Question

P=\(\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\) (\(x\ge0;x\ne1\))

a) rút gọn P

b) tìm \(x\in Z\)sao cho \(P\in Z\)

Answer 1

Mình bị nhầm

b) \(P=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1+2}{\sqrt{x}-1}=1+\dfrac{2}{\sqrt{x}-1}\)

Để P\(\in Z\) thì \(\sqrt{x}-1\inƯ\left(2\right)\in\left\{\pm1;\pm2\right\}\)

Vì \(\sqrt{x}-1\ge-1\)

Vậy \(\sqrt{x}-1\in\left\{\pm1;2\right\}\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\sqrt{x}-1=-1\\\sqrt{x}-1=2\\\sqrt{x}-1=1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=0\left(tm\right)\\x=4\left(tm\right)\\x=9\left(tm\right)\end{matrix}\right.\)

Vậy x=0, x=4,x=9 thì P\(\in Z\)

Answer 2

a)

\(=\dfrac{3x+3\sqrt{x}-3-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\) với \(x\ge0;x\ne1\)

b)

P = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=1+\dfrac{2}{\sqrt{x}-1}\)

Vì 1 \(\in Z\) nên

Để P \(\in\) Z thì \(2⋮\sqrt{x}-1=>\sqrt{x}-1\in\) Ư₍₂₎ = { -2;-1;1;2 }

=> \(\sqrt{x}\) = { -1;0;2;3 }

=> x ={0;4;9} thỏa mãn đkxđ

Vậy, ...............

Answer 3

a) \(P=\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}=\dfrac{3x+3\sqrt{x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}=\dfrac{3x+3\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{3x+3\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{x-1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{x+\sqrt{x}+2\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

b) Ta có \(P=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1+2}{\sqrt{x}-1}=1+\dfrac{2}{\sqrt{x}-1}\)

Để P\(\in Z\) thì \(\sqrt{x}+1\inƯ\left(2\right)\in\left\{\pm1;\pm2\right\}\)

Mà \(\sqrt{x}+1>0\) nên \(\sqrt{x}+1\in\left\{1;2\right\}\Leftrightarrow\)\(\left[{}\begin{matrix}\sqrt{x}+1=1\\\sqrt{x}+1=2\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\left(tm\right)\\x=1\left(ktm\right)\end{matrix}\right.\)

Vậy x=0 thì P\(\in Z\)