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CÁCH 1 : A = \(\dfrac{235}{11}-\left(\dfrac{8}{5}+\dfrac{81}{11}\right)\)
A = \(\dfrac{235}{11}-\left(\dfrac{88}{55}+\dfrac{405}{55}\right)\)
A = \(\dfrac{235}{11}-\dfrac{493}{55}\)
A = \(\dfrac{1175}{55}+\dfrac{493}{55}\)
A = \(\dfrac{1668}{55}\)
mình gõ hơi khó nhìn nhưng bài giải này là chính xác 100%
a,3^10.(-5)^21/(-50)^20.3^12
=3^10.(-5)^20.(-5)/(-5)^20.3^10.3^2
=-5/3^2=-5/9
câu b, bạn tách ra làm tương tự
a)\(\dfrac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}=\dfrac{\left(-5\right)}{9}\)
b)\(\dfrac{5^{11}.7^{12}+5^{11}.7^{11}}{5^{12}.7^{12}+9.5^{11}.7^{11}}=\dfrac{7.35^{11}+35^{11}}{35^{12}+9.35^{11}}=\dfrac{8.35^{11}}{44.35^{11}}=\dfrac{2}{11}\)
a: \(=\dfrac{2\cdot136-28\cdot6+62\cdot3}{30}\cdot\dfrac{7}{8}=\dfrac{290}{30}\cdot\dfrac{7}{8}=\dfrac{29}{3}\cdot\dfrac{7}{8}=\dfrac{203}{24}\)
b: \(=\dfrac{3}{11}\cdot\dfrac{4}{11}+\dfrac{3}{13}\cdot\dfrac{4}{11}-\dfrac{1}{13}\)
\(=\dfrac{4}{11}\left(\dfrac{3}{11}+\dfrac{3}{13}\right)-\dfrac{1}{13}\)
\(=\dfrac{4}{11}\cdot\dfrac{72}{143}-\dfrac{1}{13}=\dfrac{167}{1573}\)
\(A=\dfrac{3^{10}\cdot11+3^{10}\cdot5}{3^9\cdot2^4}=\dfrac{3^{10}\cdot\left(11+5\right)}{3^9\cdot16}=\dfrac{3^{10}\cdot16}{3^9\cdot16}=3\)
\(B=\dfrac{2^{10}\cdot13+2^{10}\cdot65}{2^8\cdot104}=\dfrac{2^{10}\cdot\left(13+65\right)}{2^8\cdot2^2\cdot26}=\dfrac{2^{10}\cdot78}{2^{10}\cdot26}=3\)
\(C=\dfrac{72^3\cdot54^2}{108^4}=\dfrac{\left(2^3\cdot3^2\right)^3\cdot\left(2\cdot3^3\right)^2}{\left(3^3\cdot2^2\right)^4}\\ =\dfrac{2^9\cdot3^6\cdot2^4\cdot3^6}{3^{12}\cdot2^8}=\dfrac{2^{13}\cdot3^{12}}{3^{12}\cdot2^8}=2^5=32\)
\(D=\dfrac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=\dfrac{11\cdot3^{29}-\left(3^2\right)^{15}}{2^2\cdot3^{28}}=\dfrac{11\cdot3^{29}-3^{30}}{2^2\cdot3^{28}}\\ =\dfrac{3^{29}\cdot\left(11-3\right)}{2^2\cdot3^{28}}=\dfrac{3^{29}\cdot8}{4\cdot3^{28}}=3\cdot2=6\)
Giải:
a) \(A=\dfrac{5}{13}.\dfrac{5}{7}+\dfrac{-20}{41}+\dfrac{5}{13}+\dfrac{-21}{41}\)
\(\Leftrightarrow A=\dfrac{5}{13}.\dfrac{5}{7}+\dfrac{5}{13}+\dfrac{-21}{41}+\dfrac{-20}{41}\)
\(\Leftrightarrow A=\dfrac{5}{13}\left(\dfrac{5}{7}+1\right)+\dfrac{-41}{41}\)
\(\Leftrightarrow A=\dfrac{5}{13}.\dfrac{12}{7}+\left(-1\right)\)
\(\Leftrightarrow A=\dfrac{60}{91}+\left(-1\right)=-\dfrac{31}{91}\)
Vậy ...
b) \(B=\dfrac{5}{7}.\dfrac{2}{11}+\dfrac{5}{7}.\dfrac{12}{11}-\dfrac{5}{7}.\dfrac{7}{11}\)
\(\Leftrightarrow B=\dfrac{5}{7}\left(\dfrac{2}{11}+\dfrac{12}{11}-\dfrac{7}{11}\right)\)
\(\Leftrightarrow B=\dfrac{5}{7}.\dfrac{7}{11}\)
\(\Leftrightarrow B=\dfrac{5}{11}\)
Vậy ...
c) \(C=\dfrac{-2}{3}+\dfrac{-5}{7}+\dfrac{2}{3}+\dfrac{-2}{7}\)
\(\Leftrightarrow C=\left(\dfrac{-2}{3}+\dfrac{2}{3}\right)+\left(\dfrac{-2}{7}+\dfrac{-5}{7}\right)\)
\(\Leftrightarrow C=0+\left(-1\right)=-1\)
Vậy ...
a. \(\dfrac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}\) = \(\dfrac{3^{10}.\left(-5\right)^{20}.\left(-5\right)}{\left(-5\right)^{20}.3^{10}.3^2}\) = \(\dfrac{-5}{3^2}\)= \(\dfrac{-5}{9}\)
b. \(\dfrac{-11^5.13^7}{11^5.13^8}\) = \(\dfrac{-11^5.13^7}{11^5.13^7.13}\)= \(\dfrac{-1}{13}\)
c. \(\dfrac{2^{10}.3^{10}-2^{10}.3^9}{2^9.3^{10}}\)= \(\dfrac{2^{10}\left(3^{10}-3^9\right)}{2^9.3^{10}}\)= \(\dfrac{2^{10}.3}{2^9.3^{10}}\)= \(\dfrac{2^9.2.3}{2^9.3.3^9}\)= \(\dfrac{2}{3^9}\)=\(\dfrac{2}{19683}\)
Bài làm
A = 3¹° . 11 + 3¹° . 5 / 39 . 2⁴
A = 3¹° . ( 11 + 5 ) / 39 . 16
A = 3¹° . 16 / 39 . 16
A = 3¹° / 39
A = 3
Vậy A = 3
Mk lm bằng đt nên k vt đc phân số nha.
A=3^10.11+3^10.5/3^9.2^4
A=3^10.(11+5)/3^9.2^4
A=3^10.16/3^9.16
A=3^10/3^9
A=3
b, \(K =\) \(\dfrac{75}{100}+\dfrac{18}{21}+\dfrac{19}{32}+\dfrac{1}{4}+\dfrac{3}{21}+\dfrac{13}{32}\)
\(K = \) \(\dfrac{3}{4}+\dfrac{18}{21}+\dfrac{19}{32}+\dfrac{1}{4}+\dfrac{3}{21}+\dfrac{13}{32}\)
\(K = \) \(\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+\left(\dfrac{18}{21}+\dfrac{3}{21}\right)+\left(\dfrac{19}{32}+\dfrac{13}{32}\right)\)
\(K = \) \(1 + 1 + 1\)
\(K = \) \(3\)
a)\(\dfrac{2}{3}-\dfrac{3}{5}:\left(-1\dfrac{1}{5}\right)+\left(\dfrac{-2}{3}\right)\cdot\dfrac{3}{8}\)
\(=\dfrac{2}{3}-\dfrac{3}{5}\cdot\dfrac{-5}{6}+\left(\dfrac{-1}{4}\right)=\dfrac{5}{12}+\dfrac{1}{2}=\dfrac{11}{12}\)
b)\(17\dfrac{11}{9}-\left(6\dfrac{3}{13}+7\dfrac{11}{19}\right)+\left(10\dfrac{3}{13}-5\dfrac{1}{4}\right)=\dfrac{164}{9}-\left(\dfrac{81}{13}+\dfrac{144}{19}\right)+\left(\dfrac{133}{13}-\dfrac{21}{4}\right)=\dfrac{164}{9}-\dfrac{3411}{247}+\dfrac{259}{52}=\dfrac{6425}{684}\)
c)\(\left(\dfrac{-3}{2}\right)^2-\left[-2\dfrac{1}{3}-\left(\dfrac{3}{4}+\dfrac{1}{3}\right):2\dfrac{3}{5}\right]\cdot\left(\dfrac{-3}{4}\right)=\dfrac{9}{4}-\left[\dfrac{-7}{3}-\dfrac{13}{12}\cdot\dfrac{5}{13}\right]\cdot\left(\dfrac{-3}{4}\right)=\dfrac{9}{4}-\left(\dfrac{-11}{4}\right)\cdot\left(\dfrac{-3}{4}\right)=\dfrac{3}{16}\)
d)\(\dfrac{21}{33}:\dfrac{11}{5}-\dfrac{13}{33}:\dfrac{11}{5}+\dfrac{25}{33}:\dfrac{11}{5}+\dfrac{6}{11}=\dfrac{5}{11}\cdot\left(\dfrac{21}{33}-\dfrac{13}{33}+\dfrac{25}{33}\right)+\dfrac{6}{11}=\dfrac{5}{11}\cdot1+\dfrac{6}{11}=1\)
\(a)\dfrac{2}{3}-\dfrac{3}{5}:\left(-1\dfrac{1}{5}\right)+\left(\dfrac{-2}{3}\right).\dfrac{3}{8}\)
\(=\dfrac{2}{3}-\dfrac{3}{5}:\left(\dfrac{-6}{5}\right)+\left(\dfrac{-2}{3}\right).\dfrac{3}{8}\)
\(=\dfrac{2}{3}-\dfrac{-1}{2}+\left(\dfrac{-2}{3}\right).\dfrac{3}{8}\)
\(=\dfrac{2}{3}-\dfrac{-1}{2}+\dfrac{-1}{4}\)
\(=\dfrac{7}{6}+\dfrac{-1}{4}\)
\(=\dfrac{11}{12}\)
`A = (3^11 . 11 + 3^11 . 21)/(3^9 . 2^5)`
`= (3^11 . (11 + 21))/(3^9 . 2^5)`
`= (3^11 . 32)/(3^9 . 2^5)`
`= (3^2 . 3^9 . 2^5)/(3^9 . 2^5)`
`= 3^2 `
`= 9`
=396718580736