}.5^6}\)
\(=\frac{3^{14}.5^4-3^{13}.5^4}{3^{12}.5^6+3^{12}.5^6}=\frac{3^{13}.5^4.\left(3-1\right)}{3^{12}.5^6\left(1+1\right)}=\frac{3^{13}.5^4}{3^{12}.5^6}=\frac{3}{5^2}=\frac{3}{25}\)
b)\(\frac{4^6.9^5+6^9.120}{-8^4.3^{12}-6^{11}}=\frac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.2^3.3.5}{-\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}=\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{-2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{-2^{12}.3^{12}-2^{11}.3^{11}}=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{-\left(2^{12}.3^{12}+2^{11}.3^{11}\right)}=\frac{2^{12}.3^{10}\left(1+5\right)}{-\left[2^{11}.3^{11}\left(2.3+1\right)\right]}=\frac{2.6}{-\left(3.7\right)}=\frac{4}{-7}\)
\(a,7^6+7^5-7^4⋮55\)
\(7^4\left(7^2+7-1\right)⋮55\)
\(7^4\times55⋮55\left(dpcm\right)\)
\(8^{12}-2^{33}-2^{30}\)
\(=8^{12}-\left(2^3\right)^{11}-\left(2^3\right)^{10}\)
\(=8^{12}-8^{11}-8^{10}\)
\(=8^{10}\left(8^2-8-1\right)\)
\(=8^{10}\times55⋮55\left(dpcm\right)\)
\(=\frac{9^{15}.6^{30}}{27^{21}.8^{11}}\)
\(=\frac{\left(3^2\right)^{15}.3^{30}.2^{30}}{\left(3^3\right)^{21}.\left(2^3\right)^{11}}\)
\(=\frac{3^{30}.3^{30}.2^{30}}{3^{63}.2^{33}}\)
\(=\frac{3^{60}.2^{30}}{2^{63}.3^{33}}=\frac{1}{2^3.3^3}=\frac{1}{216}\)
làm mẫu một bài còn lại tương tự nha bn =)
\(\frac{9^{15}.6^{30}}{27^{21}.8^{11}}=\frac{\left(3^2\right)^{15}.\left(2.3\right)^{30}}{\left(3^3\right)^{21}.\left(2^3\right)^{11}}=\frac{3^{60}.2^{30}}{3^{63}.2^{33}}=\frac{1}{3^3.2^3}\)
\(\frac{45^{12}.49^7}{35^{13}.27^8}=\frac{\left(5.3^2\right)^{12}.\left(7^2\right)^7}{\left(5.7\right)^{13}.\left(3^3\right)^8}=\frac{5^{12}.3^{24}.7^{14}}{5^{13}.7^{13}.3^{24}}=\frac{7}{5}\)
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