\(\dfrac{2^{14}.3^...">
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23 tháng 9 2021

a) \(\dfrac{2^{14}.3^{12}}{6^{11}}\)

\(=\dfrac{2^2.2^{12}.3^{12}}{6^{11}}\)

\(=\dfrac{4.6^{12}}{6^{11}}\)

\(=4.6\)

\(=24\)

23 tháng 9 2021

a) \(\dfrac{2^{14}.3^{12}}{6^{11}}=\dfrac{2^{14}.3^{12}}{2^{11}.3^{11}}=2^3.3\)

b) \(\dfrac{6^{18}}{9^9.8^5}=\dfrac{\left(2.3\right)^{18}}{\left(3^2\right)^8.\left(2^3\right)^5}\dfrac{2^{18}.3^{18}}{3^{18}.2^{15}}=2^3\)

17 tháng 8 2017

a) (0,25)^7 . 4^7 => (0,25.4)^7 = 1^7

b) 27^5 : 9^6 => (3^3)^5 : (3^2)^6 = 3^15 : 3^12 =3^3

c) (1/5)^11 . 5^10 => 5^-11 . 5^10 = 0.2 = 1/5 = 5^-1

chúc bạn hc tốt banhqua

a: \(=\dfrac{3^3\cdot2^6}{3^{-4}\cdot2^6}=3^7\)

b: \(=\left(\dfrac{3}{7}\cdot\dfrac{5}{3}\right)^6\cdot\dfrac{5}{3}\cdot\dfrac{3}{7}:\left(\dfrac{7^3}{5^4}\right)^{-2}\)

\(=\left(\dfrac{5}{7}\right)^6\cdot\dfrac{5}{7}\cdot\left(\dfrac{5}{7}\right)^6\cdot5^2\)

\(=\left(\dfrac{5}{7}\right)^{13}\cdot5^2\)

c: \(=5^4\cdot2.5^{-5}\cdot125\cdot0.04\)

\(=5^4\cdot5\cdot\left(\dfrac{5}{2}\right)^{-5}\)

\(=5^5\cdot\left(\dfrac{2}{5}\right)^5=2^5\)

a: \(=\dfrac{3^3\cdot2^6}{3^{-4}\cdot2^6}=3^7\)

b: \(=\left(\dfrac{3}{7}\right)^5\cdot\left(\dfrac{3}{7}\right)\cdot\dfrac{5^6}{3^6}:\left(\dfrac{625}{343}\right)^2\)

\(=\dfrac{3^6}{7^6}\cdot\dfrac{5^6}{3^6}:\dfrac{5^8}{7^6}\)

\(=\dfrac{1}{5^2}\)

c: \(=5^{4+3}\cdot\left(\dfrac{5}{2}\right)^{-5}\cdot\dfrac{1}{25}\)

\(=5^5\cdot\left(\dfrac{2}{5}\right)^5=2^5\)

22 tháng 7 2017

1. Tính:

a. \(\dfrac{\text{−1 }}{\text{4 }}+\dfrac{\text{5 }}{\text{6 }}=\dfrac{-3}{12}+\dfrac{10}{12}=\dfrac{7}{12}\)

b. \(\dfrac{\text{5 }}{\text{12 }}+\dfrac{\text{-7 }}{8}=\dfrac{10}{24}+\dfrac{-21}{24}=\dfrac{-11}{24}\)

c. \(\dfrac{-7}{6}+\dfrac{-3}{10}=\dfrac{-35}{30}+\dfrac{-9}{30}=\dfrac{-44}{30}=\dfrac{-22}{15}\)

d.\(\dfrac{-3}{7}+\dfrac{5}{6}=\dfrac{-18}{42}+\dfrac{35}{42}=\dfrac{17}{42}\)

2. Tính :

a. \(\dfrac{2}{14}-\dfrac{5}{2}=\dfrac{2}{14}-\dfrac{35}{14}=\dfrac{-33}{14}\)

b.\(\dfrac{-13}{12}-\dfrac{5}{18}=\dfrac{-39}{36}-\dfrac{10}{36}=\dfrac{49}{36}\)

c.\(\dfrac{-2}{5}-\dfrac{-3}{11}=\dfrac{-2}{5}+\dfrac{3}{11}=\dfrac{-22}{55}+\dfrac{15}{55}=\dfrac{-7}{55}\)

d. \(0,6--1\dfrac{2}{3}=\dfrac{6}{10}--\dfrac{5}{3}=\dfrac{3}{5}+\dfrac{5}{3}=\dfrac{9}{15}+\dfrac{25}{15}=\dfrac{34}{15}\)

3. Tính :

a.\(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-4}{156}+\dfrac{-3}{156}=\dfrac{-7}{156}\)

b.\(\dfrac{-6}{9}-\dfrac{12}{16}=\dfrac{2}{3}-\dfrac{3}{4}=\dfrac{8}{12}-\dfrac{9}{12}=\dfrac{-17}{12}\)

c. \(\dfrac{-3}{7}-\dfrac{-2}{11}=\dfrac{-3}{7}+\dfrac{2}{11}=\dfrac{-33}{77}+\dfrac{14}{77}=\dfrac{-19}{77}\)

d.\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...\dfrac{1}{8.9}+\dfrac{1}{9.10}\)

\(=\dfrac{1}{1}+\dfrac{1}{10}\)

\(=\dfrac{10}{10}-\dfrac{1}{10}\)

= \(\dfrac{9}{10}\)

Chế Kazuto Kirikaya thử tham khảo thử đi !!!

23 tháng 7 2017

Mấy câu trên kia dễ rồi mình chữa mình câu \(c\) bài \(3\) thôi nhé Kazuto Kirikaya

d) \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=1-\dfrac{1}{10}\)

\(=\dfrac{9}{10}\)

13 tháng 9 2021

\(\frac{8^{11}.3^{17}}{27^{10}.9^{15}}=\frac{8^{11}.3^{17}}{3^{30}.3^{30}}=\frac{8^{11}}{3^{13}.3^{30}}=\frac{8^{11}}{3^{43}}\)

\(\frac{\left(5^4-5^3\right)^3}{125^4}=\frac{[\left(5-1\right).5^3]^3}{5^{12}}=\frac{\left(4.5^3\right)^3}{5^{12}}=\frac{64.5^9}{5^{12}}=\frac{64}{5^3}=\left(\frac{4}{5}\right)^3\)

\(\frac{4^{20}-2^{20}+6^{20}}{6^{20}-3^{20}+9^{20}}=\frac{2^{40}-2^{20}+6^{20}}{6^{20}-3^{20}+3^{40}}=\frac{2^{20}.\left(2^{20}-1+3^{30}\right)}{3^{20}.\left(2^{20}-2+3^{20}\right)}=\frac{2^{20}}{3^{20}}=\left(\frac{2}{3}\right)^{20}\)

16 tháng 9 2018

làm bừa thui,ai tích mình mình tích lại

Số số hạng là : 

Có số cặp là :

50 : 2 = 25 ( cặp )

Mỗi cặp có giá trị là :

99 - 97 = 2 

Tổng dãy trên là :

25 x 2 = 50

Đáp số : 50

16 tháng 9 2018

a)\(12^3.3^{-4}.64\)

\(=3^3.2^6.3^{-3}.2^6=2^{12}\)

b) \(\left(\frac{3}{7}\right)^5.\left(\frac{7}{3}\right)^{-1}.\left(\frac{5}{3}\right)^6:\left(\frac{343}{625}\right)^2\)

\(=\frac{3^5.7^{-1}}{7^5.3^{-1}}.\left(\frac{5}{3}\right)^6:\frac{7^6}{5^8}\)

\(=\frac{3^6}{7^6.}.\frac{5^6}{3^6}.\frac{5^8}{7^6}\)

\(=\frac{5^{14}}{7^{12}}\)

16 tháng 9 2018

1.

a) x : \(\left(\dfrac{3}{4}\right)^3\) =\(\left(\dfrac{3}{4}\right)^3\)

x = \(\left(\dfrac{3}{4}\right)^3.\left(\dfrac{3}{4}\right)^3\)

x = \(\dfrac{3}{4}^{3+3}\)

x = \(\dfrac{3}{4}^6\)

x = \(\dfrac{729}{4096}\)

b) \(\left(\dfrac{2}{5}\right)^5.x=\left(\dfrac{2}{5}\right)^8\)

x = \(\left(\dfrac{2}{5}\right)^8:\left(\dfrac{2}{5}\right)^5\)

x = \(\dfrac{2}{5}^{8-5}\)

x = \(\dfrac{2}{5}^3\)

x = \(\dfrac{8}{5}\)

2.

(0,36)\(^8\) \([\left(0,6\right)^3]^8\) = (0,6)\(^{3.8}\) = ( 0,6)\(^{24}\)

( 0,216)\(^4\) = \([\left(0,6\right)^3]^4\) = (0.6)\(^{3.4}\) = ( 0,6)\(^{12}\)

18 tháng 9 2018

\(x:\left(\dfrac{3}{4}\right)^3=\left(\dfrac{3}{4}\right)^2\)

\(x=\left(\dfrac{3}{4}\right)^2.\left(\dfrac{3}{4}\right)^3\) <=> \(x=\left(\dfrac{3}{4}\right)^{2+3}\)

=> \(x=\left(\dfrac{3}{4}\right)^5\)

b, \(\left(\dfrac{2}{5}\right)^5.x=\left(\dfrac{2}{5}\right)^8\)

\(x=\left(\dfrac{2}{5}\right)^8:\left(\dfrac{2}{5}\right)^5\Leftrightarrow x=\left(\dfrac{2}{5}\right)^{8-5}\)

=>\(x=\left(\dfrac{2}{5}\right)^3\)

bài 2 : Với bài này ta cần áp dụng quy tắc: \(\left(x^m\right)^n=x^{m.n}\)

\(0,36^8=\left[\left(0,6\right)^2\right]^8=\left(0,6\right)^{16}\)

\(0,216^4=\left[\left(0,6\right)^3\right]^4=\left(0,6\right)^{12}\)

16 tháng 9 2017

a, \(\left(4.2\right)^5:\left(2^3.\dfrac{1}{16}\right)=8^5:\left(2^3.\dfrac{1^4}{2^4}\right)=\left(2^3\right)^5:\dfrac{2^3.1^4}{2^4}=2^{15}:\dfrac{1}{2}=2^{15}.2=2^{16}\)

\(b,\dfrac{2^2.4.32}{2^2.2^5}=\dfrac{2^2.2^4.2^5}{2^2.2^5}=2^4=16\)

16 tháng 9 2017

\(a,\dfrac{\left(4.2\right)^5}{2^3.\dfrac{1}{16}}=\dfrac{\left(2^3\right)^5}{2^3.2^{-4}}=\dfrac{2^{15}}{2^{-1}}=2^{16}\)

b,\(\dfrac{2^2.4.32}{2^2.2^5}=\dfrac{2^2.2^2.2^5}{2^2.2^5}=2^2=4\)

1 tháng 8 2017

làm bài 3 BĐT

theo bảng xét dấu

còn bài 1,2 ở trên là 1.1 và 1.2 đều trg bài 1.2

bài 1.2 (tức bài 2 ở trên )làm a,b,c,d

\còn bài 2( tức bài 2 ở trên) làm hết

1 tháng 8 2017

thanks