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a; \(\dfrac{6}{x}\) < \(\dfrac{x}{7}\) < \(\dfrac{8}{x}\)
vì \(x\) \(\in\) N* ta có: 6.7 < \(x.x\) < 7.8
42 < \(x^2\) < 56
\(x^2\) = 49
\(x\) = \(\pm\) 7
Vì \(x\) \(\in\) N*; \(x\) = 7
b; \(\dfrac{x}{11}\) < \(\dfrac{12}{x}\) < \(\dfrac{x}{9}\)
9.12< \(x^2\) < 11.12
108 < \(x^2\) < 132
\(x^2\) = 121
\(\left[{}\begin{matrix}x=-11\\x=11\end{matrix}\right.\)
Vì \(x\in\) N*
\(x\) = 11
\(\left|x-\dfrac{2}{5}\right|-\dfrac{1}{2}=\dfrac{1}{3}.\dfrac{1}{4}-\dfrac{1}{5}\)
\(\Rightarrow\left|x-\dfrac{2}{5}\right|-\dfrac{1}{2}=\dfrac{-7}{60}\)
\(\Rightarrow\left|x-\dfrac{2}{5}\right|=\dfrac{23}{60}\)
\(\Rightarrow x-\dfrac{2}{5}=\dfrac{23}{60}\) hoặc \(x-\dfrac{2}{5}=\dfrac{-23}{60}\)
\(\Rightarrow x=\dfrac{47}{60}\) hoặc \(x=\dfrac{1}{60}\)
Vậy \(x\in\left\{\dfrac{47}{60};\dfrac{1}{60}\right\}\)
a) Gọi ƯCLN(n + 3;n + 4) = d
=> \hept{n+3⋮dn+4⋮d⇒n+4−(n+3)⋮d⇒1⋮d⇒d=1\hept{n+3⋮dn+4⋮d⇒n+4−(n+3)⋮d⇒1⋮d⇒d=1
=> n + 3 ; n + 4 là 2 số nguyên tố cùng nhau
=> n+3n+4n+3n+4là phân số tối giản
b) Gọi ƯCLN(3n + 3 ; 9n + 8) = d
Ta có : \hept{3n+3⋮d9n+8⋮d⇒\hept⎧⎨⎩3(3n+3)⋮d9n+8⋮d⇒\hept{9n+9⋮d9n+8⋮d⇒9n+9−(9n+8)⋮d⇒1⋮d⇒d=1\hept{3n+3⋮d9n+8⋮d⇒\hept{3(3n+3)⋮d9n+8⋮d⇒\hept{9n+9⋮d9n+8⋮d⇒9n+9−(9n+8)⋮d⇒1⋮d⇒d=1
=> 3n + 3 ; 9n + 8 là 2 số nguyên tố cùng nhau
=> 3n+39n+83n+39n+8phân số tối giản
a) Gọi ƯCLN(n + 3;n + 4) = d
=> \hept{n+3⋮dn+4⋮d⇒n+4−(n+3)⋮d⇒1⋮d⇒d=1\hept{n+3⋮dn+4⋮d⇒n+4−(n+3)⋮d⇒1⋮d⇒d=1
=> n + 3 ; n + 4 là 2 số nguyên tố cùng nhau
=> n+3n+4n+3n+4là phân số tối giản
b) Gọi ƯCLN(3n + 3 ; 9n + 8) = d
Ta có : \hept{3n+3⋮d9n+8⋮d⇒\hept⎧⎨⎩3(3n+3)⋮d9n+8⋮d⇒\hept{9n+9⋮d9n+8⋮d⇒9n+9−(9n+8)⋮d⇒1⋮d⇒d=1\hept{3n+3⋮d9n+8⋮d⇒\hept{3(3n+3)⋮d9n+8⋮d⇒\hept{9n+9⋮d9n+8⋮d⇒9n+9−(9n+8)⋮d⇒1⋮d⇒d=1
=> 3n + 3 ; 9n + 8 là 2 số nguyên tố cùng nhau
=> 3n+39n+83n+39n+8phân số tối giản
1) \(2\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\)
\(\Leftrightarrow2\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}+\dfrac{3}{2}\)
\(\Leftrightarrow2\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}\)
\(\Leftrightarrow\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}:2\)
\(\Leftrightarrow\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{8}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=-\dfrac{7}{8}\\\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=-\dfrac{7}{8}+\dfrac{1}{3}\\\dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=-\dfrac{13}{24}\\\dfrac{1}{2}x=\dfrac{29}{24}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\left(-\dfrac{13}{24}\right):\dfrac{1}{2}\\x=\dfrac{29}{24}:\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{13}{12}\\x=\dfrac{29}{12}\end{matrix}\right.\)
2) \(\dfrac{3}{4}-2\left|2x-\dfrac{2}{3}\right|=2\)
\(\Leftrightarrow2\left|2x-\dfrac{2}{3}\right|=\dfrac{3}{4}-2\)
\(\Leftrightarrow2\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{8}\)
\(\Leftrightarrow\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{8}:2\)
\(\Leftrightarrow\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{2}{3}=\dfrac{-5}{16}\\2x-\dfrac{2}{3}=\dfrac{5}{16}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{-5}{16}+\dfrac{2}{3}\\2x=\dfrac{5}{16}+\dfrac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{17}{48}\\2x=\dfrac{47}{48}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{48}:2\\x=\dfrac{47}{48}:2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{96}\\x=\dfrac{47}{96}\end{matrix}\right.\)
Ta có:
\(A=\dfrac{3n+2}{n-1}=\dfrac{\left(3n-3\right)+5}{n-1}=\dfrac{3n-3}{n-1}+\dfrac{5}{n-1}=\dfrac{3\left(n-1\right)}{n-1}+\dfrac{5}{n-1}=3+\dfrac{5}{n-1}\)
Để \(A\in Z\Rightarrow\dfrac{5}{n-1}\in Z\Rightarrow5⋮n-1\) hay \(n-1\in U\left(5\right)=\left\{\pm1;\pm2\right\}\)
Lập bảng giá trị:
| \(n-1\) | \(1\) | \(-1\) | \(5\) | \(-5\) |
| \(n\) | \(2\) | \(0\) | \(6\) | \(-4\) |
Vậy với \(n\in\left\{-4;0;2;6\right\}\) thì \(\dfrac{3n+2}{n-1}\in Z\)
Để \(A\in Z\) thì \(3n+2⋮n-1\)
\(\Rightarrow3\left(n-1\right)+5\) \(⋮n-1\)
Vì \(3\left(n-1\right)⋮n-1\)
\(\Rightarrow5⋮n-1\)
\(\Rightarrow n-1\inƯ\left(5\right)\)
mà \(Ư\left(5\right)=\left\{\pm1;\pm5\right\}\)
Ta có bảng sau:
| \(n-1\) | 1 | \(-1\) | 5 | \(-5\) |
| \(n\) | 2 | 0 | 6 | \(-4\) |
| Kết luận | nhận | nhận | nhận | nhận |
Vậy \(n\in\left\{-4;0;2;6\right\}\).
a: 17/200>17/314
b: 11/54=22/108<22/37
c: 141/893=3/19
159/901=3/17
mà 3/19<3/17
nên 141/893<159/901
\(\dfrac{-11}{-32}>\dfrac{16}{49}\)
\(\dfrac{-2020}{-2021}>\dfrac{-2021}{2022}\)
giải thích giúp mik dc ko ạ