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Câu 14)
\(a,\\ =-\dfrac{3}{8}+\dfrac{8}{17}+\dfrac{-5}{8}-\dfrac{3}{5}+\dfrac{9}{17}\\ =\left(\dfrac{-3}{8}+\dfrac{-5}{8}\right)+\left(\dfrac{8}{17}+\dfrac{9}{17}\right)-\dfrac{3}{5}\\ =\left(-1\right)+1-\dfrac{3}{5}=0-\dfrac{3}{5}=\dfrac{-3}{5}\\ b,\\ =\dfrac{7}{15}.\dfrac{-15}{14}+\left(\dfrac{27}{16}-\dfrac{1}{8}\right):\dfrac{5}{8}\)
\(=\dfrac{-1}{2}+\dfrac{25}{16}.\dfrac{8}{5}=\dfrac{-1}{2}+\dfrac{5}{2}=2\\ c,\\ =\dfrac{2}{2}-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{4}+.....+\dfrac{2}{99}-\dfrac{2}{100}\\ =1-\dfrac{1}{50}=\dfrac{49}{50}\)
Câu 15
\(a,2x+\dfrac{-1}{4}=\dfrac{3}{2}\\ 2x=\dfrac{3}{2}-\dfrac{-1}{4}=\dfrac{7}{4}\\ x=\dfrac{7}{4}:2=\dfrac{7}{8}\\ b,\dfrac{15}{x}=\dfrac{-3}{4}\\ x=\dfrac{15.4}{-3}=-20\)
Giải:
Tên tam giác |
Tên 3 đỉnh |
Tên 3 góc |
Tên 3 cạnh |
ABI |
A,B,I |
|
AB, BI, IA |
AIC |
A,I,C |
|
AI, IC, CA |
ABC |
A,B,C |
|
AB, BC, CA |
Bài 16:
\(e,\left(6n+11\right)⋮\left(2n+3\right)\\ \Rightarrow\left[3\left(2n+3\right)+2\right]⋮\left(2n+3\right)\\ \Rightarrow2⋮\left(2n+3\right)\\ \Rightarrow2n+3\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Rightarrow n\in\varnothing\left(n\in N\right)\\ g,\left(3n+5\right)⋮\left(2n+1\right)\\ \Rightarrow\left(6n+10\right)⋮\left(2n+1\right)\\ \Rightarrow\left[3\left(2n+1\right)+7\right]⋮\left(2n+1\right)\\ \Rightarrow7⋮\left(2n+1\right)\\ \Rightarrow2n+1\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\\ \Rightarrow n\in\left\{0;3\right\}\left(n\in N\right)\)