tìm x nguyên để \(\sqrt{x}+1\) \(⋮\) \(\sqrt{x}-3\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
19 tháng 10 2021
a: Ta có: \(A=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}-1\)
\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-4-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-1\)
\(=\dfrac{x-2\sqrt{x}-x+1}{x-1}\)
\(=\dfrac{-2\sqrt{x}+1}{x-1}\)
14 tháng 1 2022
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\notin\left\{4;1\right\}\end{matrix}\right.\)
Ta có: \(A=\dfrac{x-4\sqrt{x}+3-\left(2x-4\sqrt{x}-\sqrt{x}+2\right)+x+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{2x-4\sqrt{x}+5-2x+5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
17 tháng 12 2023
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\notin\left\{1;4\right\}\end{matrix}\right.\)
\(A=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x-2}{x-3\sqrt{x}+2}\)
\(=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)-\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x-4\sqrt{x}+3-2x+5\sqrt{x}-2+x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}=\dfrac{1}{\sqrt{x}-2}\)
b: Để A>2 thì A-2>0
=>\(\dfrac{1-2\left(\sqrt{x}-2\right)}{\sqrt{x}-2}>0\)
=>\(\dfrac{5-2\sqrt{x}}{\sqrt{x}-2}>0\)
=>\(\dfrac{2\sqrt{x}-5}{\sqrt{x}-2}< 0\)
TH1: \(\left\{{}\begin{matrix}2\sqrt{x}-5>0\\\sqrt{x}-2< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\sqrt{x}>\dfrac{5}{2}\\\sqrt{x}< 2\end{matrix}\right.\)
=>\(x\in\varnothing\)
TH2: \(\left\{{}\begin{matrix}2\sqrt{x}-5< 0\\\sqrt{x}-2>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\sqrt{x}< \dfrac{5}{2}\\\sqrt{x}>2\end{matrix}\right.\)
=>\(2< \sqrt{x}< \dfrac{5}{2}\)
=>4<x<25/4
c: Để A là số nguyên thì \(1⋮\sqrt{x}-2\)
=>\(\sqrt{x}-2\in\left\{1;-1\right\}\)
=>\(\sqrt{x}\in\left\{3;1\right\}\)
=>\(x\in\left\{1;9\right\}\)
kết hợp ĐKXĐ, ta được: x=9
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 1 2024
\(P=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{-\left(\sqrt{x}-3\right)}{\sqrt{x}+1}=\dfrac{3}{\sqrt{x}+3}\)
\(P\in Z\Rightarrow\sqrt{x}+3=Ư\left(3\right)=\left\{-3;-1;1;3\right\}\)
Mà \(\sqrt{x}+3\ge3;\forall x\ge0\)
\(\Rightarrow\sqrt{x}+3=3\)
\(\Rightarrow\sqrt{x}=0\Rightarrow x=0\)
17 tháng 10 2021
a: Ta có: \(A=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\dfrac{15\sqrt{x}-11-\left(3x+7\sqrt{x}-6\right)-\left(2x+\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
DK
2
HP
27 tháng 8 2021
\(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\dfrac{4}{\sqrt{x}-3}\)
A nguyên khi và chỉ khi:
\(\sqrt{x}-3\inƯ_4=\left\{\pm1;\pm2;\pm4\right\}\)
\(\Leftrightarrow\sqrt{x}\inƯ_4=\left\{1;2;4;5;7\right\}\)
\(\Leftrightarrow x\inƯ_4=\left\{1;4;16;25;49\right\}\)
27 tháng 8 2021
Để \(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\) là số nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-3\)
\(\Leftrightarrow\sqrt{x}-3\in\left\{1;-1;2;-2;4\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{4;2;5;1;7\right\}\)
hay \(x\in\left\{16;4;25;1;49\right\}\)
23 tháng 8 2021
Bài 1:
Để biểu thức nhận giá trị nguyên thì \(3\sqrt{x}+1⋮2\sqrt{x}-1\)
\(\Leftrightarrow6\sqrt{x}+2⋮2\sqrt{x}-1\)
\(\Leftrightarrow2\sqrt{x}-1\in\left\{1;-1;5\right\}\)
\(\Leftrightarrow2\sqrt{x}\in\left\{2;0;6\right\}\)
hay \(x\in\left\{4;0;36\right\}\)
15 tháng 8 2021
1: Ta có: \(A=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}-9-\left(x-9\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
Để \(A=-\dfrac{1}{\sqrt{x}}\) thì \(x+\sqrt{x}=-\sqrt{x}+3\)
\(\Leftrightarrow x+2\sqrt{x}-3=0\)
\(\Leftrightarrow\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow x=1\left(nhận\right)\)
2: Để A nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-3\)
\(\Leftrightarrow\sqrt{x}-3\in\left\{-1;1;2;-2;4;-4\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{2;4;5;1;7\right\}\)
\(\Leftrightarrow x\in\left\{16;25;1;49\right\}\)
\(\sqrt{x}+1⋮\sqrt{x-3}\left(đk:x\ge0\right)\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)-\left(\sqrt{x}-3\right)⋮\sqrt{x}-3\)
\(\Leftrightarrow4⋮\sqrt{x-3}\)
\(\Leftrightarrow4⋮\sqrt{x-3}\Rightarrow\sqrt{x-3}\inƯ\left(4\right)\)
\(\Leftrightarrow\sqrt{x-3}\in\left\{1;2;3;4;-1;-2;4\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{4;5;7;2;1;-1\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{4;5;7;2;1\right\}\)
\(\Leftrightarrow x\in\left\{16;25;49;4;1\right\}\)
P/S: Bn loại các TH x thuộc Z ko t/m nhé
Để \(\sqrt{x}+1⋮\sqrt{x}-3\)
\(\Leftrightarrow\sqrt{x}-3+4⋮\sqrt{x}-3\)
\(\Leftrightarrow4⋮\sqrt{x}-3\)
\(\Leftrightarrow\sqrt{x}-3\inƯ\left(4\right)\)
\(\Leftrightarrow\sqrt{x}-3\in\left\{-1,1,-2,2,-4,4\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{2,4,1,5,-1,7\right\}\)
Mà : \(x\inℤ\Rightarrow\sqrt{x}\) phải là một số chính phương
\(\Rightarrow\sqrt{x}\in\left\{4,1\right\}\)
\(\Leftrightarrow x\in\left\{2,1\right\}\)