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\(Q=\frac{2\sqrt{x}}{x-\sqrt{x}+1}\)là một số nguyên
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26 tháng 10 2020
\(\left(\frac{1}{x+2\sqrt{x}}-\frac{1}{\sqrt{x}+2}\right)\div\frac{1-\sqrt{x}}{x+4\sqrt{x}+4}\)
ĐKXĐ : \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)
\(=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}+2\right)}-\frac{1}{\sqrt{x}+2}\right)\div\frac{1-\sqrt{x}}{\left(\sqrt{x}+2\right)^2}\)
\(=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\right)\div\frac{1-\sqrt{x}}{\left(\sqrt{x}+2\right)^2}\)
\(=\frac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\times\frac{\left(\sqrt{x}+2\right)^2}{1-\sqrt{x}}\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}}\)
=> đpcm
12 tháng 8 2021
a, \(A=\left(\frac{1}{\sqrt{x}+2}-\frac{1}{\sqrt{x}-2}\right):\frac{-\sqrt{x}}{x-2\sqrt{x}}\)
\(A=\left(\frac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\frac{-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(A=\frac{\sqrt{x}-2-\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\frac{-\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}}\)
\(A=\frac{4}{\sqrt{x}+2}\)
b, \(A=\frac{4}{\sqrt{x}+2}=\frac{2}{3}\)
=> 2cawn x + 4 = 12
=> 2.căn x = 8
=> căn x = 4
=> x = 16 (thỏa mãn)
c, có A = 4/ căn x + 2 và B = 1/căn x - 2
=> A.B = 4/x - 4
mà AB nguyên
=> 4 ⋮ x - 4
=> x - 4 thuộc Ư(4)
=> x - 4 thuộc {-1;1;-2;2;-4;4}
=> x thuộc {3;5;2;6;0;8} mà x > 0 và x khác 4
=> x thuộc {3;5;2;6;8}
d, giống c thôi
4 tháng 8 2019
\(a,Q=\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}};x>0;x\ne1;x\ne4\)
\(=\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\left(\frac{x-\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}-\frac{x+\sqrt{x}-2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{x-\sqrt{x}+2\sqrt{x}-2-x-\sqrt{x}+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\cdot\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\frac{1}{\sqrt{x}}\)
\(=\frac{2}{x-1}\)
4 tháng 8 2019
\(a,\)\(Q=\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1}\right).\)\(\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\frac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right).\)\(\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+1\right)^2}.\frac{\sqrt{x}+1}{\sqrt{x}}-\frac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{2}{x-1}\)\(\left(đpcm\right)\)
\(b,Q=\frac{2}{x-1}\)
\(Q\in Z\Leftrightarrow\frac{2}{x-1}\in Z\Rightarrow x-1\inƯ_2\)
Mà \(Ư_2=\left\{\pm1;\pm2\right\}\)
TH1 : \(x-1=-1\Rightarrow x=0\)
TH2 : \(x-1=1\Rightarrow x=2\)
TH3 : \(x-1=-2\Rightarrow x=-1\)
TH4 :\(x-1=2\Rightarrow x=3\)
\(\Rightarrow\)x nguyên lớn nhất là 3 để Q là số nguyên
TB
27 tháng 7 2018
a, \(A=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\) (ĐKXĐ: \(x\ne1,x\ge0\))
\(=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)
\(=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)
\(=\frac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
b, \(A-\frac{1}{3}\Leftrightarrow\frac{\sqrt{x}}{x+\sqrt{x}+1}-\frac{1}{3}\)\(=\frac{3\sqrt{x}-x-\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}=\frac{-x+2\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}=-\frac{-\left(x-2\sqrt{x}+1\right)}{3\left(x+\sqrt{x}+1\right)}=-\frac{\left(\sqrt{x}+1\right)^2}{3\left(x+\sqrt{x}+1\right)}< 0\)
\(\Rightarrow A-\frac{1}{3}< 0\Leftrightarrow A< \frac{1}{3}\)
c, ĐKXĐ: \(x\ge0,x\ne1\)
Ta có: x = \(19-8\sqrt{3}\)(TMĐK) \(\Leftrightarrow\sqrt{x}=\sqrt{19-8\sqrt{3}}\Leftrightarrow\sqrt{x}=\sqrt{\left(4-\sqrt{3}\right)^2}\Leftrightarrow\sqrt{x}=4-\sqrt{3}\)
Thay \(\sqrt{x}=4-\sqrt{3}\)vào A ta có:
\(A=\frac{4-\sqrt{3}}{\left(4-\sqrt{3}\right)^2+4-\sqrt{3}+1}=\frac{4-\sqrt{3}}{19-8\sqrt{3}+4-\sqrt{3}+1}=\frac{4-\sqrt{3}}{24-9\sqrt{3}}\)
Vậy với \(x=19-8\sqrt{3}\)thì \(A=\frac{4-\sqrt{3}}{24-9\sqrt{3}}\)