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a,
\(\dfrac{4^2\cdot4^3}{2^{10}}=\dfrac{4^5}{2^{10}}=\dfrac{\left(2^2\right)^5}{2^{10}}=\dfrac{2^{10}}{2^{10}}=1\)
b,
\(\dfrac{\left(0,6\right)^5}{\left(0,2\right)^6}=\dfrac{\left(0,2\cdot3\right)^5}{\left(0,2\right)^5\cdot0,2}=\dfrac{\left(0,2\right)^5\cdot3^5}{\left(0,2\right)^5\cdot0,2}=\dfrac{243}{0,2}=\dfrac{243}{\dfrac{1}{5}}=243\cdot5=1215\)
c,
\(\dfrac{2^7\cdot9^3}{6^5\cdot8^2}=\dfrac{2^7\cdot\left(3^2\right)^3}{\left(2\cdot3\right)^5\cdot\left(2^3\right)^2}=\dfrac{2^6\cdot2\cdot3^6}{2^5\cdot3^5\cdot2^6}=\dfrac{3}{2^4}=\dfrac{3}{16}\)
d,
\(\dfrac{6^3+3\cdot6^2+3^3}{-13}=\dfrac{\left(2\cdot3\right)^3+3\cdot\left(2\cdot3\right)^2+3^3}{-13}=\dfrac{2^3\cdot3^3+3\cdot2^2\cdot3^2+3^3}{-13}=\dfrac{2^3\cdot3^3+2^2\cdot3^3+3^3}{-13}\dfrac{3^3\left(2^3+2^2+1\right)}{-13}=\dfrac{3^3\cdot13}{-13}=-3^3=-27\)
\(A=\dfrac{12^{15}\cdot3^4-4^5\cdot3^9}{27^3\cdot2^{10}-32^3\cdot3^9}\\ =\dfrac{\left(2^2\cdot3\right)^{15}\cdot3^4-\left(2^2\right)^5\cdot3^9}{\left(3^3\right)^3\cdot2^{10}-\left(2^5\right)^3\cdot3^9}\\ =\dfrac{2^{30}\cdot3^{15}\cdot3^4-2^{10}\cdot3^9}{3^9\cdot2^{10}-2^{15}\cdot3^9}\\ =\dfrac{3^9\cdot2^{10}\left(2^{20}\cdot3^{10}\right)}{3^9\cdot2^{10}\left(1-2^5\right)}\\ =\dfrac{\left(2^2\right)^{10}\cdot3^{10}}{1-32}\\ =\dfrac{\left(2^2\cdot3\right)^{10}}{-31}\\ =\dfrac{-12^{10}}{31}\)
\(B=\dfrac{3}{1^2\cdot2^2}+\dfrac{5}{2^2\cdot3^2}+...+\dfrac{99}{49^2\cdot50^2}\\ =\dfrac{2^2-1^2}{1^2\cdot2^2}+\dfrac{3^2-2^2}{2^2\cdot3^2}+...+\dfrac{50^2-49^2}{49^2\cdot50^2}\\ =\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{49^2}-\dfrac{1}{50^2}\\ =1-\dfrac{1}{2500}\\ =\dfrac{2499}{2500}\)
a) \(\dfrac{4^2.4^3}{(2^2)^5}=\dfrac{4^2.4^3}{4^5}=\dfrac{4^3}{4^3}=1\)
b) = 1215
c) = \(\dfrac{3}{16}\)
d) = (-27)
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: \(=6-\dfrac{2}{3}+\dfrac{1}{2}-5-\dfrac{5}{3}+\dfrac{3}{2}-3+\dfrac{7}{3}-\dfrac{5}{2}\)
\(=\left(6-5-3\right)+\left(-\dfrac{2}{3}-\dfrac{5}{3}+\dfrac{7}{3}\right)+\left(\dfrac{1}{2}+\dfrac{3}{2}-\dfrac{5}{2}\right)\)
\(=-2-\dfrac{1}{2}=-\dfrac{5}{2}\)
b: \(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}=\dfrac{2^{10}\cdot3^8\cdot\left(-2\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=-\dfrac{1}{3}\)
a) \(\dfrac{4^2.4^3}{2^{10}}\)
Hướng dẫn:
- Đưa các lũy thừa trên tử số về cơ số có dạng giống mẫu số
\(=\dfrac{\left(2^2\right)^2.\left(2^2\right)^3}{2^{10}}\)
- Dùng tính chất \(\left(a^n\right)^m=a^{n.m}\) để làm
\(=\dfrac{2^4.2^6}{2^{10}}\)
- Gộp các lũy thừa cùng cơ số lại, dùng tính chất \(a^m.a^n=a^{m+n}\)
\(=\dfrac{2^{10}}{2^{10}}\)
- Chia tử và mẫu cho nhau, dùng tính chất \(a^m:a^n=a^{m-n}\)
\(=1\)
b) \(\dfrac{2^7.9^3}{6^5.8^2}\)
Hướng dẫn:
- Đưa lũy thừa ở tử và mẫu về cơ số nhỏ nhất ( Đưa về cơ số 2 và 3 )
\(=\dfrac{2^7.\left(3^2\right)^3}{\left(2.3\right)^5.\left(2^3\right)^2}\)
- Dùng tính chất \(\left(a^m\right)^n=a^{m.n}\) và \(\left(a.b\right)^m=a^m.b^m\)
\(=\dfrac{2^7.3^6}{2^5.3^5.2^6}\)
- Dùng tính chất \(a^m.a^n=a^{m+n}\) để gộp các lũy thừa có cùng cơ số
\(=\dfrac{2^7.3^6}{2^{11}.3^5}\)
- Chia tử và mẫu cho nhau theo cách rút gọn những số giống nhau ở trên tử và mẫu
\(=\dfrac{3}{2^4}\)
\(=\dfrac{3}{16}\)
c) \(\dfrac{5^4.20^4}{25^5.4^5}\)
Hướng dẫn:
- Đưa các lũy thừa của tử và mẫu về cơ số nhỏ nhất ( Cơ số 2 và 5 )
\(=\dfrac{5^4.\left(2^2.5\right)^4}{\left(5^2\right)^5.\left(2^2\right)^5}\)
- Dùng tính chất \(\left(a^m\right)^n=a^{m.n}\) và \(\left(a.b\right)^m=a^m.b^m\)
\(=\dfrac{5^4.\left(2^2\right)^4.5^4}{5^{10}.2^{10}}\)
- Dùng tính chất \(a^m.a^n=a^{m+n}\)
\(=\dfrac{5^8.2^8}{5^{10}.2^{10}}\)
- Rút gọn
\(=\dfrac{1}{5^2.2^2}\)
\(=\dfrac{1}{100}\)
\(\dfrac{6^2.6^3}{3^5}=\dfrac{3^2.2^2.3^3.2^3}{3^5}=\dfrac{2^5.3^5}{3^5}=2^5=32.\)
\(\dfrac{25^2.4^2}{5^5.\left(-2\right)^5}=\dfrac{5^4.2^4}{\left(-10\right)^5}=\dfrac{10^4}{10^4.\left(-10\right)}=\dfrac{-1}{10}.\)
\(\dfrac{6^2.6^3}{3^5}=\dfrac{6^{2+3}}{3^5}=\dfrac{6^5}{3^5}=2^5=32\)
\(\dfrac{25^2.4^2}{5^5.\left(-2\right)^5}=\dfrac{\left(5^2\right)^2.\left(2^2\right)^2}{\left[5.\left(-2\right)\right]^5}=\dfrac{5^4.2^4}{-10^5}=\dfrac{10^4}{-10^5}=\dfrac{-1}{10}\)
\(\dfrac{0,125^5.2,4^5}{-0,3^5.0,001^3}=\dfrac{\left(0,125.2,4\right)^5}{-0,3^5.0,001^3}=\dfrac{0,3^5}{-0,3^5.0,001^3}=-\dfrac{1}{0,001^3}=-1000000000\)
\(\left(2\dfrac{3}{4}+\dfrac{1}{2}\right)=\dfrac{11}{4}+\dfrac{1}{2}=\dfrac{11}{4}+\dfrac{2}{4}=\dfrac{13}{4}\)
Ta có:
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2+10^2}\)
\(=\left(\dfrac{1}{1^2}-\dfrac{1}{2^2}\right)+\left(\dfrac{1}{2^2}-\dfrac{1}{3^2}\right)+...+\left(\dfrac{1}{9^2}-\dfrac{1}{10^2}\right)\)
\(=\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)
\(=\dfrac{1}{1^2}-\dfrac{1}{10^2}\)
\(=1-\dfrac{1}{100}\)
Vì \(1-\dfrac{1}{100}< 1\)
Nên \(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2+10^2}< 1\) (Đpcm)
\(vt:\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+...+\dfrac{19}{9^2+10^2}\)
=\(\dfrac{1}{1}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+..+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)
=\(\dfrac{1}{1}-\dfrac{1}{10^2}\)
=>A<1
a) \(\left(0,25\right)^3\cdot32=0,015625\cdot32=0,5\)
b) \(\left(-0,125\right)^3\cdot80^4=\dfrac{-1}{512}\cdot40960000=80000\)
c) \(\dfrac{8^2\cdot4^5}{2^{20}}=\dfrac{2^{3^2}\cdot2^{2^5}}{2^{20}}=\dfrac{2^6\cdot2^{10}}{2^{20}}=\dfrac{2^{16}}{2^{20}}=\dfrac{1}{2^4}=\dfrac{1}{16}\)
d) \(\dfrac{81^{11}\cdot3^{17}}{27^{10}\cdot9^{15}}=\dfrac{3^{4^{11}}\cdot3^{17}}{3^{3^{10}}\cdot3^{2^{15}}}=\dfrac{3^{44}\cdot3^{17}}{3^{30}\cdot3^{30}}=\dfrac{3^{61}}{3^{60}}=3\)
\(\dfrac{4^2\cdot4^3}{2^{10}}=\dfrac{\left(2^2\right)^2\cdot\left(2^2\right)^3}{2^{10}}=\dfrac{2^4\cdot2^6}{2^{10}}=1\)
\(=\dfrac{4^5}{2^{10}}\) \(=\dfrac{\left(2^2\right)^5}{2^{10}}\) \(=\dfrac{2^{10}}{2^{10}}=1\)