Tìm x sao cho \(\dfrac{3\sqrt{x}}{\sqrt{x}-3}\) là số nguyên
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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
19 tháng 10 2021
a: Ta có: \(A=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}-1\)
\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-4-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-1\)
\(=\dfrac{x-2\sqrt{x}-x+1}{x-1}\)
\(=\dfrac{-2\sqrt{x}+1}{x-1}\)
TT
1
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 3 2023
\(A=\dfrac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\dfrac{4}{\sqrt{x}-3}\)
Để A nguyên \(\Rightarrow4⋮\left(\sqrt{x}-3\right)\Rightarrow\sqrt{x}-3=Ư\left(4\right)\)
Mà \(\sqrt{x}-3\ge-3\Rightarrow\sqrt{x}-3=\left\{-2;-1;1;2;4\right\}\)
\(\Rightarrow\sqrt{x}=\left\{1;2;4;5;7\right\}\)
\(\Rightarrow x=\left\{1;4;16;25;49\right\}\)
AM
26 tháng 12 2021
\(A=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{3-11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)\(A=\dfrac{2x-6\sqrt{x}+x+\sqrt{x+}3\sqrt{x}+3+3-11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)\(A=\dfrac{3x-13\sqrt{x}+6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
T
12 tháng 9 2023
\(a,P=B:A\)
\(=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\right):\left(\dfrac{2}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}\right)\left(ĐKXĐ:x\ge0;x\ne9\right)\)
\(=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\right):\left[\dfrac{2\left(\sqrt{x}+3\right)+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]\)
\(=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\right):\left[\dfrac{3\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\cdot\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{3\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+3}{3}\)
\(b,\) Để \(P=\dfrac{\sqrt{x}+3}{3}\) có giá trị nguyên
thì \(\sqrt{x}+3⋮3\)
\(\Leftrightarrow\sqrt{x}+3\in B\left(3\right)\)
\(\Leftrightarrow\sqrt{x}\in B\left(3\right)\)
Kết hợp với điều kiện, ta được:
\(P\) nguyên khi \(x=m^2\left(m\in Z;m⋮3;m\ne3\right)\)
#Toru
12 tháng 9 2023
a:
ĐKXĐ: x>=0; x<>9
\(A=\dfrac{2\sqrt{x}+6+\sqrt{x}-3}{\left(x-9\right)}=\dfrac{3\sqrt{x}+3}{x-9}\)
\(P=B:A=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\cdot\dfrac{x-9}{3\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+3}{3}\)
b: P nguyên khi \(\sqrt{x}+3⋮3\)
=>\(\sqrt{x}\in B\left(3\right)\)
=>\(x=k^2\left(k\in Z;k⋮3\right)\)
T
30 tháng 8 2023
Ta có: \(P=A\cdot B\) (ĐK: \(x>0;x\ne4\))
\(=\left(\dfrac{3\sqrt{x}-6}{x-2\sqrt{x}}+\dfrac{\sqrt{x}-3}{\sqrt{x}}-\dfrac{1}{2-\sqrt{x}}\right)\left(\dfrac{\sqrt{x}-2}{\sqrt{x}+9}\right)\)
\(=\left[\dfrac{3\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-3}{\sqrt{x}}+\dfrac{1}{\sqrt{x}-2}\right]\left(\dfrac{\sqrt{x}-2}{\sqrt{x}+9}\right)\)
\(=\left(\dfrac{3+\sqrt{x}-3}{\sqrt{x}}+\dfrac{1}{\sqrt{x}-2}\right)\left(\dfrac{\sqrt{x}-2}{\sqrt{x}+9}\right)\)
\(=\left(1+\dfrac{1}{\sqrt{x}-2}\right)\left(\dfrac{\sqrt{x}-2}{\sqrt{x}+9}\right)\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}+9}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+9}\)
Với x > 0; x ≠ 4 thì \(\sqrt{P}< \dfrac{1}{3}\Leftrightarrow P< \dfrac{1}{9}\)
\(\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+9}< \dfrac{1}{9}\)
\(\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+9}-\dfrac{1}{9}< 0\)
\(\Leftrightarrow\dfrac{9\left(\sqrt{x}-1\right)}{9\left(\sqrt{x}+9\right)}-\dfrac{\sqrt{x}+9}{9\left(\sqrt{x}+9\right)}< 0\)
\(\Leftrightarrow\dfrac{9\sqrt{x}-9-\sqrt{x}-9}{9\sqrt{x}+81}< 0\)
\(\Leftrightarrow\dfrac{8\sqrt{x}-18}{9\sqrt{x}+18}< 0\)
Ta thấy: \(9\sqrt{x}+18>0\forall x\)
\(\Rightarrow8\sqrt{x}-18< 0\)
\(\Rightarrow\sqrt{x}< \dfrac{18}{8}\)
\(\Rightarrow\sqrt{x}< \dfrac{9}{4}\Leftrightarrow x< \dfrac{81}{16}\)
Kết hợp với điều kiện, ta được: \(0< x\le5\)\(;x\ne4\)
\(\Rightarrow x\in\left\{1;2;3;5\right\};x\in Z\) thì \(\sqrt{P}< \dfrac{1}{3}\)
#Urushi
Lời giải:
\(\frac{3\sqrt{x}}{\sqrt{x}-3}=\frac{3(\sqrt{x}-3)+9}{\sqrt{x}-3}=3+\frac{9}{\sqrt{x}-3}\)
Để biểu thức đã cho nguyên thì $\frac{9}{\sqrt{x}-3}$ nguyên. Đặt $\frac{9}{\sqrt{x}-3}=t$ với $t$ nguyên.
$\sqrt{x}=\frac{9}{t}+3$
Do $\sqrt{x}\geq 0$ nên $\frac{9}{t}+3\geq 0\Leftrightarrow \frac{3(3+t)}{t}\geq 0$
$\Leftrightarrow t>0$ hoặc $t\leq -3$
$x=(\frac{9}{t}+3)^2$ với $t$ là số nguyên thỏa mãn $t>0$ hoặc $t\leq -3$
A =\(\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)
A = \(\dfrac{3\sqrt{x}-3+3}{\sqrt{x}-3}\)= 3 -\(\dfrac{3}{\sqrt{x}-3}\)
Để A nguyên thì : 3 \(⋮\)\(\sqrt{x}\) - 3
\(\sqrt{x}\) - 3 \(\in\) \(\left\{-3;-1;1;3\right\}\)
\(\sqrt{x}\) - 3 = -3 \(\Rightarrow\) \(x\) = 0
\(\sqrt{x}\) - 3 = -1 \(\Rightarrow\) \(x\) = 4
\(\sqrt{x}\) - 3 = 1 \(\Rightarrow\) \(x\) = 16
\(\sqrt{x}\) - 3 = 3 \(\Rightarrow\) \(x\)= 36
kết luận \(x\)\(\in\) \(\left\{0;4;16;36\right\}\)