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Ta có: \(B=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)+5\left(\sqrt{x}+1\right)+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x+2\sqrt{x}-3+5\sqrt{x}+5+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+6}{\sqrt{x}-1}\)
do đó \(P=\frac{\sqrt{x}-1}{\sqrt{x}+1}.\frac{\sqrt{x}-6}{\sqrt{x}-1}=\frac{\sqrt{x}-6}{\sqrt{x}+1}=1-\frac{7}{\sqrt{x}+1}\)
Vì \(x\ge0\Rightarrow0< \frac{7}{\sqrt{x}+1}\le7\)
Để P nguyên thì \(\frac{7}{\sqrt{x}+1}\in Z\)
do đó \(\frac{7}{\sqrt{x}+1}\in\left\{1,2,3,4,5,6,7\right\}\)
Đến đây xét từng TH là ra
rút gọn B ta có B=\(\frac{\sqrt{x}+6}{\sqrt{x}-1}\)\(\Rightarrow\)\(AB=\frac{\sqrt{x}+6}{\sqrt{x}+1}\in Z\)
=\(1+\frac{5}{\sqrt{x}+1}\)
Vì 1\(\in Z\) nên để P thuộc Z thì \(\frac{5}{\sqrt{x}+1}\in Z\)
\(\Rightarrow\left(\sqrt{x}+1\right)\inƯ\left(5\right)=\pm1;\pm5\)
Đến đây thì ez rồi
a/ Ta có: \(x+2\sqrt{x}+1=\left(\sqrt{x}+1\right)^2\)
Và: \(x-1=\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\)
=> \(P=\left[\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\frac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right].\frac{\sqrt{x}+1}{\sqrt{x}}\)
=> \(P=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2.\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
=> \(P=\frac{x+2\sqrt{x}-\sqrt{x}-2-x-\sqrt{x}+2\sqrt{x}+2}{\left(\sqrt{x}+1\right).\left(\sqrt{x}-1\right)}.\frac{1}{\sqrt{x}}=\frac{2\sqrt{x}}{\left(\sqrt{x}+1\right).\left(\sqrt{x}-1\right)}.\frac{1}{\sqrt{x}}\)
=> \(P=\frac{2}{\left(\sqrt{x}+1\right).\left(\sqrt{x}-1\right)}=\frac{2}{x-1}\)
b/ Thay \(x=\frac{\sqrt{3}}{2+\sqrt{3}}\) => \(P=\frac{2}{\frac{\sqrt{3}}{2+\sqrt{3}}-1}=\frac{2\left(2+\sqrt{3}\right)}{\sqrt{3}-2-\sqrt{3}}\)
=> \(P=-\left(2+\sqrt{3}\right)\)
c/ \(P=\frac{2}{x-1}=-\frac{4}{\sqrt{x}+1}\) <=> \(\frac{1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=-\frac{2}{\sqrt{x}+1}\)
<=> \(\frac{1}{\sqrt{x}-1}=-2\)
<=> \(1=-2\sqrt{x}+2\)
<=> \(2\sqrt{x}=1=>\sqrt{x}=\frac{1}{2}=>x=\frac{1}{4}\)
\(C=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{\sqrt{x}+1}-\frac{2}{x-1}\)
\(C=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(C=\frac{\sqrt{x}\left(\sqrt{x}+1\right)-2\left(\sqrt{x}-1\right)-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(C=\frac{x+\sqrt{x}-2\sqrt{x}+2-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(C=\frac{\sqrt{x}\left(\sqrt{x}+1-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(C=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(C=\frac{\sqrt{x}}{\sqrt{x}+1}\)
P/s tham khảo nha
\(A=\left(\sqrt{x}-\frac{x+2}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}-4}{1-x}\right)\) \(ĐKXĐ:x\ge0;x\ne1;x\ne4\)
\(A=\left[\frac{\sqrt{x}\left(\sqrt{x}+1\right)-x-2}{\sqrt{x}+1}\right]:\left[\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}-4}{x-1}\right]\)
\(A=\frac{x+\sqrt{x}-x-2}{\sqrt{x}+1}:\left[\frac{x-\sqrt{x}+\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\)
\(A=\frac{\sqrt{x}-2}{\sqrt{x}+1}:\frac{x-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{\sqrt{x}-2}{\sqrt{x}+1}.\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(A=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
vậy \(A=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
b)theo bài ra: \(A=\frac{1}{\sqrt{x}}\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+2}=\frac{1}{\sqrt{x}}\)
\(\Leftrightarrow\left(\sqrt{x}-1\right).\sqrt{x}=\sqrt{x}+2\)
\(\Leftrightarrow x-\sqrt{x}-\sqrt{x}-2=0\)
\(\Leftrightarrow x-2\sqrt{x}-2=0\)
\(\Leftrightarrow x-2\sqrt{x}+1-3=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2-\left(\sqrt{3}\right)^2=0\)
\(\Leftrightarrow\left(\sqrt{x}-1-\sqrt{3}\right)\left(\sqrt{x}-1+\sqrt{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-1-\sqrt{3}=0\\\sqrt{x}-1+\sqrt{3}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=\sqrt{3}+1\\\sqrt{x}=1-\sqrt{3}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\left(\sqrt{3}+1\right)^2\\x=\left(1-\sqrt{3}\right)^2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3+2\sqrt{3}+1\\x=3-2\sqrt{3}+1\end{cases}}\)
vậy......
\(P=\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{2\sqrt{x}-1}{\sqrt{x}-1}+\frac{x-2}{x-3\sqrt{x}+2}\)
ĐK : \(\hept{\begin{cases}x\ge0\\x\ne1\\x\ne4\end{cases}}\)
\(=\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{2\sqrt{x}-1}{\sqrt{x}-1}+\frac{x-2}{x-\sqrt{x}-2\sqrt{x}+2}\)
\(=\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{2\sqrt{x}-1}{\sqrt{x}-1}+\frac{x-2}{\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{2\sqrt{x}-1}{\sqrt{x}-1}+\frac{x-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}-\frac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}+\frac{x-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-4\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}-\frac{2x-5\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}+\frac{x-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-4\sqrt{x}+3-2x+5\sqrt{x}-2+x-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}=\frac{1}{\sqrt{x}-2}\)
b) Để P < 1
=> \(\frac{1}{\sqrt{x}-2}< 1\)
<=> \(\frac{1}{\sqrt{x}-2}-1< 0\)
<=> \(\frac{1}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}-2}< 0\)
<=> \(\frac{1-\sqrt{x}+2}{\sqrt{x}-2}< 0\)
<=> \(\frac{3-\sqrt{x}}{\sqrt{x}-2}< 0\)
Xét hai trường hợp :
1. \(\hept{\begin{cases}3-\sqrt{x}>0\\\sqrt{x}-2< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}-\sqrt{x}>-3\\\sqrt{x}< 2\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}< 3\\\sqrt{x}< 2\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 9\\x< 4\end{cases}}\Leftrightarrow x< 4\)
2. \(\hept{\begin{cases}3-\sqrt{x}< 0\\\sqrt{x}-2>0\end{cases}}\Leftrightarrow\hept{\begin{cases}-\sqrt{x}< -3\\\sqrt{x}>2\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}>3\\\sqrt{x}>2\end{cases}}\Leftrightarrow\hept{\begin{cases}x>9\\x>4\end{cases}}\Leftrightarrow x>9\)
Kết hợp với ĐK => Với \(\orbr{\begin{cases}x\in\left\{0;2;3\right\}\\x>9\end{cases}}\)thì thỏa mãn đề bài
\(P=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}\)
\(=\frac{x+1+3\sqrt{x}-3-6\sqrt{x}+4}{x-1}\)
=x-2 căn x +1/x-1
=(căn x-1)^2/(căn x-1)(căn x+1)
=căn x-1/căn x+1
b, Để căn x-1/căn x+1
=> căn x-1/căn x+1
=căn +1-2/căn x+1
=(căn x+1/căn x+1)+(-2/căn x+1)
=1+ (-2)/căn +1
=>căn x+1 thuộc Ư(-2)={+-1;+-2}
=> x=0 (loại)
=> x vô lý loại
=> x=1
=> x vô lý loại
Vậy để P nghiệm nguyên =>x=4
\(P=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}\)
\(P=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1}+\frac{3\left(\sqrt{x}-1\right)}{x-1}-\frac{6\sqrt{x}-4}{x-1}\)
\(P=\frac{x+\sqrt{x}}{x-1}+\frac{3\sqrt{x}-3}{x-1}-\frac{6\sqrt{x}-4}{x-1}\)
\(P=\frac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{x-1}\)
\(P=\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(P=\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(P=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
b) Theo câu a) \(P=\frac{\sqrt{x}-1}{\sqrt{x}+1}\) với \(x\ge0;x\ne1\)
có \(P=\frac{\sqrt{x}-1}{\sqrt{x}+1}=\frac{\left(\sqrt{x}+1\right)-2}{\sqrt{x}+1}=1-\frac{2}{\sqrt{x}+1}\)
vì \(1\in Z\)nên để \(P\)nhận giá trị nguyên thì \(\frac{2}{\sqrt{x}+1}\)nhận giá trị nguyên
vì \(x\in Z\)nên \(\sqrt{x}\)có thể là số nguyên hoặc số vô tỉ
- nếu \(\sqrt{x}\)là số vô tỉ thì \(\frac{2}{\sqrt{x}+1}\)ko nhận giá trị nguyên ( Trường hợp này ko xảy ra)
- nếu \(\sqrt{x}\)là số nguyên thì \(\sqrt{x}+1\)nhận giá trị nguyên
để \(\frac{2}{\sqrt{x}+1}\)nhận giá trị nguyên thì \(2⋮\sqrt{x}+1\Leftrightarrow\sqrt{x}+1\inƯ_{\left(2\right)}\Leftrightarrow\sqrt{x}+1\in\left\{-+1;-+2\right\}\)
vì \(\sqrt{x}+1\ge1\forall x\ge0\) nên
\(\sqrt{x}+1\in\left\{1;2\right\}\)
+ \(\sqrt{x}+1=1\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\)
\(\sqrt{x}+1=2\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)
kết hợp với điều kiện \(x\ge0;x\ne1\)và \(x\in Z\)
Ta có \(x=0\)thì \(P\)nhận giá trị nguyên