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Bạn ghi nhỏ lại nhé. Hơn nũa bạn nên tách riêng từng câu hỏi, làm vầy nhiều lắm
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a)\(\frac{11}{125}-\frac{17}{18}-\frac{5}{7}+\frac{4}{9}+\frac{7}{14}\)
=\(\frac{11}{125}-\left(\frac{17}{18}-\frac{4}{9}\right)+\left(\frac{17}{14}-\frac{5}{7}\right)\)
=\(\frac{11}{125}-\frac{1}{2}+\frac{1}{2}\)
=\(\frac{11}{125}+\left(\frac{1}{2}-\frac{1}{2}\right)\)
=\(\frac{11}{125}\)
b)................................đề bài......................................................
= ( 1 - 1 ) + ( 2 - 2 ) + ( 3 - 3 ) + ( -1/2 + -1/2 ) + ( -2/3 + -1/3) + ( -3/4 + -1/4) + 4
= 0 + 0 + 0 + (-1) + (-1) + (-1) + 4
= 1
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mh biết làm bài này rùi bn có cần mi2h đang cho bn ko?
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Ta có : \(A=\frac{6n-1}{3n+2}=\frac{2\left(3n+2\right)-5}{3n+2}=2-\frac{5}{3n+2}\)
Để A là số nguyên thì \(5⋮3n+2\)
hay \(3n+2\inƯ_5=\left\{\pm1;\pm5\right\}\)
3n+2 | 1 | -1 | 5 | -5 |
3n | -1 | -3 | 3 | -7 |
n | \(\frac{-1}{3}\) | -1 | 1 | \(\frac{-7}{3}\) |
Vậy để A nguyên thì \(n\in\left\{\frac{-1}{3};-1;1;\frac{-7}{3}\right\}\)
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a: \(\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}}{1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}:\dfrac{13+\dfrac{13}{2}+\dfrac{13}{3}+\dfrac{13}{4}}{17-\dfrac{17}{2}+\dfrac{17}{3}-\dfrac{17}{4}}\)
\(=\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}}{1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}\cdot\dfrac{17\left(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}{13\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\right)}=\dfrac{17}{13}\)
b: \(\dfrac{0.125-\dfrac{1}{5}+\dfrac{1}{7}}{0.375-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0.2}{\dfrac{3}{4}+0.5-\dfrac{3}{10}}\)
\(=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{\dfrac{3}{8}-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{4}+\dfrac{3}{6}-\dfrac{3}{10}}\)
\(=\dfrac{1}{3}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{2}\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}\right)}=\dfrac{1}{3}+\dfrac{2}{3}=1\)
2585
còn cách giải bạn tham khảo câu hỏi tương tự nha bạn
\(2A=\left(1+3+3^2+....+3^{n-1}\right)+\left(n-1\right)=B+\left(n-1\right)\)
\(3B=\left(3+3^2+3^3....+3^n\right)\)
\(2B=3^n-1\)
\(A=\frac{3^n-1}{4}+\frac{\left(n-1\right)}{2}\)