A= \(\frac{x-2}{\sqrt{x}+1}\) với x \(\ge\)0. Tìm x \(\in\)Z để C \(\in\)Z
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BA
30 tháng 7 2019
\(ĐKXĐ:\)
\(\hept{\begin{cases}x-9\ne0\\\sqrt{x}-2\ne0\\\sqrt{x}+3\ne0;x\ge0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ne9\\x\ne4\\x\ge0\end{cases}}\)
Vậy...................................................
BA
30 tháng 7 2019
\(A=\left(\frac{x-3\sqrt{x}}{x-9}-1\right):\left(\frac{9-x}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}+3}\right)\)
\(=\left(\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-1\right):\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}+3}\right)\)
\(=\frac{\sqrt{x}-\sqrt{x}-3}{\left(\sqrt{x}+3\right)}:\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{-3}{\sqrt{x}+3}:\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\frac{x-4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{-3}{\sqrt{x}+3}:\frac{9-x+x-9-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-3}{\sqrt{x}+3}:\frac{-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-3}{\sqrt{x}+3}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{4-x}\)
\(=\frac{3\left(2-\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
\(=\frac{3}{\left(2+\sqrt{x}\right)}\)
18 tháng 1 2018
\(M=\frac{\sqrt{x}-2}{\sqrt{x}+1}\) \(ĐKXĐ:x\ne1\)
\(M=\frac{\sqrt{x}+1-3}{\sqrt{x}+1}\)
\(M=1-\frac{3}{\sqrt{x}+1}\)
để \(x\in Z\)thì \(M\in Z\)
mà \(1\in Z\) nên \(\frac{3}{\sqrt{x}+1}\in Z\)
\(\Leftrightarrow\sqrt{x}+1\inƯ\left(3\right)\)
\(\Leftrightarrow\sqrt{x}+1\in\left\{\pm1;\pm3\right\}\)
+ \(\sqrt{x}+1=-1\)
\(\Leftrightarrow\sqrt{x}=-2\Leftrightarrow x\in\varnothing\)
+ \(\sqrt{x}+1=3\)
\(\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\) ( thỏa mãn )
+ \(\sqrt{x}+1=-3\)
\(\Leftrightarrow\sqrt{x}=-4\Leftrightarrow x\in\varnothing\)
vậy \(x=4\)
29 tháng 6 2019
Để \(A=\frac{5}{\sqrt{x}-1}\)nguyên thì \(5⋮\sqrt{x}-1\)
\(\Rightarrow\sqrt{x}-1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{2;0;6;-4\right\}\)
\(\Leftrightarrow x\in\left\{4;0;36\right\}\)( thỏa )
Để \(B=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-3\)
\(\Leftrightarrow\sqrt{x}-3+4⋮\sqrt{x}-3\)
Vì \(\sqrt{x}-3⋮\sqrt{x}-3\)
\(\Rightarrow4⋮\sqrt{x}-3\)
\(\Rightarrow\sqrt{x}-3\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{4;2;5;1;7;-1\right\}\)
\(\Leftrightarrow x\in\left\{16;4;25;1;49\right\}\)
Máy kia tương tự đi ăn cơm đây :>
HT
29 tháng 6 2019
Mk lm nốt câu C
\(C=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+2}{\sqrt{x}-1}=1+\frac{3}{\sqrt{x}-1}\)
Để C nguyên\(\Leftrightarrow\sqrt{x}-1\inƯ_{\left(3\right)}=\left\{\pm3;\pm1\right\}\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-1=3\left(tm\right)\\\sqrt{x}-1=-3\left(l\right)\\\sqrt{x}-1=-1\left(tm\right)\\\sqrt{x}-1=1\left(tm\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=16\\x=0\\x=4\end{matrix}\right.\)
16 tháng 11 2017
a/ \(\frac{x-2}{x+2\sqrt{x}}-\frac{1}{\sqrt{x}}+\frac{2}{\sqrt{x}+2}\)
\(=\frac{x-2}{x+2\sqrt{x}}-\frac{\sqrt{x}+2}{x+2\sqrt{x}}+\frac{2\sqrt{x}}{x+2\sqrt{x}}\)
\(=\frac{x+\sqrt{x}-4}{x+2\sqrt{x}}\)
b/ \(\frac{x+\sqrt{x}-4}{x+2\sqrt{x}}=\frac{4+2\sqrt{3}+\sqrt{\left(\sqrt{3}+1\right)^2}-4}{4+2\sqrt{3}+2\sqrt{\left(\sqrt{3}+1\right)^2}}\)
\(=\frac{4+2\sqrt{3}+\sqrt{3}+1-4}{4+2\sqrt{3}+2\sqrt{3}+2}=\frac{1+3\sqrt{3}}{6+4\sqrt{3}}\)
ĐK: $x\ge 0$
$A=\dfrac{x-2}{\sqrt x+1}\\=\dfrac{x-1-1}{\sqrt x+1}\\=\dfrac{(\sqrt x-1)(\sqrt x+1)-1}{\sqrt x+1}\\=\sqrt x-1-\dfrac{1}{\sqrt x+1}$
Để $A$ nguyên thì $x$ phải là số chính phương và $1\vdots \sqrt x+1$ hay $\sqrt x+1\in Ư(1)=\{\pm 1\}$
mà $\sqrt x+1\ge 1(x\ge 0)$
$\to \sqrt x+1=1$
$\Leftrightarrow \sqrt x=0\\\Leftrightarrow x=0(TM)$
Vậy $x=0$ thì $A$ nguyên