Cho \(A=\frac{5\sqrt{x}}{2\sqrt{x}+1}\)\(\left(x>0;x\ne4\right)\)
Tìm x sao cho A nhận giá trị là một số nguyên.
Helppppppppp..... Plzzzzzzzzzzzz
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
16 tháng 10 2020
1: Rút gọn biểu thức
a) Ta có: \(5\sqrt{\frac{1}{5}}+\frac{1}{3}\sqrt{45}+\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(=5\cdot\frac{1}{\sqrt{5}}+\frac{1}{3}\cdot3\sqrt{5}+\left|2-\sqrt{5}\right|\)
\(=\sqrt{5}+\sqrt{5}+\sqrt{5}-2\)(Vì \(2< \sqrt{5}\))
\(=3\sqrt{5}-2\)
b) Ta có: \(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\)
\(=\frac{\left(5+\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}+\frac{\left(5-\sqrt{5}\right)^2}{\left(5+\sqrt{5}\right)\left(5-\sqrt{5}\right)}\)
\(=\frac{30+10\sqrt{5}+30-10\sqrt{5}}{25-5}\)
\(=\frac{60}{20}=3\)
2:
Sửa đề: \(A=\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
a) Ta có: \(A=\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{x-1-\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}\)
\(=\frac{\sqrt{x}-2}{3\sqrt{x}}\)
b) ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\notin\left\{1;4;\frac{14\pm6\sqrt{5}}{4}\right\}\end{matrix}\right.\)
Để \(A>\frac{1}{6}\) thì \(A-\frac{1}{6}>0\)
\(\Leftrightarrow\frac{\sqrt{x}-2}{3\sqrt{x}}-\frac{1}{6}>0\)
\(\Leftrightarrow\frac{2\sqrt{x}-4}{6\sqrt{x}}-\frac{\sqrt{x}}{6\sqrt{x}}>0\)
\(\Leftrightarrow\frac{\sqrt{x}-4}{6\sqrt{x}}>0\)
mà \(6\sqrt{x}>0\forall x\) thỏa mãn ĐKXĐ
nên \(\sqrt{x}-4>0\)
\(\Leftrightarrow\sqrt{x}>4\)
hay x>16
Kết hợp ĐKXĐ, ta được: x>16
Vậy: Để \(A>\frac{1}{6}\)thì x>16
KT
18 tháng 7 2018
a) \(A=\left(\sqrt{6}+\sqrt{10}\right).\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{2}\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=2\sqrt{2}\)
\(B=\frac{1}{\sqrt{x}-2}-\frac{1}{\sqrt{x}+2}+1\)
\(=\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+1\)
\(=\frac{4}{x-4}+1\)
\(=\frac{4}{x-4}+\frac{x-4}{x-4}=\frac{x}{x-4}\)
MC
21 tháng 8 2019
\(A=\left(\frac{2}{\sqrt{x}-2}+\frac{3}{2\sqrt{x}+1}-\frac{5\sqrt{x}-7}{2x-3\sqrt{x}-2}\right):\)\(\frac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)
\(=\left(\frac{2}{\sqrt{x}-2}+\frac{3}{2\sqrt{x}+1}-\frac{5\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\right)\)\(:\frac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)
\(=\frac{2\left(2\sqrt{x}+1\right)+3\left(\sqrt{x}-2\right)-5\sqrt{x}+7}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\)\(:\frac{2\sqrt{x}+3}{5\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\)\(.\frac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\frac{2\sqrt{x}+3}{2\sqrt{x}+1}.\frac{5\sqrt{x}}{2\sqrt{x}+3}=\frac{5\sqrt{x}}{2\sqrt{x}+1}\)
\(A\in Z\Leftrightarrow\frac{5\sqrt{x}}{2\sqrt{x}+1}\in Z\Leftrightarrow\frac{10\sqrt{x}}{2\sqrt{x}+1}\in Z\)
\(\Rightarrow\frac{10\sqrt{x}+5-5}{2\sqrt{x}+1}\in Z\Leftrightarrow5-\frac{5}{2\sqrt{x}+1}\in Z\)
\(\Rightarrow\frac{5}{2\sqrt{x}+1}\in Z\Rightarrow2\sqrt{x}+1\inƯ_5\)
Mà \(Ư_5=\left\{\pm1;\pm5\right\}\)
Nhưng \(2\sqrt{x}+1\ge1\)
\(\Rightarrow\orbr{\begin{cases}2\sqrt{x}+1=1\\2\sqrt{x}+1=5\end{cases}\Rightarrow\orbr{\begin{cases}2\sqrt{x}=0\\2\sqrt{x}=4\end{cases}}}\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}=2\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}}\)
Vậy \(x\in\left\{0;4\right\}\)