Bài 1: 

a: \(Q=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\left(x+\sqrt{x}\right)\)

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

\(=\dfrac{2x}{x-1}\)

a: \(Q=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}+2}-3=\sqrt{x}-3\)

b: Để \(Q=2\) thì \(\sqrt{x}=5\)

hay x=25

a: \(Q=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\cdot\left(x+\sqrt{x}\right)\)

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

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

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

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

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

b: Để Q là số nguyên thì \(2x⋮x-1\)

=>\(2x-2+2⋮x-1\)

=>\(2⋮x-1\)

=>\(x-1\in\left\{1;-1;2;-2\right\}\)

=>\(x\in\left\{2;0;3;-1\right\}\)

Kết hợp ĐKXĐ, ta được: \(x\in\left\{0;2;3\right\}\)

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:

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)

      \(=\frac{\sqrt{\left(a-4\right)+4\sqrt{a-4}+4}+\sqrt{\left(a-4\right)-4\sqrt{a-4}+4}}{\sqrt{\left(1-\frac{4}{a}\right)^2}}\)

     \(=\frac{\sqrt{\left(\sqrt{a-4}+2\right)^2}+\sqrt{\left(\sqrt{a-4}-2\right)^2}}{1-\frac{4}{a}}\)

      \(=\sqrt{a-4}+2+\sqrt{a-4}-2:\frac{a-4}{a}\)

     \(=2\sqrt{a-4}\cdot\frac{a}{a-4}\)

     \(=\frac{2a}{\sqrt{a-4}}\)

a: ĐKXĐ: x>=0; x<>25

Sửa đề: \(Q=\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{10\sqrt{x}}{x-25}-\dfrac{5}{\sqrt{x}+5}\)

\(=\dfrac{x+5\sqrt{x}-10\sqrt{x}-5\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\dfrac{x-10\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\dfrac{\sqrt{x}-5}{\sqrt{x}+5}\)

b: Q=-3/7

=>\(\dfrac{\sqrt{x}-5}{\sqrt{x}+5}=-\dfrac{3}{7}\)

=>7căn x-35=-3căn x-15

=>10căn x=20

=>x=4

c: Q nguyên

=>căn x+5-10 chia hết cho căn x+5

=>căn x+5 thuộc {5;10}

=>căn x thuộc {0;5}

Kết hợp ĐKXĐ, ta được: x=0

a) \(Q=\dfrac{\sqrt[]{x}}{\sqrt[]{x}-5}-\dfrac{10\sqrt[]{x}}{x-25}-\dfrac{5}{\sqrt[]{x}-5}\left(1\right)\)

Q có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x-25\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne25\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow Q=\dfrac{\sqrt[]{x}-5}{\sqrt[]{x}-5}-\dfrac{10\sqrt[]{x}}{x-25}\)

\(\Leftrightarrow Q=1-\dfrac{10\sqrt[]{x}}{x-25}\)

\(\Leftrightarrow Q=\dfrac{x+10\sqrt[]{x}-25}{x-25}\)

\(\Leftrightarrow Q=\dfrac{\left(\sqrt[]{x}-5\right)^2}{\left(\sqrt[]{x}-5\right)\left(\sqrt[]{x}+5\right)}=\dfrac{\sqrt[]{x}-5}{\sqrt[]{x}+5}\)

b) \(Q=-\dfrac{3}{7}\)

\(\Leftrightarrow\dfrac{\sqrt[]{x}-5}{\sqrt[]{x}+5}=-\dfrac{3}{7}\)

\(\Leftrightarrow7\left(\sqrt[]{x}-5\right)=-3\left(\sqrt[]{x}+5\right)\)

\(\Leftrightarrow7\sqrt[]{x}-35=-3\sqrt[]{x}-15\)

\(\Leftrightarrow10\sqrt[]{x}=20\)

\(\Leftrightarrow\sqrt[]{x}=2\Leftrightarrow x=4\)

c) \(Q\in Z\Leftrightarrow\dfrac{\sqrt[]{x}-5}{\sqrt[]{x}+5}\in Z\) \(\left(x\in Z^+\right)\)

\(\Leftrightarrow\sqrt[]{x}-5⋮\sqrt[]{x}+5\)

\(\Leftrightarrow\sqrt[]{x}-5-\left(\sqrt[]{x}-5\right)⋮\sqrt[]{x}-5\)

\(\Leftrightarrow\sqrt[]{x}-5-\sqrt[]{x}-5⋮\sqrt[]{x}+5\)

\(\Leftrightarrow-10⋮\sqrt[]{x}+5\)

\(\Leftrightarrow\sqrt[]{x}+5\in U\left(10\right)=\left\{1;2;5;10\right\}\)

\(\Leftrightarrow x\in\left\{0;25\right\}\)

a) ĐKXĐ: \(x>0;x\ne4\)

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

\(=\left[\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\right]:\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

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

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

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

b) Để biểu thức \(Q\) có giá trị âm thì \(\dfrac{3\sqrt{x}}{\sqrt{x}-2}< 0\)

\(\Rightarrow\sqrt{x}-2< 0\) (vì \(3\sqrt{x}>0\forall x>0;x\ne4\))

\(\Leftrightarrow\sqrt{x}< 2\Leftrightarrow0\le x< 4\) 

Kết hợp với điều kiện xác định của \(x\), ta được: \(0< x< 4\)

\(\text{#}\mathit{Toru}\)

bn ơi thiếu đề hả?

\(A=\frac{a}{a-1}-\frac{a}{a+1}+\frac{2}{a^2-1}\left(ĐK:a\ne\pm1\right)\)

\(=\frac{a\left(a+1\right)-a\left(a-1\right)}{\left(a-1\right)\left(a+1\right)}+\frac{2}{a^2-1}\)

\(=\frac{a^2+a-a^2+a+2}{a^2-1}=\frac{2}{a-1}\left(Q.E.D\right)\)

Để A nguyên suy ra 2/a-1 nguyên

\(< =>2⋮a-1< =>a\in\left\{2;3;-1;0\right\}\)

Để \(A\ge1< =>\frac{2}{a-1}\ge1< =>2\ge a-1< =>a\le3\)

mấy bài khác để từ từ mình làm dần hoặc bạn khác làm

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