Bài 1 :

a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)

\(A=\left(1-\frac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)

\(\Leftrightarrow A=\frac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}:\frac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(\Leftrightarrow A=\frac{1}{\sqrt{x}+1}:\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(\Leftrightarrow A=\frac{1}{\sqrt{x}+1}:\frac{1}{\sqrt{x}-2}\)

\(\Leftrightarrow A=\frac{\sqrt{x}-2}{\sqrt{x}+1}\)

b) Để \(A< -1\)

\(\Leftrightarrow\frac{\sqrt{x}-2}{\sqrt{x}+1}< -1\)

\(\Leftrightarrow\sqrt{x}-2< -\sqrt{x}-1\)

\(\Leftrightarrow2\sqrt{x}< 1\)

\(\Leftrightarrow\sqrt{x}< \frac{1}{2}\)

\(\Leftrightarrow x< \frac{1}{4}\)

Vậy để \(A< -1\Leftrightarrow x< \frac{1}{4}\)

a. ĐK: \(x\ge0,x\ne49\)

\(M=\frac{3\left(\sqrt{x}+7\right)-\left(\sqrt{x}-7\right)}{\left(\sqrt{x}-7\right)\left(\sqrt{x}+7\right)}:\frac{2\sqrt{x}+6}{x-49}\)

\(=\frac{2\sqrt{x}+28}{x-49}.\frac{x-49}{2\sqrt{x}+6}=\frac{2\sqrt{x}+28}{2\sqrt{x}+6}\)

b. M nguyên \(\Leftrightarrow\frac{2\sqrt{x}+28}{2\sqrt{x}+6}\in Z\Rightarrow\frac{2\sqrt{x}+6+22}{2\sqrt{x}+6}\in Z\Rightarrow1+\frac{22}{2\sqrt{x}+6}\in Z\Rightarrow\frac{22}{2\sqrt{x}+6}\in Z\Rightarrow\left(2\sqrt{x}+6\right)\inƯ\left(22\right)\)

Đến đây đã rất dễ dàng rồi nhé ^^

đề không cho tìm x NGUYÊN để m nguyên mà chỉ tìm các điểm x để  m nguyên thôi

Hồ Thị Hải Yến: Đúng rồi em, ta chỉ cần tìm x để Z nguyên thôi, x không cần nguyên. Chú ý một điều là \(2\sqrt{x}+6\ge6\) nên e chỉ cần chú ý các ước lớn hơn 6 của 22 thôi nhé :)

\(\(A=\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1}\right):\frac{\sqrt{x}}{\sqrt{x}+1}\left(x\ge0;x\ne1\right)\)\)

\(\(=\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\frac{\sqrt{x}}{\sqrt{x}+1}\)\)

\(\(=\frac{\left(\sqrt{x}-1\right).\left(\sqrt{x}+2\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}:\frac{\sqrt{x}}{\sqrt{x}+1}\)\)

\(\(=\frac{x+2\sqrt{x}-\sqrt{x}-2-\left(x+\sqrt{x}-2\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}:\frac{\sqrt{x}}{\sqrt{x}+1}\)\)

\(=\frac{x+2\sqrt{x}-\sqrt{x}-2-x-\sqrt{x}+2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}:\frac{\sqrt{x}}{\sqrt{x}+1}\)

\(=\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\sqrt{x}+1}{\sqrt{x}}\)

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

Vậy \(A=\frac{2}{x-1}vs\left(x\ge0;x\ne1\right)\)

_Ko chắc , đag bận nên còn phần b , tí mk giải nối_

_Minh ngụy_

\(ĐK:x\ge0;x\ne1\)

\(a,A=\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1}\right):\frac{\sqrt{x}}{\sqrt{x}+1}\)

\(=\left(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right):\frac{\sqrt{x}}{\sqrt{x}+1}\)

\(=\left(\frac{x-\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}-\frac{x+\sqrt{x}-2\sqrt{x}-2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right):\frac{\sqrt{x}}{\sqrt{x}+1}\)

\(=\frac{x-\sqrt{x}+2\sqrt{x}-2-x-\sqrt{x}+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)

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

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

Vậy với \(x\ge0;x\ne1\)thì \(A=\frac{2}{x-1}\)

\(b,\)Ta có:\(A=\frac{2}{x-1}\)

Để A nhận giá trị nguyên \(\Leftrightarrow2⋮x-1\)

Vì \(x\in Z\Rightarrow x-1\inƯ_{\left(2\right)}=\left\{\pm1;\pm2\right\}\)

Ta có bảng sau:

Vậy để A nhận giá trị nguyên \(x\in\left\{2;0;3\right\}\)

ô, tôi nhớ là đã sửa kết quả thành \(\frac{2}{x-1}\)và chỗ sai dấu từ 30 mấy phút trc mà sao ở đây vẫn chưa đổi??

b) Ta có :\(A=\frac{2}{x-1}\)

_____-Làm tiếp như " bạn j có cái tên ngược là đc_____

_Minh ngụy_

a) Đk \(x>0\)và \(x\ne4\)

=\(\left(\frac{\sqrt{x}-2+\sqrt{x}+2}{x-4}\right)\).\(\frac{\sqrt{x}-2}{\sqrt{x}}\)

=\(\frac{2\sqrt{x}}{x-4}\).\(\frac{\sqrt{x}-2}{\sqrt{x}}\)

=\(\frac{2}{\sqrt{x}+2}\)

b) Để \(\frac{2}{\sqrt{x}+2}>\frac{1}{2}\)

\(\Leftrightarrow\frac{4-\sqrt{x}-2}{2\left(\sqrt{x}+2\right)}\)\(>0\)

\(\Leftrightarrow\frac{-\sqrt{x}+2}{2\left(\sqrt{x}+2\right)}\)\(>0\)

Vì \(2\left(\sqrt{x}+2\right)>0\)

mà\(\frac{-\sqrt{x}+2}{2\left(\sqrt{x}+2\right)}\)\(>0\)

nên \(-\sqrt{x}+2>0\)\(\Leftrightarrow x< 4\)

Vậy vs \(0< x< 4\)thì \(A>\frac{1}{2}\)

c) Để B=\(\frac{14}{3\left(\sqrt{x}+2\right)}\)đạt giá trị nguyên thì

\(3\left(\sqrt{x}+2\right)\inƯ\left(14\right)=\left(1;2;7\right)\)

+)\(3\left(\sqrt{x}+2\right)=1\)\(\Leftrightarrow\sqrt{x}+2=\frac{1}{3}\)\(\Rightarrow\)vô lý

+)\(3\left(\sqrt{x}+2\right)=2\)\(\Leftrightarrow\sqrt{x}+2=\frac{2}{3}\)\(\Rightarrow\)vô lý

+)\(3\left(\sqrt{x}+2\right)=7\)\(\Leftrightarrow\sqrt{x}+2=\frac{7}{3}\)\(\Leftrightarrow\sqrt{x}=\frac{1}{3}\)\(\Rightarrow x=\frac{1}{9}\)(TM đk)

Vậy vs \(x=\frac{1}{9}\)thì B có giá trị nguyên

Bài 1 : Với : \(x>0;x\ne1\)

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

Thay vào ta được : \(P=x=25\)

Bài 2 : 

a, Với \(x\ge0;x\ne1\)

\(A=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{\sqrt{x}+1}-\frac{2}{x-1}=\frac{x+\sqrt{x}-2\sqrt{x}+2-2}{x-1}\)

\(=\frac{x-\sqrt{x}}{x-1}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}}{\sqrt{x}+1}\)

Thay x = 9 vào A ta được : \(\frac{3}{3+1}=\frac{3}{4}\)

Bài 3 : \(x\ge0;x\ne1\)

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

\(=\left(\frac{2+\sqrt{x}}{x-1}\right).\left(\sqrt{x}+1\right)=\frac{\sqrt{x}+2}{\sqrt{x}-1}\)

b, Ta có : \(P=\frac{\sqrt{x}+2}{\sqrt{x}-1}=\frac{5}{4}\Rightarrow4\sqrt{x}+8=5\sqrt{x}-5\)

\(\Leftrightarrow\sqrt{x}=13\Leftrightarrow x=169\)(tmđk )

ĐK để phân thức XĐ : x khác 1 và x> 0

 Đặt \(B=\left(\frac{\left(\sqrt{x}+2\right)\left(x-1\right)-\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+1\right)}{\left(x+2\sqrt{x}+1\right)\left(x-1\right)}\right)\) ( Đây là mình vừa đặt vừa làm mẫu thức chung nhe)

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

=>\(B=\frac{2\sqrt{x}+2x}{\left(\sqrt{x}+1\right)^2\left(x-1\right)}=\frac{2\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(x-1\right)}=\frac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-1\right)}\)

A = \(B:\frac{\sqrt{x}}{\sqrt{x+1}}=\frac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-1\right)}\cdot\frac{\sqrt{x}+1}{\sqrt{x}}=\frac{2}{x-1}\)

B, Bạn tự làm ý B nhe

HD để A nguyên => x - 1 thuộc ước của 2 mà 2 có các ước là +-1 và +-2

(+) với x-1 = 2 => x = 3

............................

1/ 

a/ ĐKXĐ: \(x\ge0\) và \(x\ne\frac{1}{9}\)

 b/  \(P=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right]:\left(\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)

    \(=\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}.\frac{3\sqrt{x}+1}{3}\)

      \(=\frac{3x+3\sqrt{x}}{3\sqrt{x}-1}.\frac{1}{3}=\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)

c/ \(P=\frac{6}{5}\Rightarrow\frac{x+\sqrt{x}}{3\sqrt{x}-1}=\frac{6}{5}\Rightarrow6\left(3\sqrt{x}-1\right)=5\left(x+\sqrt{x}\right)\)

                  \(\Rightarrow5x-13\sqrt{x}+6=0\Rightarrow\left(5\sqrt{x}-3\right)\left(\sqrt{x}-2\right)=0\)

                   \(\Rightarrow\orbr{\begin{cases}\sqrt{x}=\frac{3}{5}\\\sqrt{x}=2\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{9}{25}\\x=4\end{cases}}}\)

                                                      Vậy x = 9/25 , x = 4

1) a) ĐKXĐ :  \(0\le x\ne\frac{1}{9}\)

b) \(P=\left(\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}\right):\left(1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)

\(=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}-\frac{3\sqrt{x}-1}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}+\frac{8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right]:\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\)

\(=\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\frac{3\sqrt{x}+1}{3}=\frac{3x+3\sqrt{x}}{3\left(3\sqrt{x}-1\right)}=\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)

c) \(P=\frac{6}{5}\Leftrightarrow18\sqrt{x}-6=5x+5\sqrt{x}\Leftrightarrow5x-13\sqrt{x}+6=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{9}{25}\\x=4\end{cases}}\)

2)a) \(P=\left(1-\frac{2\sqrt{a}}{a+1}\right):\left(\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}-a-1}\right)\)

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

b) \(19-8\sqrt{3}=\left(\sqrt{3}-4\right)^2\Rightarrow P=\sqrt{\left(\sqrt{3}-4\right)^2}-1=4-\sqrt{3}-1=3-\sqrt{3}\)

c) P < 1 <=> \(\sqrt{a}-1< 1\Leftrightarrow a< 4\)

Kết hợp với điều kiện : \(P< 1\Leftrightarrow\hept{\begin{cases}0< a< 4\\a\ne1\end{cases}}\)

a) \(ĐKXĐ:\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)

    \(M=\frac{\sqrt{x}}{\sqrt{x}-x}-\frac{\sqrt{x}+2}{1-x}\)

\(\Leftrightarrow M=\frac{1}{1-\sqrt{x}}-\frac{\sqrt{x}+2}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\)

\(\Leftrightarrow M=\frac{1+\sqrt{x}-\sqrt{x}-2}{1-x}\)

\(\Leftrightarrow M=\frac{-1}{1-x}\)

\(\Leftrightarrow M=\frac{1}{x-1}\)

b) Để M nhận giá trị nguyên

\(\Leftrightarrow\frac{1}{x-1}\inℤ\)

\(\Leftrightarrow x-1\inƯ\left(1\right)=\left\{\pm1\right\}\)

\(\Leftrightarrow x\in\left\{0;2\right\}\)

Mà \(x>0\)

Vậy để M nguyên \(\Leftrightarrow x=2\)