Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có \(x\inℕ\Rightarrow x\ge0\Rightarrow x+1\ge1>0\Rightarrow\frac{1}{x+1}\le\frac{1}{1}=1\)
Dấu = xảy ra \(\Leftrightarrow x=0\left(tm\right)\)
Vậy GTLN 1/x+1 =1 tại x=0
Để \(\frac{3n+4}{n-1}\)là số nguyên thì:
\(3n+4⋮n-1\)
Mà \(3\left(n-1\right)⋮n-1\)
nên \(3n+4-3\left(n-1\right)⋮n-1\\ \Rightarrow7⋮n-1\)
\(\Rightarrow\left(n-1\right)\inƯ\left(7\right)=\left\{1;-1;7;-7\right\}\)
\(\Rightarrow n\in\left\{2;0;8;-6\right\}\)
Bài kia bạn nhân 3n+1 lên 2 lần rồi làm tương tự
\(\dfrac{x+3}{x-1}=\dfrac{x-1+4}{x-1}=\dfrac{x-1}{x-1}+\dfrac{4}{x-1}=1+\dfrac{4}{x-1}\)
Để đạt GT nguyên thì \(\dfrac{4}{x-1}\in Z\)
\(\Rightarrow x-1\inƯ_{\left(4\right)}=\left\{-4;-2;-1;1;2;4\right\}\\ \Rightarrow x\in\left\{-3;-1;0;2;3;5\right\}\)
\(\dfrac{x-1+4}{x-1}=1+\dfrac{4}{x-1}\Rightarrow x-1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
x-1 | 1 | -1 | 2 | -2 | 4 | -4 |
x | 2 | 0 | 3 | -1 | 5 | -3 |
x + 3 chia hết x - 1
x + 3 - ( x - 1 ) chia hết x - 1
2 chia hết x - 1
Do đó x - 1 thuộc Ư (2) = ( 1,-1,2,-2)
x - 1 = 1 suy ra x = 2
x - 1 = -1 suy ra x = 0
x - 1 = 2 suy ra x = 3
x - 1 = -2 suy ra x = -1
Vậy x = 2, 0, 3, -1
Khó vãi lìn.Ai mà giải được,toán lớp 6cow màaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
\(\frac{x+6}{x+1}=\frac{x+1+5}{x+1}=1+\frac{5}{x+1}\in N\)
\(\Leftrightarrow x+1\inƯ\left(5\right)\Leftrightarrow x+1\in\left\{1;5\right\}\)
\(\Leftrightarrow x\in\left\{0;4\right\}\)