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cứ phan tích cho hết đi là đc 9^6. 9^10 = (3^2)^6...................
tự làm đi
1. Tính:
a. \(\dfrac{9^6.9^{10}}{3^{32}}=\dfrac{\left(3^2\right)^6.\left(3^2\right)^{10}}{3^{32}}=\dfrac{3^{12}.3^{20}}{3^{32}}=\dfrac{3^{32}}{3^{32}}=1\)
b. \(\dfrac{25^8.25^{10}}{5^{34}}=\dfrac{\left(5^2\right)^8.\left(5^2\right)^{10}}{5^{34}}=\dfrac{5^{16}.5^{20}}{5^{34}}=\dfrac{5^{36}}{5^{34}}=5^{36}:5^{34}=5^2=25\)
c. \(\dfrac{7^{56}}{49^9.49^{20}}=\dfrac{7^{56}}{\left(7^2\right)^9.\left(7^2\right)^{20}}=\dfrac{7^{56}}{7^{18}.7^{40}}=\dfrac{7^{56}}{7^{58}}=7^{56}:7^{58}=\dfrac{7^{56}}{7^{56}.7^2}=\dfrac{1}{7^2}=\dfrac{1}{49}\)
d. \(\dfrac{4^2.4^3}{2^{10}}=\dfrac{\left(2^2\right)^2.\left(2^2\right)^3}{2^{10}}=\dfrac{2^4.3^6}{2^{10}}=\dfrac{2^{10}}{2^{10}}=1\)
e. \(\dfrac{2^{17}.25^5}{10^8.8^3}=\dfrac{2^{17}.\left(5^2\right)^5}{\left(2.5\right)^8.\left(2^3\right)^3}=\dfrac{2^{17}.5^{10}}{2^8.5^8.2^9}=\dfrac{2^{17}.5^{10}}{2^{17}.5^8}=\dfrac{5^{10}}{5^8}=5^{10}:5^8=5^2=25\)
f. \(\dfrac{3^{15}.25^4}{15^6.27^3}=\dfrac{3^{15}.\left(5^2\right)^4}{\left(3.5\right)^6.\left(3^3\right)^3}=\dfrac{3^{15}.5^8}{5^6.3^6.3^9}=\dfrac{3^{15}.5^8}{5^6.3^6.3^9}=\dfrac{5^8}{5^6}=5^8:5^6=5^2=25\)
2. Tính lũy thừa âm:
a. 3-2 = \(\dfrac{1}{3^2}\) = \(\dfrac{1}{9}\)
b. 2-3 = \(\dfrac{1}{2^3}\) = \(\dfrac{1}{8}\)
3. Tính :
a. \(\dfrac{\left(0,8\right)^4}{\left(0,4\right)^3}=\dfrac{\left(0,8\right)^3.0,8}{\left(0,4\right)^3}=\left(\dfrac{0,8}{0,4}\right)^3.0,8=2^3.0,8=8.0,8=6,4\)
b. \(\dfrac{\left(0,8\right)^3}{\left(0,4\right)^4}=\dfrac{\left(0,8\right)^3}{\left(0,4\right)^3:0,4}=\left(\dfrac{0,8}{0,4}\right)^3.\dfrac{1}{0,4}=2^3.2,5=8.2,5=20\)
c. \(\dfrac{\left(0,6\right)^5}{\left(0,2\right)^6}=\dfrac{\left(0,6\right)^5}{\left(0,2\right)^5.\left(0,2\right)}=\left(\dfrac{\left(0,6\right)}{\left(0,2\right)}\right)^5.\dfrac{1}{0,2}=3^5.\dfrac{1}{0,2}=\dfrac{3^5}{0,2}=1215\)
P/s : Chế Mai Ngọc Trâm thử tham khảo thử đi!!!
a: \(=\left(1+\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+\dfrac{1}{2}\)
\(=1+1+\dfrac{1}{2}=2+\dfrac{1}{2}=\dfrac{5}{2}\)
b: \(=\left(\dfrac{1}{25}+\dfrac{5}{25}+\dfrac{25}{25}\right):\left(\dfrac{1}{25}-\dfrac{5}{25}-\dfrac{25}{25}\right)\)
\(=\dfrac{31}{25}:\dfrac{-29}{25}=\dfrac{-31}{29}\)
c: \(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{\dfrac{3}{5}-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-\dfrac{4}{25}-\dfrac{4}{125}-\dfrac{4}{625}}\)
=1/4+3/4
=1
\(=\dfrac{\left(\dfrac{53}{4}-\dfrac{59}{27}-\dfrac{65}{6}\right)\cdot230.04+46.75}{\left(\dfrac{13}{10}+\dfrac{10}{3}\right):\left(\dfrac{37}{3}-\dfrac{100}{7}\right)}\)
\(=\dfrac{53.25+46.75}{\dfrac{139}{30}:\dfrac{-41}{21}}=100:\left(\dfrac{139}{30}\cdot\dfrac{-21}{41}\right)\)
\(=100\cdot\dfrac{-400}{973}=\dfrac{-40000}{973}\)
a) \(\dfrac{3}{7}x-\dfrac{2}{3}x=\dfrac{10}{21}\)
<=> \(x\left(\dfrac{3}{7}-\dfrac{2}{3}\right)=\dfrac{10}{21}\)
<=> \(x.\left(-\dfrac{5}{21}\right)=\dfrac{10}{21}\)
<=> \(x=-2\)
b) \(\dfrac{7}{35}:\left(x-\dfrac{1}{3}\right)=-\dfrac{2}{25}\)
<=> \(x-\dfrac{1}{3}=-\dfrac{5}{2}\)
<=> \(x=-\dfrac{13}{6}\)
a, \(\dfrac{3}{7}x-\dfrac{2}{3}x=\dfrac{10}{21}\)
\(\Leftrightarrow x\left(\dfrac{3}{7}-\dfrac{2}{3}\right)=\dfrac{10}{21}\)
\(\Leftrightarrow x.\dfrac{-5}{21}=\dfrac{10}{21}\)
\(\Leftrightarrow x=-2\)
Vậy ...
b, \(\dfrac{7}{35}:\left(x-\dfrac{1}{3}\right)=\dfrac{-2}{25}\)
\(\Leftrightarrow x-\dfrac{1}{3}=\dfrac{-5}{2}\)
\(\Leftrightarrow x=\dfrac{-13}{6}\)
Vậy ...
1: \(A=\dfrac{-25}{27}-\dfrac{31}{42}+\dfrac{7}{27}+\dfrac{3}{42}=\dfrac{-2}{3}-\dfrac{2}{3}=\dfrac{-4}{3}\)
2: \(B=\dfrac{10.3-\left(9.5-4.5\right)\cdot2}{1.2-1.5}=\dfrac{10.3-10}{-0.3}=-1\)
c: \(=\dfrac{3}{49}\left(\dfrac{19}{2}-\dfrac{5}{2}\right)-\left(\dfrac{1}{20}-\dfrac{5}{20}\right)^2\cdot\left(\dfrac{-7}{14}-\dfrac{193}{14}\right)\)
\(=\dfrac{3}{49}\cdot7-\dfrac{1}{25}\cdot\dfrac{-200}{14}\)
\(=\dfrac{3}{7}+\dfrac{8}{14}=1\)
\(A=\sqrt{\dfrac{11}{25}+\dfrac{25}{25}}-2\sqrt{5}\cdot\left(\dfrac{1}{20}\sqrt{5}-\dfrac{1}{30}\sqrt{10}\right)+\dfrac{1}{6}\)
\(=\dfrac{6}{5}-\dfrac{1}{2}+\dfrac{1}{15}\sqrt{50}+\dfrac{1}{6}\)
\(=\dfrac{13}{15}+\dfrac{1}{3}\sqrt{2}\)
\(E=\dfrac{\left(13\dfrac{1}{4}-2\dfrac{5}{27}-10\dfrac{5}{6}\right).230\dfrac{1}{25}+46\dfrac{3}{4}}{\left(1\dfrac{3}{7}+\dfrac{10}{3}\right):\left(12\dfrac{1}{3}-14\dfrac{2}{7}\right)}\)
\(E=\dfrac{\left(\dfrac{53}{4}-\dfrac{59}{27}-\dfrac{65}{6}\right).\dfrac{5751}{25}+\dfrac{187}{4}}{\dfrac{100}{21}:\left(-\dfrac{41}{21}\right)}\)
\(E=\dfrac{\dfrac{25}{108}.\dfrac{5751}{25}+\dfrac{187}{4}}{-\dfrac{100}{41}}\)
\(E=\dfrac{\dfrac{213}{4}+\dfrac{187}{4}}{-\dfrac{100}{41}}\)
\(E=\dfrac{100}{-\dfrac{100}{41}}\)
\(E=-41\)
\(A=\dfrac{25^{10}+1}{25^{10}-1}=\dfrac{25^{10}-1+2}{25^{10}-1}=\dfrac{25^{10}-1}{25^{10}-1}+\dfrac{2}{25^{10}-1}=1+\dfrac{2}{25^{10}-1}\left(1\right)\)
\(B=\dfrac{25^{10}-1}{25^{10}-3}=\dfrac{25^{10}-3+2}{25^{10}-3}=\dfrac{25^{10}-3}{25^{10}-3}+\dfrac{2}{25^{10}-3}=1+\dfrac{2}{25^{10}-3}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow A< B\)
Ta có : \(\dfrac{25^{10}+1}{25^{10}-1}=\dfrac{25^{10}-1+2}{25^{10}-1}=\dfrac{25^{10}-1}{25^{10}-1}+\dfrac{2}{25^{10}-1}\)\(=1+\dfrac{2}{25^{10}-1}\)
Ta có : \(\dfrac{25^{10}-1}{25^{10}-3}=\dfrac{25^{10}-3+2}{25^{10}-3}=\dfrac{25^{10}-3}{25^{10}-3}+\dfrac{2}{25^{10}-3}=1+\dfrac{2}{25^{10}-3}\)
Vì \(25^{10}-1>25^{10}-3\)
\(\Rightarrow\dfrac{2}{25^{10}-1}< \dfrac{2}{25^{10}-3}\)
\(\Rightarrow1+\dfrac{2}{25^{10}-1}< 1+\dfrac{2}{25^{10}-3}\)
\(\Rightarrow A< B\)
Vậy \(A< B\)