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Tính hợp lí:
a) \(\left(-0,4\right)+\dfrac{3}{8}+\left(-0,6\right)\)
\(=\left[\left(-0,4\right)+\left(-0,6\right)\right]+\dfrac{3}{8}\)
\(=-1+\dfrac{3}{8}\)
\(=\dfrac{\left(-8\right)+3}{8}\)
\(=\dfrac{-5}{8}\)
b) \(\dfrac{4}{5}-1,8+0,375+\dfrac{5}{8}\)
\(=\dfrac{4}{5}-\dfrac{9}{5}+\dfrac{3}{8}+\dfrac{5}{8}\)
\(=-1+1\)
\(=0\\\)
c) \(\dfrac{7}{3}.\left(-2,5\right).\dfrac{6}{7}\)
\(=\dfrac{7}{3}.\dfrac{-5}{2}.\dfrac{6}{7}\)
\(=\dfrac{7}{3}.\dfrac{6}{7}.\dfrac{-5}{2}\)
\(=2.\dfrac{-5}{2}\)
\(=-5\)
d) \(\dfrac{7}{12}.\left(-2,34\right)-\dfrac{7}{12}.\left(-0,34\right)\)
\(=\dfrac{7}{12}.\left[\left(-2,34\right)+0,34\right]\)
\(=\dfrac{7}{12}.\left(-2\right)\)
\(=\dfrac{-7}{6}\)
e) \(\dfrac{-8}{3}.\dfrac{2}{11}-\dfrac{8}{3}:\dfrac{11}{9}\)
\(=\dfrac{8}{3}.\dfrac{-2}{11}-\dfrac{8}{3}.\dfrac{9}{11}\)
\(=\dfrac{8}{3}.\left(\dfrac{-2}{11}-\dfrac{9}{11}\right)\)
\(=\dfrac{8}{3}.-1\)
\(=\dfrac{-8}{3}\)
Chúc bạn học tốt
Từ 100 đến 2023 có : (2023-100+1):2= 962 số tự nhiên là số chẵn.
Ta có số tự nhiên chăn nhỏ nhất trong khoảng đó là 100
Và số lớn nhất là 2022
Số lượng số tự nhiên chẵn có trong khoảng đó là:
\(\left(2022-100\right):2+1=962\) (số)
Các phần tử của tập hợp A là:
\(A=\left\{10;11;12;...;99\right\}\)
Số lượng phần tử:
\(\left(99-10\right):1+1=90\) (phần tử)
Các phần tử của tập hợp A là:
Số lượng phần tử:
(phần tử)
\(\dfrac{5}{4}+\left(2x-\dfrac{1}{2}\right)=\dfrac{5}{6}\\ =>2x-\dfrac{1}{2}=\dfrac{5}{6}-\dfrac{5}{4}\\ =>2x-\dfrac{1}{2}=\dfrac{10}{12}-\dfrac{15}{12}\\ =>2x-\dfrac{1}{2}=-\dfrac{5}{12}\\ =>2x=-\dfrac{5}{12}+\dfrac{1}{2}\\ =>2x=-\dfrac{5}{12}+\dfrac{6}{12}\\ =>2x=\dfrac{1}{12}\\ =>x=\dfrac{1}{12}:2\\ =>x=\dfrac{1}{12}\cdot\dfrac{1}{2}\\ =>x=\dfrac{1}{24}\)
__
\(\dfrac{3}{2}-\left(x+\dfrac{1}{4}\right)=\dfrac{5}{8}\\ =>x+\dfrac{1}{4}=\dfrac{3}{2}-\dfrac{5}{8}\\ =>x+\dfrac{1}{4}=\dfrac{12}{8}-\dfrac{5}{8}\\ =>x+\dfrac{1}{4}=\dfrac{7}{8}\\ =>x=\dfrac{7}{8}-\dfrac{1}{4}\\ =>x=\dfrac{7}{8}-\dfrac{2}{8}\\ =>x=\dfrac{5}{8}\)
__
\(\dfrac{x}{3}=\dfrac{12}{x}\\ =>x^2=3\cdot12\\ =>x^2=36\\ =>x^2=6^2\\ =>x=\pm6\)
Tìm x:
a) \(\dfrac{5}{4}+\left(2x-\dfrac{1}{2}\right)=\dfrac{5}{6}\)
\(=>2x-\dfrac{1}{2}=\dfrac{5}{6}-\dfrac{5}{4}\)
\(=>2x-\dfrac{1}{2}=\dfrac{-5}{12}\)
\(=>2x=\dfrac{-5}{12}+\dfrac{1}{2}\)
\(=>2x=\dfrac{1}{12}\)
\(=>x=\dfrac{1}{12}:2\)
\(=>x=\dfrac{1}{24}\)
b) \(\dfrac{3}{2}-\left(x+\dfrac{1}{4}\right)=\dfrac{5}{8}\)
\(=>x+\dfrac{1}{4}=\dfrac{3}{2}-\dfrac{5}{8}\)
\(=>x+\dfrac{1}{4}=\dfrac{7}{8}\)
\(=>x=\dfrac{7}{8}-\dfrac{1}{4}\)
\(=>x=\dfrac{5}{8}\)
c) \(\dfrac{x}{3}=\dfrac{12}{x}\)
Ta có: \(x.x=3.12\)
\(\Rightarrow x^2=36\)
Vậy x = 6 hoặc x = -6
Chúc bạn học tốt
1. Tính hợp lí
a) \(0,7+\dfrac{-7}{19}-\left(-0,3\right)\)
\(=\dfrac{7}{10}+\dfrac{-7}{19}+\dfrac{3}{10}\)
\(=\left(\dfrac{7}{10}+\dfrac{3}{10}\right)+\dfrac{-7}{19}\)
\(=1+\dfrac{-7}{19}\)
\(=\dfrac{12}{19}\)
b) \(\dfrac{5}{3}.\left(-2,5\right):\dfrac{5}{6}\)
\(=\dfrac{5}{3}.\dfrac{-5}{2}.\dfrac{6}{5}\)
\(=\left(\dfrac{5}{3}.\dfrac{6}{5}\right).\dfrac{-5}{2}\)
\(=2.\dfrac{-5}{2}\)
\(=-5\)
c) \(0,6.\dfrac{-5}{17}-\dfrac{3}{5}.\dfrac{12}{17}\)
\(=\dfrac{3}{5}.\dfrac{-5}{17}-\dfrac{3}{5}.\dfrac{12}{17}\)
\(=\dfrac{3}{5}.\left(\dfrac{-5}{17}-\dfrac{12}{17}\right)\)
\(=\dfrac{3}{5}.-1\)
\(=\dfrac{-3}{5}\)
d) \(\dfrac{7}{4}.\dfrac{5}{2}-\dfrac{7}{4}.\dfrac{3}{2}\)
\(=\dfrac{7}{4}.\left(\dfrac{5}{2}-\dfrac{3}{2}\right)\)
\(=\dfrac{7}{4}.1\)
\(=\dfrac{7}{4}\)
Chúc bạn học tốt
\(\dfrac{1}{3}-\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{30}-\dfrac{1}{42}-\dfrac{1}{56}-\dfrac{1}{72}-\dfrac{1}{90}-\dfrac{1}{110}=x-\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - \(\dfrac{1}{3.4}\) - \(\dfrac{1}{4.5}\) - \(\dfrac{1}{5.6}\) - \(\dfrac{1}{6.7}\) - \(\dfrac{1}{7.8}\)- \(\dfrac{1}{8.9}\) - \(\dfrac{1}{9.10}\) - \(\dfrac{1}{10.11}\) = \(x\) - \(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - (\(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\)+ \(\dfrac{1}{7.8}\) + \(\dfrac{1}{8.9}\) + \(\dfrac{1}{9.10}\) + \(\dfrac{1}{10.11}\) =\(x\)-\(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) +...+ \(\dfrac{1}{9}\) - \(\dfrac{1}{10}\) + \(\dfrac{1}{10}\) - \(\dfrac{1}{11}\)) = \(x\) - \(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{11}\)) = \(x\) - \(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{11}\) = \(x\) - \(\dfrac{5}{13}\)
\(x-\dfrac{5}{13}=\dfrac{1}{11}\)
\(x\) = \(\dfrac{1}{11}\) + \(\dfrac{5}{13}\)
\(x\) = \(\dfrac{68}{143}\)
`@` `\text {Ans}`
`\downarrow`
`82.8^n=41984`
`=> 8^n = 41984 \div 82`
`=> 8^n = 512`
`=> 8^n = 8^3`
`=> n=3`
Vậy, `n = 3`.
\(\dfrac{2\cdot3\cdot5^2}{3^2\cdot5^3}\)
\(=2\cdot\dfrac{3}{3^2}\cdot\dfrac{5^2}{5^3}\)
\(=2\cdot\dfrac{1}{3}\cdot\dfrac{1}{5}\)
\(=\dfrac{2}{15}\)
\(\dfrac{2.3.5^2}{3^2.5^2}=\dfrac{2}{3}\)