a) \(=\dfrac{\sqrt{x}+1+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}\)
b) \(M=\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}.\dfrac{\sqrt{x}+1}{2\sqrt{x}+1}+\dfrac{x-\sqrt{x}-5}{\sqrt{x}+3}=\dfrac{x-2}{\sqrt{x}+3}=\dfrac{x-9+7}{\sqrt{x}+3}\)
\(M=\sqrt{x}-3+\dfrac{7}{\sqrt{x}+3}\)
M nguyên \(< =>\sqrt{x}+3\inƯ\left(7\right)=>\sqrt{x}+3=7< =>\sqrt{x}=4< =>x=16\)
a: \(A=\dfrac{\sqrt{x}+1+\sqrt{x}}{x-1}:\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{2\sqrt{x}+1}{x-1}\cdot\dfrac{\sqrt{x}-1}{1}=\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}\)
b: \(M=1+\dfrac{x-\sqrt{x}-5}{\sqrt{x}+3}=\dfrac{x-\sqrt{x}-5+\sqrt{x}+3}{\sqrt{x}+3}=\dfrac{x-2}{\sqrt{x}+3}\)
Để M là số nguyên thì \(x-9+7⋮\sqrt{x}+3\)
\(\Leftrightarrow\sqrt{x}+3\in\left\{1;-1;7;-7\right\}\)
=>căn x+3=7
=>x=16
a) A =\(\left[\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]:\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}-1}\right)\)
A =\(\dfrac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{1}{\sqrt{x}-1}\)
A =\(\dfrac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)\)
A =\(\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}\)
b) M =\(\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}.\dfrac{\sqrt{x}+1}{2\sqrt{x}+1}+\dfrac{x-\sqrt{x}-5}{\sqrt{x}+3}\)
M =\(1+\dfrac{x-\sqrt{x}-5}{\sqrt{x}+3}\)
M =\(\dfrac{\sqrt{x}+3}{\sqrt{x}+3}+\dfrac{x-\sqrt{x}-5}{\sqrt{x}+3}\)
M =\(\dfrac{x-2}{\sqrt{x}+3}\)
M =\(\dfrac{x-9+7}{\sqrt{x}+3}\)
M =\(\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)+7}{\sqrt{x}+3}\)
M =\(\sqrt{x}-3+\dfrac{7}{\sqrt{x}+3}\)
Để \(M\in Z\) ⇒\(\left\{{}\begin{matrix}\sqrt{x}\in Z\\\sqrt{x}+3\inƯ\left(7\right)\end{matrix}\right.\)
⇒\(\left\{{}\begin{matrix}x\in\left\{0;1;4;9;16;...\right\}\\\sqrt{x}+3\in\left\{-7;-1;1;7\right\}\end{matrix}\right.\)
Vì \(\sqrt{x}+3\ge3\)
⇒ \(\sqrt{x}+3=7\)
⇔ \(\sqrt{x}=4\)
⇔ \(x=16\)
Vậy để M nguyên thì \(x=16\)
