Question

Cho biểu thức P=\(\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{x\sqrt{x}-1}\right).\dfrac{3\sqrt{x}-3}{x+\sqrt{x}}\)

Với x>0,x khác 1

a) rút gọn p

b) Tính giá trị của P với x thỏa mãn căn (x-2)=2

c)Tím x để P nguyên

Answer 1

Điều kiên x khác 1, x>0 \(\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{x\sqrt{x}-1}\right).\dfrac{3\sqrt{x}-3}{x+\sqrt{x}}\\ =\dfrac{x\sqrt{x}-1-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x\sqrt{x}-1\right)}.\dfrac{3\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}\left(x-1\right)}{x\sqrt{x}-1}.\dfrac{3}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{x\sqrt{x}-1}\dfrac{3}{\sqrt{x}+1}=\dfrac{3\sqrt{x}-3}{x\sqrt{x}-1}\)

Answer 2

P = \(\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{x\sqrt{x}-1}\right).\dfrac{3\sqrt{x}-3}{x+\sqrt{x}}\)

P = \(\left(\dfrac{1}{\sqrt{x}-1}.\dfrac{3\left(\sqrt{x}-1\right)}{x+\sqrt{x}}\right)-\left(\dfrac{1}{x\sqrt{x}-1}.\dfrac{3\sqrt{x}-3}{x+\sqrt{x}}\right)\)

P = \(\dfrac{3}{x+\sqrt{x}}-\dfrac{3\sqrt{x}-3}{\left(x\sqrt{x}-1\right)\left(x+\sqrt{x}\right)}\)

P = \(\dfrac{3\left(x\sqrt{x}-1\right)-\left(3\sqrt{x}-3\right)}{\left(x\sqrt{x}-1\right)\left(x+\sqrt{x}\right)}\)

P = \(\dfrac{3x\sqrt{x}-3\sqrt{x}}{\left(x\sqrt{x}-1\right)\left(x+\sqrt{x}\right)}\) = \(\dfrac{3\sqrt{x}\left(x-1\right)}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\sqrt{x}}\)

P = \(\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\sqrt{x}}\) = \(\dfrac{3\sqrt{x}\left(\sqrt{x}-1\right)}{\left(x\sqrt{x}-1\right)\sqrt{x}}\)

P = \(\dfrac{3x-3\sqrt{x}}{x^2-\sqrt{x}}\)

Answer 3

a/ ĐKXĐ: \(x\ge0,x\ne1\)

\(P=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{x\sqrt{x}-1}\right).\dfrac{3\sqrt{x}-3}{x+\sqrt{x}}\)

= \(\dfrac{x+\sqrt{x}+1-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{3\left(\sqrt{x}-1\right)}{x+\sqrt{x}}\)

= \(\dfrac{\left(x+\sqrt{x}\right).3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\left(x+\sqrt{x}\right)}\)

= \(\dfrac{3}{x+\sqrt{x}+1}\)

Answer 4

b/

Ta có: \(\sqrt{x-2}=2\left(đk:x\ge2\right)\)

\(\Leftrightarrow x-2=4\)

\(\Leftrightarrow x=6\) ( tmđk)

Thay x=6 vào P ta được:

\(P=\dfrac{3}{6+\sqrt{6}+1}=\dfrac{3}{7+\sqrt{6}}=\dfrac{3\left(7-\sqrt{6}\right)}{\left(7+\sqrt{6}\right)\left(7-\sqrt{6}\right)}\)

= \(\dfrac{21-3\sqrt{6}}{43}\)

Vậy với x thỏa mãn \(\sqrt{x-2}=2\) thì \(P=\dfrac{21-3\sqrt{6}}{43}\)

Answer 5

c/ Với \(x\ge0,x\ne1\)

Để P nhận giá trị nguyên \(\Leftrightarrow\dfrac{3}{x+\sqrt{x}+1}\in Z\Rightarrow x+\sqrt{x}+1\inƯ_{\left(3\right)}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\sqrt{x}+1=3\\x+\sqrt{x}+1=-3\\x+\sqrt{x}+1=1\\x+\sqrt{x}+1=-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x+\sqrt{x}-2=0\\x+\sqrt{x}+4=0\\x+\sqrt{x}=0\\x+\sqrt{x}+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)=0\\\Delta=-15< 0\\\sqrt{x}\left(\sqrt{x}+1\right)=0\\\Delta=-7< 0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)=0\\\sqrt{x}\left(\sqrt{x}+1\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-1=0\\\sqrt{x}+2=0\\\sqrt{x}=0\\\sqrt{x}+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=-2\left(ktm\right)\\\sqrt{x}=0\\\sqrt{x}=-1\left(ktm\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\left(ktmđk\right)\\x=0\left(tmđk\right)\end{matrix}\right.\)

Vậy để P nguyên thì x=0