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a. \(P=\dfrac{3a+\sqrt{9a}-3}{a+\sqrt{a}-2}-\dfrac{\sqrt{a}+1}{\sqrt{a}+2}+\dfrac{\sqrt{a}-2}{1-\sqrt{a}}\) \(\left(a\ge0,a\ne1\right)\)
\(P=\dfrac{3a+3\sqrt{a}-3-\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)
\(P=\dfrac{3a+3\sqrt{a}-3-a+1-a+4}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)
\(P=\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}=\dfrac{a+\sqrt{a}+2\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)+2\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)
\(P=\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}=\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\)
b. \(P=\dfrac{\sqrt{a}+1}{\sqrt{a}-1}=\dfrac{\sqrt{a}-1+2}{\sqrt{a}-1}=1+\dfrac{2}{\sqrt{a}-1}\)
Để: \(P\in Z\Rightarrow\sqrt{a}-1\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{a}=2\\\sqrt{a}=0\\\sqrt{a}=3\\\sqrt{a}=-1\left(l\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=4\\a=0\\a=9\end{matrix}\right.\)
Vậy: \(a\in\left\{4;0;9\right\}\)
a =4 .bạn xem MÌNH trả lời câu hỏi của NGUYỄN THỊ DIỆP
ĐKXĐ: \(a\ne1\)
a/ P = \(\dfrac{3a+3\sqrt{a}-3-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)+\sqrt{a}-1-\left(a+\sqrt{a}-2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)
= \(\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)
= \(\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)
= \(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\)
b/ \(|P|=1\Leftrightarrow P=\pm1\)
* Với P = 1 thì \(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}=1\Leftrightarrow\sqrt{a}+1=\sqrt{a}-1\) (loại)
* Với P = -1 thì \(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}=-1\Leftrightarrow\sqrt{a}+1=1-\sqrt{a}\Leftrightarrow2\sqrt{a}=0\Leftrightarrow a=0\left(tm\right)\)
c/ P = \(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}=\dfrac{\sqrt{a}-1+2}{\sqrt{a}-1}=1+\dfrac{2}{\sqrt{a}-1}\)
Để P \(\in N\) thì \(2⋮\sqrt{a}-1\) \(\Leftrightarrow\sqrt{a}-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)
Đối chiếu với đk: a \(\ne1\) ta thấy a = 0; 4 và 9
Vậy để P \(\in N\) thì a = 0; a = 4; a = 9.
Bài 1
a) \(P=\frac{3a+\sqrt{9a}-3}{a+\sqrt{a}-2}-\frac{\sqrt{a}+1}{\sqrt{a}+2}+\frac{\sqrt{a}-2}{1-\sqrt{a}}\) (ĐK : x\(\ge0\) ; x\(\ne\) 1)
\(=\frac{3a+\sqrt{9a}-3}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\frac{\sqrt{a}+1}{\sqrt{a}+2}-\frac{\sqrt{a}-2}{\sqrt{a}-1}\)
\(=\frac{3a+\sqrt{9a}-3-\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{3a+\sqrt{9a}-3-a+1-a+4}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{\sqrt{a}+1}{\sqrt{a}-1}\)
b) \(P=\frac{\sqrt{a}+1}{\sqrt{a}-1}=\frac{\sqrt{a}-1+2}{\sqrt{a}-1}=1+\frac{2}{\sqrt{a}-1}\)
Vậy để P là số nguyên thì: \(\sqrt{a}-1\inƯ\left(2\right)\)
Mà Ư(2)={-1;1;2;-1}
=> \(\sqrt{a}-1\in\left\{1;-1;2;-2\right\}\)
Ta có bảng sau:
| \(\sqrt{a}-1\) | 1 | -1 | 2 | -2 |
| a | 4 | 0 | 9 | \(\sqrt{a}=-1\) (ktm) |
vậy a={0;4;9} thì P nguyên
Bài 2
\(P=\frac{\sqrt{a+4\sqrt{a-4}}+\sqrt{a-4\sqrt{a-4}}}{\sqrt{1-\frac{8}{a}+\frac{16}{a^2}}}\)(ĐK:a\(\ge\)8)
\(=\frac{\sqrt{\left(a-4\right)+4\sqrt{a-4}+4}+\sqrt{\left(a-4\right)-4\sqrt{a-4}+4}}{\sqrt{\left(1-\frac{4}{a}\right)^2}}\)
\(=\frac{\sqrt{\left(\sqrt{a-4}+2\right)^2}+\sqrt{\left(\sqrt{a-4}-2\right)^2}}{1-\frac{4}{a}}\)
\(=\sqrt{a-4}+2+\sqrt{a-4}-2:\frac{a-4}{a}\)
\(=2\sqrt{a-4}\cdot\frac{a}{a-4}\)
\(=\frac{2a}{\sqrt{a-4}}\)
Ta có \(\left(\sqrt{a}+2\right)\left(1-\sqrt{a}\right)=a+\sqrt{a}-2\)
\(=\frac{3\text{a}+3\sqrt{a}-3}{a+\sqrt{a}-2}-\frac{\sqrt{a}+1}{\sqrt{a}+2}-\frac{\sqrt{a}-2}{\sqrt{a}-1}\)
\(=\frac{3\text{a}+3\sqrt{a}-3-a+1+a-4}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{3\text{a}+3\sqrt{a}-6}{a+\sqrt{a}-2}\)
\(=\frac{3\left(a+\sqrt{a}-2\right)}{a+\sqrt{a}-2}\)
\(=3\)
b/ Ta có 3 là số nguyên nên biểu thức P luôn nguyên với mọi x
TICK CHO MÌNH NHA
a) điều kiện \(a\ge0;a\ne1\)
\(P=\dfrac{\sqrt{a}-2}{1-\sqrt{a}}-\dfrac{1+\sqrt{a}}{2+\sqrt{a}}+\dfrac{3a-3+\sqrt{9a}}{a+\sqrt{a}-2}\)
\(P=\dfrac{2-\sqrt{a}}{\sqrt{a}-1}-\dfrac{1+\sqrt{a}}{2+\sqrt{a}}+\dfrac{3a-3+3\sqrt{a}}{\left(\sqrt{a}-1\right)\left(2+\sqrt{a}\right)}\)
\(P=\dfrac{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)-\left(1+\sqrt{a}\right)\left(\sqrt{a}-1\right)+3a-3+3\sqrt{a}}{\left(\sqrt{a}-1\right)\left(2+\sqrt{a}\right)}\)
\(P=\dfrac{4-a-\left(a-1\right)+3a-3+3\sqrt{a}}{\left(\sqrt{a}-1\right)\left(2+\sqrt{a}\right)}\)
\(P=\dfrac{4-a-a+1+3a-3+3\sqrt{a}}{\left(\sqrt{a}-1\right)\left(2+\sqrt{a}\right)}\) \(=\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(2+\sqrt{a}\right)}\)
\(P=\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}=\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\)
b) điều kiện \(a\in Z;x\ge0;x\ne1\)
ta có : \(P=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1+2}{\sqrt{x}-1}=1+\dfrac{2}{\sqrt{x}-1}\) nguyên
\(\Leftrightarrow\sqrt{x}-1\) thuộc ước của 2 là : \(\pm1;\pm2\)
ta có : * \(\sqrt{x}-1=1\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tmđk\right)\)
* \(\sqrt{x}-1=-1\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\left(tmđk\right)\)
* \(\sqrt{x}-1=2\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\left(loại\right)\)
\(\sqrt{x}-1=-2\Leftrightarrow\sqrt{x}=-1\left(vôlí\right)\)
vậy \(x=4;x=0\)
1: \(P=\dfrac{3a+3\sqrt{a}-3-a+1-a+4}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}=\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\)
2: Để P nguyên thì \(\sqrt{a}-1+2⋮\sqrt{a}-1\)
\(\Leftrightarrow\sqrt{a}-1\in\left\{1;-1;2\right\}\)
hay \(a\in\left\{4;0;9\right\}\)