Tính nhanh nếu có thể:
a) \(\dfrac{15}{12}+\dfrac{5}{13}-\dfrac{3}{12}-\dfrac{18}{13}\)
b) \(\dfrac{5^4.20^4}{25^5.4^5}\)
c) \(\dfrac{8^{10}+4^{10}}{8^4+4^{11}}\)
Tính nhanh nếu có thể:
a) \(\dfrac{15}{12}+\dfrac{5}{13}-\dfrac{3}{12}-\dfrac{18}{13}\)
b) \(\dfrac{5^4.20^4}{25^5.4^5}\)
c) \(\dfrac{8^{10}+4^{10}}{8^4+4^{11}}\)
a) \(\frac{15}{12}+\frac{5}{13}-\frac{3}{12}-\frac{18}{13}\)
\(=\left(\frac{15}{12}-\frac{3}{12}\right)+\left(\frac{5}{13}-\frac{18}{13}\right)\)
\(=1+\left(-1\right)\)
\(=0\)
b) \(\frac{5^4.20^4}{25^5.4^5}=\frac{\left(20.5\right)^4}{\left(25.4\right)^5}=\frac{100^4}{100^5}=\frac{1}{100}\)
c) \(\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}=\frac{2^{12}.\left(2^{18}+2^8\right)}{2^{12}.\left(1+2^{10}\right)}=\frac{2^{18}+2^8}{1+2^{10}}=256\)
Tính (Tính hợp lí nếu có thể)
a) \(\dfrac{-7}{12}\)-\(\dfrac{3}{36}\)
b) (4-\(\dfrac{5}{12}\)):2+\(\dfrac{5}{24}\)
c) \(\dfrac{8}{9}\)+\(\dfrac{1}{9}\).\(\dfrac{2}{13}\)+\(\dfrac{1}{9}\).\(\dfrac{11}{13}\)
d) \(\dfrac{3}{4}\).\(\dfrac{8}{9}\).\(\dfrac{15}{16}\). ... .\(\dfrac{9999}{10000}\)
e) \(\dfrac{3}{1.4}\)+\(\dfrac{3}{4.7}\)+\(\dfrac{3}{7.10}\)+...+\(\dfrac{3}{97.100}\)
*Lưu ý: Mong các anh chị trình bày chi tiết để em có thể hiểu bài, em xin các anh chị đừng viết mỗi kết quả xong em chả biết một cái gì ;-;
a: =-21/36-3/36=-24/36=-2/3
b: =43/12*1/2+5/24=43/24+5/24=2
c: =8/9+1/9=1
e: =1-1/4+1/4-1/7+...+1/97-1/100
=1-1/100=99/100
Bài 1: Thực hiện các phép tính (Tính nhanh nếu có thể)
a) \(\dfrac{-5}{9}\) - \(\dfrac{-5}{12}\) b) \(\dfrac{-5}{12}\) : \(\dfrac{15}{4}\) c) \(\dfrac{1}{13}\) x \(\dfrac{8}{13}\) + \(\dfrac{5}{13}\) x \(\dfrac{1}{13}\) - \(\dfrac{14}{13}\)
Bài 2: Tìm \(x\), biết:
a) \(x=\dfrac{1}{5}+\dfrac{-3}{7}\) b) \(\dfrac{3}{5}-\dfrac{4}{7}\)\(:\)\(x=\dfrac{-9}{10}\) c) \(x-\left(\dfrac{-3}{4}\right)=\dfrac{-2}{3}-\dfrac{1}{2}\)
d) \(\dfrac{-5}{9}-x=\dfrac{1}{3}+\dfrac{7}{18}\)
Bài 3: Cho S = \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{10^2}\). Chứng minh rằng: \(S>\dfrac{9}{22}\)
Tính rồi rút gọn (theo mẫu):
Mẫu: \(\dfrac{9}{10}-\dfrac{4}{10}=\dfrac{9-4}{10}=\dfrac{5}{10}=\dfrac{1}{2}\) |
a) \(\dfrac{15}{8}-\dfrac{13}{8}\) b) \(\dfrac{7}{15}-\dfrac{2}{15}\) c) \(\dfrac{11}{12}-\dfrac{2}{12}\) d) \(\dfrac{19}{7}-\dfrac{5}{7}\)
a: \(\dfrac{15}{8}-\dfrac{13}{8}=\dfrac{15-13}{8}=\dfrac{2}{8}=\dfrac{1}{4}\)
b: \(\dfrac{7}{15}-\dfrac{2}{15}=\dfrac{7-2}{15}=\dfrac{5}{15}=\dfrac{1}{3}\)
c: \(\dfrac{11}{12}-\dfrac{2}{12}=\dfrac{11-2}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)
d: \(\dfrac{19}{7}-\dfrac{5}{7}=\dfrac{19-5}{7}=\dfrac{14}{7}=2\)
\(\dfrac{4^5.9^4}{8^3.27^3}\);\(\dfrac{4^{20}.3^{35}}{2^{37}.27^{12}}\)\(;\dfrac{5^4.20^4}{25^5.4^5};\dfrac{2^{15}.9^4}{6^6.8^3}\)
\(\dfrac{4^5\cdot9^4}{8^3\cdot27^3}=\dfrac{\left(2^2\right)^5\cdot\left(3^2\right)^4}{\left(2^3\right)^3\cdot\left(3^3\right)^3}=\dfrac{2^{10}\cdot3^8}{2^9\cdot3^9}=\dfrac{2}{3}\)
\(\dfrac{4^{20}\cdot3^{35}}{2^{37}\cdot27^{12}}=\dfrac{\left(2^2\right)^{20}\cdot3^{35}}{2^{37}\cdot\left(3^3\right)^{12}}=\dfrac{2^{40}\cdot3^{35}}{2^{37}\cdot3^{36}}=\dfrac{2^3}{3}\)
\(\dfrac{5^4\cdot20^4}{25^5\cdot4^5}=\dfrac{5^4\cdot5^4\cdot4^4}{5^5\cdot5^5\cdot4^5}=\dfrac{1}{5^2\cdot4}=\dfrac{1}{100}\)
\(\dfrac{2^{15}\cdot9^4}{6^6\cdot8^3}=\dfrac{2^{15}\cdot\left(3^2\right)^4}{2^6\cdot3^6\cdot\left(2^3\right)^3}=\dfrac{2^{15}\cdot3^8}{2^6\cdot3^6\cdot2^9}=3^2\)
Tính hợp lý:
\(a.\dfrac{3}{17}+\dfrac{-5}{13}+\dfrac{-18}{35}+\dfrac{14}{17}+\dfrac{17}{-35}+\dfrac{-8}{13}\)
\(b.\dfrac{-3}{8}\cdot\dfrac{1}{6}+\dfrac{3}{-8}\cdot\dfrac{5}{6}+\dfrac{-10}{16}\)
\(c.\dfrac{-4}{11}\cdot\dfrac{5}{15}\cdot\dfrac{11}{-4}\)
a: \(=\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{-5}{13}-\dfrac{8}{13}\right)+\left(\dfrac{-18}{35}-\dfrac{17}{35}\right)\)
=1-1-1
=-1
b: \(=\dfrac{-3}{8}\left(\dfrac{1}{6}+\dfrac{5}{6}\right)+\dfrac{-5}{8}=\dfrac{-3}{8}-\dfrac{5}{8}=-1\)
c: \(=\dfrac{4}{4}\cdot\dfrac{5}{15}\cdot\dfrac{11}{11}=\dfrac{1}{3}\)
a) \(=\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(-\dfrac{5}{13}+-\dfrac{8}{13}\right)+\left(-\dfrac{18}{35}+-\dfrac{17}{35}\right)=1+-1+-1=-1\)
b) \(=-\dfrac{3}{8}\cdot\left(\dfrac{1}{6}+\dfrac{5}{6}\right)-\dfrac{10}{16}=-\dfrac{3}{8}-\dfrac{10}{16}=-1\)
c) \(=\left(-\dfrac{4}{11}\cdot-\dfrac{11}{4}\right)\cdot\dfrac{5}{15}=1\cdot\dfrac{1}{3}=\dfrac{1}{3}\)
a)\(=\left(-\dfrac{5}{13}+\dfrac{-8}{13}\right)+\left(-\dfrac{18}{35}-\dfrac{17}{35}\right)+\left(\dfrac{3}{14}+\dfrac{14}{17}\right)=-1-1+1=-1\)
b)\(=\dfrac{-3}{8}.\left(\dfrac{1}{6}+\dfrac{5}{6}\right)-\dfrac{10}{16}=-\dfrac{3}{8}.1-\dfrac{10}{16}=-\dfrac{6}{16}-\dfrac{10}{16}=-\dfrac{16}{16}=-1\)
c)\(\dfrac{-4.5.11}{11.5.3.-4}=\dfrac{1}{3}\)
bài 1: tính
a) \(\dfrac{5^4.20^4}{25^5.4^5}\) b)3,5-\(\left(-\dfrac{2}{7}\right)\) c)\(\left(\dfrac{11}{12}:\dfrac{33}{16}\right).\dfrac{3}{5}\) d)\(15.\left(-\dfrac{2}{3}\right)^2.-\dfrac{7}{3}\) e)\(\left(\dfrac{9}{25}-2.8\right):\left(3\dfrac{4}{5}+0,2\right)\) g)\(\dfrac{21}{47}+\dfrac{9}{45}+\dfrac{26}{47}+\dfrac{4}{5}\) h)\(\dfrac{15}{12}+\dfrac{5}{13}-\dfrac{3}{12}-\dfrac{18}{13}\) j)12.\(\left(-\dfrac{2}{3}\right)^2+\dfrac{4}{3}\) k)\(\dfrac{13}{25}+\dfrac{6}{41}-\dfrac{38}{25}+\dfrac{35}{41}-\dfrac{1}{2}\) l)12,5.\(\left(-\dfrac{5}{7}\right)+1,5.\left(-\dfrac{5}{7}\right)\) m)\(\sqrt{\dfrac{64}{25}}.3\dfrac{1}{2}-\dfrac{3}{5}.3\dfrac{1}{2}\)
Toàn câu dễ nên bạn tự làm đi.
Trong lúc bạn đánh xong bài này thì bạn có thể làm xong rồi đó.
Đừng có ỷ lại vào người khác ,động não lên.
a: \(=\dfrac{5^8\cdot2^8}{5^{10}\cdot2^{10}}=\dfrac{1}{100}\)
b: =7/2+2/7
=49/14+4/14
=53/14
c: \(=\dfrac{11}{12}\cdot\dfrac{16}{33}\cdot\dfrac{3}{5}=\dfrac{1}{3}\cdot\dfrac{3}{5}\cdot\dfrac{4}{3}=\dfrac{1}{5}\cdot\dfrac{4}{3}=\dfrac{4}{15}\)
d: \(=15\cdot\dfrac{4}{9}\cdot\dfrac{-7}{3}=\dfrac{15}{27}\cdot\left(-28\right)=-28\cdot\dfrac{5}{9}=-\dfrac{140}{9}\)
g: \(=\left(\dfrac{21}{47}+\dfrac{26}{47}\right)+\left(\dfrac{9}{45}+\dfrac{36}{45}\right)=1+1=2\)
j: =12*4/9+4/3
=16/3+4/3
=20/3
Tính:
a) \(\dfrac{13}{14}\)-\(\dfrac{-7}{8}\)+\(\dfrac{-3}{2}\)
b) \(\dfrac{5}{17}\)+\(\dfrac{-15}{34}\).\(\dfrac{2}{5}\)
c) \(\dfrac{1}{5}\):\(\dfrac{1}{10}\)-\(\dfrac{1}{3}\).(\(\dfrac{6}{5}\)-\(\dfrac{2}{4}\))
d) \(\dfrac{-3}{4}\):(\(\dfrac{12}{-5}\)-\(\dfrac{-7}{10}\))
*Lưu ý: Không viết luôn kết quả, giải chi tiết.
\(a,\dfrac{13}{14}\cdot\dfrac{-7}{8}+\dfrac{-3}{2}\)
\(=-\dfrac{13}{16}+\dfrac{-3}{2}\)
\(=-\dfrac{13}{16}+\dfrac{-24}{16}\)
\(=-\dfrac{37}{16}\)
\(b,\dfrac{5}{17}+\dfrac{-15}{34}\cdot\dfrac{2}{5}\)
\(=\dfrac{5}{17}+\dfrac{-3}{17}\)
\(=\dfrac{2}{17}\)
\(c,\dfrac{1}{5}:\dfrac{1}{10}-\dfrac{1}{3}\cdot\left(\dfrac{6}{5}-\dfrac{2}{4}\right)\)
\(=2-\dfrac{1}{3}\cdot\dfrac{7}{10}\)
\(=2-\dfrac{7}{30}\)
\(=\dfrac{53}{30}\)
\(d,\dfrac{-3}{4}:\left(\dfrac{12}{-5}-\dfrac{-7}{10}\right)\)
\(=\dfrac{-3}{4}:\dfrac{-17}{10}\)
\(=\dfrac{15}{34}\)
Tính hợp lí
1) \((\)\(\dfrac{7}{-18}\)+\(\dfrac{-5}{12}\)) -\(\dfrac{13}{-18}\)
2) \(\dfrac{-13}{17}\)+(\(\dfrac{13}{-21}\)+\(\dfrac{-4}{17}\))
3) ( \(\dfrac{13}{-10}\)-\(\dfrac{-4}{13}\))+\(\dfrac{11}{-10}\)
4) \(\dfrac{13}{17}\)\(\times\)\(\dfrac{4}{-5}\)+\(\dfrac{13}{17}\)\(\times\)\(\dfrac{-3}{4}\)
5) \(\dfrac{-5}{12}\)\(\times\)\(\dfrac{7}{-17}\)\(\times\)\(\dfrac{9}{-20}\)
6) \(\dfrac{5}{9}\)\(\times\)\(\dfrac{11}{23}\)+ \(\dfrac{11}{23}\)\(\times\)\(\dfrac{17}{9}\)\(-\)\(\dfrac{13}{9}\)\(\times\)\(\dfrac{11}{23}\)
\(1.\dfrac{-7}{18}+\dfrac{-5}{12}-\dfrac{-13}{18}\text{=}\left(\dfrac{-7}{18}-\dfrac{-13}{18}\right)+\dfrac{-5}{12}\text{=}\dfrac{1}{3}+\dfrac{-5}{12}\text{=}\dfrac{-1}{12}\)
\(2.\dfrac{-13}{17}+\dfrac{-13}{21}+\dfrac{-4}{17}\text{=}\left(\dfrac{-13}{17}+\dfrac{-4}{17}\right)+\dfrac{-13}{21}\text{=}-1+\dfrac{-13}{21}\text{=}\dfrac{-34}{21}\)
\(3.\dfrac{-13}{10}-\dfrac{-4}{13}+\dfrac{-11}{10}\text{=}\dfrac{-12}{5}-\dfrac{-4}{13}\text{=}\dfrac{-136}{65}\)
\(4.\dfrac{13}{17}\times\left(\dfrac{-4}{5}+\dfrac{-3}{4}\right)\text{=}\dfrac{13}{17}\times\dfrac{-31}{20}\text{=}\dfrac{-403}{340}\)
\(5.\left(\dfrac{-5}{12}\times\dfrac{-9}{20}\right)\times\dfrac{-7}{17}\text{=}\dfrac{3}{16}\times\dfrac{-7}{17}\text{=}\dfrac{-21}{272}\)
\(6.\dfrac{11}{23}\times\left(\dfrac{5}{9}+\dfrac{17}{9}-\dfrac{13}{9}\right)\text{=}\dfrac{11}{23}\times1\text{=}\dfrac{11}{23}\)