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3 tháng 10 2018

\(\dfrac{\left(5\right)^4.\left(20\right)^4}{\left(24\right)^4.45}\)

\(=\dfrac{5^4.5^4.\left(2^2\right)^4}{\left(2^3\right)^4.3^4.3^2.5}\)

\(=\dfrac{5^8.2^8}{2^{12}.3^6.5}=\dfrac{5^7}{2^4.3^6}\)

Câu 1:

\(=\dfrac{15}{34}+\dfrac{19}{34}-1-\dfrac{15}{17}+\dfrac{1}{3}+\dfrac{3}{5}\)

\(=-\dfrac{15}{17}+\dfrac{14}{15}=\dfrac{13}{255}\)

Câu 2: 

\(=\dfrac{5^4\cdot5^4\cdot2^8}{4^4\cdot6^4\cdot3^2\cdot5}=\dfrac{5^7}{6^4\cdot3^2}\)

2 tháng 9 2023

Vì \(\dfrac{1}{11}>\dfrac{1}{18}>\dfrac{1}{21}>\dfrac{1}{24}>\dfrac{1}{27}>\dfrac{1}{29}\)

\(\Rightarrow\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}>\dfrac{1}{4}+\dfrac{1}{11}+\dfrac{1}{18}+\dfrac{1}{21}+\dfrac{1}{24}+\dfrac{1}{27}+\dfrac{1}{29}\)\(\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}=\dfrac{1}{11}.7=\dfrac{7}{11}\)

Ta có:

\(\dfrac{7}{11}=\dfrac{7.5}{11.5}=\dfrac{35}{55};\dfrac{4}{5}=\dfrac{4.11}{5.11}=\dfrac{44}{55}\)

\(Vì\) \(\dfrac{44}{55}>\dfrac{35}{55}\)

\(\Rightarrow\dfrac{4}{5}>\dfrac{7}{11}\)

\(\Rightarrow\dfrac{1}{4}+\dfrac{1}{11}+\dfrac{1}{18}+\dfrac{1}{21}+\dfrac{1}{24}+\dfrac{1}{27}+\dfrac{1}{29}< \dfrac{4}{5}\left(đpcm\right)\)

2 tháng 9 2023

Ta thấy :

\(\dfrac{1}{4}+\dfrac{1}{11}< \dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}=1-\dfrac{1}{2}\)

\(\dfrac{1}{18}+\dfrac{1}{21}< \dfrac{1}{12}+\dfrac{1}{12}=\dfrac{1}{6}=\dfrac{1}{2}-\dfrac{1}{3}\)

\(\dfrac{1}{24}+\dfrac{1}{27}< \dfrac{1}{24}+\dfrac{1}{24}=\dfrac{1}{12}=\dfrac{1}{3}-\dfrac{1}{4}\)

\(\dfrac{1}{29}< \dfrac{1}{20}=\dfrac{1}{4}-\dfrac{1}{5}\)

\(\Rightarrow\dfrac{1}{4}+\dfrac{1}{11}+\dfrac{1}{18}+\dfrac{1}{21}+\dfrac{1}{24}+\dfrac{1}{27}+\dfrac{1}{29}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}\)

\(\Rightarrow\dfrac{1}{4}+\dfrac{1}{11}+\dfrac{1}{18}+\dfrac{1}{21}+\dfrac{1}{24}+\dfrac{1}{27}+\dfrac{1}{29}< 1-\dfrac{1}{5}=\dfrac{4}{5}\)

\(\Rightarrow dpcm\)

1 tháng 4 2017

Câu hỏi của Lê Khánh Nhi - Toán lớp 7 - Học toán với OnlineMath sửa n thành x cho sửa cho nó thành lũy thừa luôn

4 tháng 1 2022

\(\dfrac{45^{10}\cdot5^{20}}{75^{15}}=\dfrac{\left(3^2\cdot5\right)^{10}\cdot5^{20}}{\left(3\cdot5^2\right)^{15}}=\dfrac{3^{20}\cdot5^{10}\cdot5^{20}}{3^{15}\cdot5^{30}}=3^5=243\\ \dfrac{6^6+6^3+3^3+3^6}{-73}=\dfrac{46656+216+27+729}{-73}=-\dfrac{47628}{73}\\ \dfrac{27^7+3^{15}}{9^9-27}=\dfrac{\left(3^3\right)^7+3^{15}}{\left(3^2\right)^9-3^3}=\dfrac{3^{21}+3^{15}}{3^{18}-3^3}=\dfrac{3^{15}\left(3^6+1\right)}{3^3\left(3^{15}-1\right)}=\dfrac{3^5\cdot730}{3^{15}-1}\\ \dfrac{8^{20}+4^{20}}{4^{25}+64^5}=\dfrac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\dfrac{2^{60}+2^{40}}{2^{50}+2^{30}}=\dfrac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}=2^{10}=1024\)

22 tháng 11 2022

\(B=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^{10}}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}=\dfrac{2^{10}\cdot3^8\left(1-3^2\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{1-9}{6}=\dfrac{-8}{6}=-\dfrac{4}{3}\)

13 tháng 7 2017

a, \(\dfrac{5^4.20^4}{25^5.4^5}=\dfrac{5^4.2^8.5^4}{5^{10}.2^{10}}=\dfrac{1}{5^2.2^2}=\dfrac{1}{25.4}=\dfrac{1}{100}\)

b, \(\dfrac{2^7.9^3}{6^5.8^2}=\dfrac{2^7.3^6}{2^5.3^5.2^6}=\dfrac{3}{2^4}=\dfrac{3}{16}\)

c, \(\dfrac{45^{10}.5^{20}}{75^5}=\dfrac{5^{10}.3^{20}.5^{20}}{3^5.5^{10}}=5^{20}.3^{15}\)

d, \(\left(0,8\right)^5=\left(0,1\right)^5.8^5=\dfrac{1}{100000}.32768=0,32768\)

e, \(\dfrac{2^{15}.9^4}{6^6.8^3}=\dfrac{2^{15}.3^8}{2^6.3^6.2^9}=3^2=9\)

d, \(\dfrac{8^{20}+4^{20}}{4^{25}+64^5}=\dfrac{2^{60}+2^{40}}{2^{50}+2^{30}}=\dfrac{2^{40}.\left(2^{20}+1\right)}{2^{30}.\left(2^{20}+1\right)}=2^{10}=1024\)

Chúc bạn học tốt!!!

13 tháng 7 2017

\(\text{a) }\dfrac{5^4\cdot20^4}{25^5\cdot4^5}=\dfrac{5^4\cdot\left(5\cdot4\right)^4}{\left(5^2\right)^5\cdot4^5}=\dfrac{5^4\cdot5^4\cdot4^4}{5^{10}\cdot4^5}=\dfrac{5^8\cdot4^4}{5^{10}\cdot4^5}=\dfrac{1}{5^2\cdot4}=\dfrac{1}{25\cdot4}=\dfrac{1}{100}\)

\(\text{b) }\dfrac{2^7\cdot9^3}{6^5\cdot8^2}=\dfrac{2^7\cdot\left(3^2\right)^3}{\left(2\cdot3\right)^5\cdot\left(2^3\right)^2}=\dfrac{2^7\cdot3^6}{2^5\cdot3^5\cdot2^6}=\dfrac{2^7\cdot3^6}{2^5\cdot2^6\cdot3^5}=\dfrac{2^7\cdot3^6}{2^{11}\cdot3^5}=\dfrac{3}{2^4}=\dfrac{3}{16}\)

\(\text{c) }\dfrac{45^{10}\cdot5^{20}}{75^5}=\dfrac{\left(5\cdot9\right)^{10}\cdot5^{20}}{\left(25\cdot3\right)^5}=\dfrac{5^{10}\cdot9^{10}\cdot5^{20}}{25^5\cdot3^5}=\dfrac{5^{10}\cdot5^{20}\cdot\left(3^2\right)^{10}}{\left(5^2\right)^5\cdot3^5}=\dfrac{5^{30}\cdot3^{20}}{5^{10}\cdot3^5}=5^{20}\cdot3^{15}\)

\(\text{d) }\left(0.8\right)^5=\left(\dfrac{8}{10}\right)^5=\left(\dfrac{4}{5}\right)^5=\dfrac{4^5}{5^5}=\dfrac{64}{3125}\)

\(\text{e) }\dfrac{2^{15}\cdot9^4}{6^6\cdot8^3}=\dfrac{2^{15}\cdot\left(3^2\right)^4}{\left(2\cdot3\right)^6\cdot\left(2^3\right)^3}=\dfrac{2^{15}\cdot3^8}{2^6\cdot3^6\cdot2^9}=\dfrac{2^{15}\cdot3^8}{2^6\cdot2^9\cdot3^6}=\dfrac{2^{15}\cdot3^8}{2^{15}\cdot3^6}=3^2=9\)

\(f\text{) }\dfrac{8^{20}+4^{20}}{4^{25}+64^5}=\dfrac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\dfrac{2^{60}+2^{40}}{2^{50}+2^{30}}=\dfrac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}=2^{10}=1024\)

21 tháng 10 2017

a. \(\dfrac{11}{24}-\dfrac{5}{41}+\dfrac{13}{24}+0,5-\dfrac{36}{41}\)

\(=\left(\dfrac{11}{24}+\dfrac{13}{24}\right)+\left(\dfrac{-5}{41}-\dfrac{36}{41}\right)+0,5\)

\(=1+\left(-1\right)+0,5\)

\(=0,5\)

b. \(-12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)

\(=-12:\left(\dfrac{-1}{12}\right)^2\)

\(=-12:\dfrac{1}{144}\)

\(=-1728\)

c. \(\dfrac{7}{23}.\left[\left(-\dfrac{8}{6}\right)-\dfrac{45}{18}\right]\)

\(=\dfrac{7}{23}.\dfrac{-23}{6}\)

\(=\dfrac{-7}{6}\)

d. \(23\dfrac{1}{4}.\dfrac{7}{5}-13\dfrac{1}{4}:\dfrac{5}{7}\)

\(=23\dfrac{1}{4}.\dfrac{7}{5}-13\dfrac{1}{4}.\dfrac{7}{5}\)

\(=\left(23\dfrac{1}{4}-13\dfrac{1}{4}\right).\dfrac{7}{5}\)

\(=10.\dfrac{7}{5}\)

\(=14\)

e. \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right).\left(0,8-\dfrac{3}{4}\right)^2\)

\(=\dfrac{17}{12}.\left(\dfrac{1}{20}\right)^2\)

\(=\dfrac{17}{12}.\dfrac{1}{400}=\dfrac{17}{4800}\)

15 tháng 10 2017

Ta có: \(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)

\(\Rightarrow\dfrac{4\left(3x-2y\right)}{4.4}=\dfrac{3\left(2z-4x\right)}{3.3}=\dfrac{2\left(4y-3z\right)}{2.2}\)

\(\Rightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}=\dfrac{12x-8y+6z-12x+8y-6z}{16+9+4}=0\)

\(\Rightarrow12x-8y=6z-12x=8y-6z=0\)

\(\Rightarrow\left\{{}\begin{matrix}12x=8y\\6z=12x\\8y=6z\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}3x=2y\\z=2x\\4y=3z\end{matrix}\right.\)

\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{3},\dfrac{z}{2}=x,\dfrac{y}{3}=\dfrac{z}{4}\)

\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{3},\dfrac{z}{4}=\dfrac{x}{2},\dfrac{y}{3}=\dfrac{z}{4}\)

\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\) (đpcm)

20 tháng 8 2017

Tham khảo tại đây nhé: Câu hỏi của Phong Tuấn Đỗ - Toán lớp 7 | Học trực tuyến

10 tháng 12 2017

M=\(\dfrac{8^{20}+4^{20}}{4^{25}+64^5}=\dfrac{4^{20}\left(2^{20}+1\right)}{4^{25}+4^{15}}=\dfrac{4^{20}\left(2^{20}+1\right)}{4^{15}\left(4^{10}+1\right)}=\dfrac{2^{20}+1}{4^{10}+1}\)

10 tháng 12 2017

T=\(\dfrac{45^{10}.5^{20}}{75^{15}}=\dfrac{9^{10}.5^{30}}{25^{15}.3^{15}}=\dfrac{3^{20}.5^{30}}{5^{30}.3^{15}}=3^5=243\)