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\(\dfrac{-5}{8}:\dfrac{15}{4}=\dfrac{-5}{8}\cdot\dfrac{4}{15}=\dfrac{-1\cdot1}{2\cdot3}=\dfrac{-1}{6}\)
\(\dfrac{-15}{17}:\dfrac{25}{-34}=\dfrac{-15}{17}\cdot\dfrac{-34}{25}=\dfrac{-3\cdot-2}{1\cdot5}=\dfrac{-6}{5}\)
\(-12:\dfrac{8}{3}=-12\cdot\dfrac{3}{8}=\dfrac{-12\cdot3}{8}=\dfrac{-3\cdot3}{2}=\dfrac{-9}{2}\)
\(\dfrac{-15}{14}:\dfrac{20}{-21}=\dfrac{-15}{14}\cdot\dfrac{-21}{20}=\dfrac{-3\cdot-3}{2\cdot4}=\dfrac{-9}{8}\)
\(-48:\dfrac{-24}{5}=-48\cdot\dfrac{5}{-24}=\dfrac{-48\cdot5}{-24}=2\cdot5=10\)
\(\dfrac{-30}{7}:\dfrac{-5}{-14}=\dfrac{-30}{7}:\dfrac{5}{14}=\dfrac{-30}{7}\cdot\dfrac{14}{5}=\dfrac{-6\cdot2}{1\cdot1}=-18\)
a) Gọi \(ƯCLN\left(a^2,a+b\right)=d\) với \(d\inℕ^∗\)
\(\Rightarrow\left\{{}\begin{matrix}a^2⋮d\\a+b⋮d\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a^2⋮d\\a^2+ab⋮d\end{matrix}\right.\)
\(\Rightarrow ab⋮d\)
Vì \(a,b\) nguyên tố cùng nhau \(\Rightarrow\left[{}\begin{matrix}a⋮d\\b⋮d\end{matrix}\right.\)
Hơn nữa, vì \(a+b⋮d\) nên nếu \(a⋮d\) thì \(b⋮d\). Nếu \(b⋮d\) thì \(a⋮d\). Như vậy \(a,b⋮d\).
Nhưng do \(a,b\) nguyên tố cùng nhau nên \(d=1\) \(\RightarrowƯCLN\left(a^2,a+b\right)=1\) hay \(a^2,a+b\) nguyên tố cùng nhau.
b) Gọi \(ƯCLN\left(ab,a+b\right)=d\)
\(\Rightarrow\left\{{}\begin{matrix}ab⋮d\\a+b⋮d\end{matrix}\right.\)
Vì a và b nguyên tố cùng nhau nên từ \(ab⋮d\Rightarrow\left[{}\begin{matrix}a⋮d\\b⋮d\end{matrix}\right.\). Đến đây kết hợp với \(a+b⋮d\) và lập luận tương tự như câu a), sẽ chứng minh được \(d=1\)
Bài 12:
a) \(\dfrac{12}{3n-1}\) là một số nguyên khi:
\(12\) ⋮ \(3n-1\)
\(\Rightarrow3n-1\inƯ\left(12\right)=\left\{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12\right\}\)
\(\Rightarrow3n\in\left\{2;0;3;-1;4;-2;5;-3;7;-5;13;-11\right\}\)
\(\Rightarrow n\in\left\{\dfrac{2}{3};0;1;-\dfrac{1}{3};\dfrac{4}{3};-\dfrac{2}{3};\dfrac{5}{3};-1;\dfrac{7}{3};-\dfrac{5}{3};\dfrac{13}{3};-\dfrac{11}{3}\right\}\)
Mà: \(n\in Z\Rightarrow n\in\left\{0;1;-1\right\}\)
b) \(\dfrac{2n+3}{7}\) là một số nguyên khi:
\(2n+3\) ⋮ 7
\(\Rightarrow2n+3\in B\left(7\right)\)
Do \(n\in Z\) nên \(2n+3\) lẻ
\(\Rightarrow2n+3\in B\left(7,\text{lẻ}\right)\)
\(\Rightarrow n\in\dfrac{B\left(7,\text{lẻ}\right)-3}{2}\)
c) \(\dfrac{2n+5}{n-3}=\dfrac{2n-6+11}{n-3}=\dfrac{2\left(n-3\right)+11}{n-3}=2+\dfrac{11}{n-3}\)
Là một số nguyên khi:
11 ⋮ \(n-3\)
\(\Rightarrow n-3\inƯ\left(11\right)=\left\{1;-1;11;-11\right\}\)
\(\Rightarrow n\in\left\{4;2;14;-8\right\}\)
Bài 8:
S = \(\dfrac{3}{10}\) + \(\dfrac{3}{11}\) + \(\dfrac{3}{12}\) + \(\dfrac{3}{13}\) + \(\dfrac{3}{14}\)
\(\dfrac{3}{10}>\dfrac{3}{11}>\dfrac{3}{12}>\dfrac{3}{13}>\dfrac{3}{14}\)
S < \(\dfrac{3}{10}\).5 = \(\dfrac{3}{2}\) = 1 + \(\dfrac{1}{2}\)
S > \(\dfrac{3}{14}.5\) = \(\dfrac{15}{14}\) = 1 + \(\dfrac{1}{14}\)
Vì \(\dfrac{1}{14}\) < \(\dfrac{1}{2}\) nên
1 < \(\dfrac{15}{14}\) < S < \(\dfrac{3}{2}\) < 2
Vậy 1 < S < 2 (đpcm)
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2008}{2009}\)
\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-...-\dfrac{1}{x+1}=\dfrac{2008}{2009}\)
\(1-\dfrac{1}{x+1}=\dfrac{2008}{2009}\)
\(\dfrac{1}{x+1}=\dfrac{1}{2009}\)
\(x+1=2009\)
\(x=2008\)
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dots+\dfrac{1}{x\left(x+1\right)}=\dfrac{2008}{2009}\)
\(\Rightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dots+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2008}{2009}\)
\(\Rightarrow1-\dfrac{1}{x+1}=\dfrac{2008}{2009}\)
\(\Rightarrow\dfrac{1}{x+1}=1-\dfrac{2008}{2009}\)
\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{2009}\)
\(\Rightarrow x+1=2009\)
\(\Rightarrow x=2009-1\)
\(\Rightarrow x=2008\)
\(\dfrac{3}{7}\cdot\dfrac{5}{4}-\dfrac{5}{7}\cdot\dfrac{1}{3}-\dfrac{1}{7}\cdot\dfrac{5}{12}\)
\(=\dfrac{15}{28}-\dfrac{5}{21}-\dfrac{5}{84}\)
\(=\dfrac{25}{84}-\dfrac{5}{84}\)
\(=\dfrac{5}{21}\)
\(\dfrac{4}{7}x-x=-\dfrac{9}{14}\)
\(\left(\dfrac{4}{7}-1\right)x=-\dfrac{9}{14}\)
\(-\dfrac{3}{7}x=-\dfrac{9}{14}\)
\(x=-\dfrac{9}{14}:\dfrac{-3}{7}\)
\(x=\dfrac{3}{2}\)
Bài 2:
d; \(\dfrac{3}{5}\) : (-5) = \(\dfrac{3}{5}\) x \(\dfrac{1}{\left(-5\right)}\) = \(\dfrac{3}{-25}\) = \(\dfrac{-3}{25}\)
e; \(\dfrac{-4}{15}\): 2 = \(\dfrac{-4}{15}\) x \(\dfrac{1}{2}\) = \(\dfrac{-4}{30}\)
f; 24 : \(\dfrac{-6}{7}\) = 24 x \(\dfrac{7}{-6}\) = \(\dfrac{28}{-1}\) = -28
Bài 3:
b; \(\dfrac{-4}{5}\) : \(\dfrac{2}{7}\) + \(\dfrac{4}{7}\)
= \(\dfrac{-4}{5}.\dfrac{7}{2}\) + \(\dfrac{4}{7}\)
= \(\dfrac{-14}{5}\) + \(\dfrac{4}{7}\)
= \(\dfrac{-98}{35}\) + \(\dfrac{20}{35}\)
= \(\dfrac{-78}{35}\)
\(\dfrac{1}{2}x-\dfrac{3}{5}x=-\dfrac{2}{3}\)
\(x\left(\dfrac{1}{2}-\dfrac{3}{5}\right)=-\dfrac{2}{3}\)
\(\dfrac{1}{10}x=-\dfrac{2}{3}\)
\(x=-\dfrac{20}{3}\)
giúp mh vs ah
\(\dfrac{3}{7}-\dfrac{1}{2}x=\dfrac{5}{3}\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{3}{7}-\dfrac{5}{3}\)
\(\Rightarrow x=2\cdot-\dfrac{26}{21}\)
\(\Rightarrow x=\dfrac{-52}{21}\)
______
\(2x-\dfrac{3}{4}=\dfrac{-5}{8}\)
\(\Rightarrow2x=\dfrac{-5}{8}+\dfrac{3}{4}\)
\(\Rightarrow2x=\dfrac{1}{8}\)
\(\Rightarrow x=\dfrac{1}{8}:2\)
\(\Rightarrow x=\dfrac{1}{16}\)
________
\(\dfrac{1}{4}x-\left(-\dfrac{7}{5}\right)=\dfrac{-5}{3}\)
\(\Rightarrow\dfrac{1}{4}x+\dfrac{7}{5}=\dfrac{-5}{3}\)
\(\Rightarrow\dfrac{1}{4}x=\dfrac{-5}{3}-\dfrac{7}{5}\)
\(\Rightarrow x=4\cdot-\dfrac{46}{15}\)
\(\Rightarrow x=-\dfrac{184}{15}\)
______
\(2\dfrac{1}{3}x-\dfrac{3}{4}=1\dfrac{1}{6}\)
\(\Rightarrow\dfrac{7}{3}x-\dfrac{3}{4}=\dfrac{7}{6}\)
\(\Rightarrow\dfrac{7}{3}x=\dfrac{7}{6}+\dfrac{3}{4}\)
\(\Rightarrow\dfrac{7}{3}x=\dfrac{23}{12}\)
\(\Rightarrow x=\dfrac{23}{12}:\dfrac{7}{3}\)
\(\Rightarrow x=\dfrac{23}{28}\)
Bạn làm nốt cho mình bài trên dc ko