Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:
a) \(x=\frac{a+1}{a+9}=\frac{a+9-8}{a+9}=\frac{a+9}{a+9}-\frac{8}{a+9}=1-\frac{8}{a+9}\)
Để \(x\in Z\)thì \(a+9\inƯ\left(8\right)=\left\{-8;-4;-2;-1;1;2;4;8\right\}\)
Vậy \(a\in\left\{-17;-13;-11;-10;-8;-7;-5;-1\right\}\)
b) \(x=\frac{a-1}{a+4}=\frac{a+4-5}{a+4}=\frac{a+4}{a+4}-\frac{5}{a+4}=1-\frac{5}{a+4}\)
Để \(x\in Z\)thì \(a+4\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)
Vậy \(a\in\left\{-9;-5;-3;1\right\}\)
Bài 2:
a) \(t=\frac{3x-8}{x-5}=\frac{3x-15}{x-5}+\frac{7}{x-5}=\frac{3\left(x-5\right)}{x-5}+\frac{7}{x-5}=3+\frac{7}{x-5}\)
Để \(t\in Z\)thì \(x-5\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)
Vậy \(x\in\left\{-2;4;6;12\right\}\)
b)\(q=\frac{2x+1}{x-3}=\frac{2x-6}{x-3}+\frac{7}{x-3}=\frac{2\left(x-3\right)}{x-3}+\frac{7}{\left(x-3\right)}=2+\frac{7}{x-3}\)
Để \(q\in Z\)thì \(x-3\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)
Vậy \(x\in\left\{-4;2;4;10\right\}\)
c)\(p=\frac{3x-2}{x+3}=\frac{3x+9}{x+3}-\frac{11}{x+3}=\frac{3\left(x+3\right)}{x+3}-\frac{11}{x+3}=3-\frac{11}{x+3}\)
Để \(p\in Z\)thì \(x+3\inƯ\left(11\right)=\left\{-11;-1;1;11\right\}\)
Vậy \(x\in\left\{-14;-4;-2;8\right\}\)
Bài 3:
Gọi \(d\inƯC\left(2m+9;14m+62\right)\)
\(\Rightarrow\hept{\begin{cases}\left(2m+9\right)⋮d\\\left(14m+62\right)⋮d\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}7\left(2m+9\right)⋮d\\\left(14m+62\right)⋮d\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\left(14m+63\right)⋮d\\\left(14m+62\right)⋮d\end{cases}}\)
\(\Rightarrow\left[\left(14m+63\right)-\left(14m+62\right)\right]⋮d\)
\(\Rightarrow1⋮d\)
\(\Rightarrow d=1\)
\(\RightarrowƯC\left(2m+9;14m+62\right)=1\)
Vậy \(x=\frac{2m+9}{14m+62}\)là p/s tối giản
a) Ta có:
\(x=\frac{a-5}{a}=\frac{a}{a}-\frac{5}{a}=1-\frac{5}{a}\)
Để x nguyên thì \(\frac{5}{a}\)nguyên
=> 5 chia hết cho a
=> \(a\inƯ\left(5\right)\)
=> \(a\in\left\{1;-1;5;-5\right\}\)
b) 5.3x+2 - 2.3x-2 = 3627
=> 3x-2.(5.34 - 2.1) = 3627
=> 3x-2.(5.81 - 2) = 3627
=> 3x-2.(405 - 2) = 3627
=> 3x-2.403 = 3627
=> 3x-2 = 3627 : 403
=> 3x-2 = 9 = 32
=> x - 2 = 2
=> x = 2 + 2 = 4
\(t=\frac{3x-8}{x-5}=\frac{3x-15+7}{x-5}=3+\frac{7}{x-5}\)
\(t\in Z\Rightarrow7⋮\left(x-5\right)\)
\(\Rightarrow x-5\in\left(1;7;-1;-7\right)\)
\(\Rightarrow x\in\left(6;12;4;-2\right)\)
Theo bài ra ,ta có:
t=\(\frac{3x-8}{x-5}\) =\(\frac{3x-15+7}{x-5}\) =\(3+\frac{7}{x-5}\)
để t \(\in\)Z thì 7\(⋮\) x-5
\(\Rightarrow\)x-5\(\in\)Ư(7)={-1;1;-7;7}
\(\Rightarrow\)x\(\in\)(-2;4;6;12)
Vậy x\(\in\)(-2;4;6;12)
x=(x-3)/(2a)
=>x2a=x-3
=>x2a-x=-3
=>x(2a-1)=-3
Vì -3;x là số nguyên => 2a-1 cũng là số nguyên=>x;2a-1 thuộc U(-3)={+-1;+-3}
Ta có bảng:
x | 1 | -1 | 3 | -3 |
2a-1 | -3 | 3 | -1 | 1 |
a | -1 | 2 | 0 | 1 |
TM | TM | TM | TM |
Vậy........
1.
a) m > 2011
b) m<2011
c) m =2011
2.
a) \(m< \frac{-11}{20}\)
b)\(m>\frac{-11}{20}\)
3. -101 chia hết cho (a+7)
4. (3x-8) chia hết cho (x-5)
5. đề sai, N chứ ko phải n, tui ngu như con bòoooooooooooooooooooooo
5) Gọi \(d\inƯC\left(2m+9;14m+62\right)\)
\(\Rightarrow\hept{\begin{cases}\left(2m+9\right)⋮d\\\left(14m+62\right)⋮d\end{cases}\Rightarrow\hept{\begin{cases}7\left(2m+9\right)⋮d\\\left(14m+62\right)⋮d\end{cases}\Rightarrow}\hept{\begin{cases}\left(14m+63\right)⋮d\\\left(14m+62\right)⋮d\end{cases}}}\)
\(\Rightarrow\left(14m+63\right)-\left(14m+62\right)⋮d\)
\(\Rightarrow1⋮d\)
\(\Rightarrow d=\left\{-1;1\right\}\)
\(\RightarrowƯC\left(2m+9;14m+62\right)=\left\{-1;1\right\}\)
Vậy \(x=\frac{2m+9}{14m+62}\)là p/s tối giản (Vì tử và mẫu của p/s có ƯC là 1)
Bài 1 :
x < 0 \(\Leftrightarrow\) 3a - 5 < -2 \(\Leftrightarrow\) 3a < 3 \(\Leftrightarrow\) a < 1
Bài 2 :
a) \(\frac{3a-5}{a}=3+\frac{5}{a}\in Z\)\(\Leftrightarrow a\inƯ\left(5\right)\)
\(\Leftrightarrow a\in\left\{-5;-1;1;5\right\}\)
b) \(\frac{2b-7}{b+2}=\frac{2b+4-11}{b+2}=2-\frac{11}{b+2}\in Z\) \(\Leftrightarrow b+2\inƯ\left(11\right)\)
\(\Leftrightarrow b+2\in\left\{-11;-1;1;11\right\}\)
\(\Leftrightarrow b\in\left\{-13;-3;-1;9\right\}\)
Để X là số nguyên thì 3 phải chia hết cho 2a-1
=> 2a-1 E Ư(3) = { -1,-3,1,3}
=> a = { 0 ;-1; 1;2}
Vậy a = 0;1;-1;2
X nguyên => 3/(2a+1) => 2a+1 thuộc ước 3 => 2a+1 thuộc {1;3;-1;-3}
(1) 2a+1=1 => a=0 (thả mãn)
(2) 2a+1=3 => a=1 (thả mãn)
(3) 2a+1=-1 => a=-1 (thả mãn)
(4) 2a+1=-3 => a=-2 (thả mãn)
vậy ....
\(B=\dfrac{x-10}{x-5}\in Z\left(x\ne5\right)\)
\(\Rightarrow x-10⋮x-5\)
\(\Rightarrow x-10-\left(x-5\right)⋮x-5\)
\(\Rightarrow x-10-x+5⋮x-5\)
\(\Rightarrow-5⋮x-5\)
\(\Rightarrow x-5\in U\left(5\right)=\left\{-1;1;-5;5\right\}\)
\(\Rightarrow x\in\left\{4;6;0;10\right\}\)