\(a,A=\dfrac{9-3x+x^2+10x+25-x^2+1}{\left(x-1\right)\left(x+5\right)}\\ A=\dfrac{7x+35}{\left(x-1\right)\left(x+5\right)}=\dfrac{7\left(x+5\right)}{\left(x-1\right)\left(x+5\right)}=\dfrac{7}{x-1}\\ b,A\in Z\\ \Leftrightarrow x-1\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\\ \Leftrightarrow x\in\left\{-6;0;2;8\right\}\left(tm\right)\\ b,A< 0\Leftrightarrow x-1< 0\left(7>0\right)\\ \Leftrightarrow x< 1;x\ne-5\\ c,\left|A\right|=3\Leftrightarrow\dfrac{7}{\left|x-1\right|}=3\Leftrightarrow\left|x-1\right|=\dfrac{7}{3}\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}+1=\dfrac{10}{3}\left(tm\right)\\x=-\dfrac{7}{3}+1=-\dfrac{4}{3}\left(tm\right)\end{matrix}\right.\)

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

Ta có: \(A=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)

\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

Thay \(x=6-2\sqrt{5}\) vào A, ta được:

\(A=\dfrac{\sqrt{5}-1-1}{\sqrt{5}-1+1}=\dfrac{\sqrt{5}-2}{\sqrt{5}}=\dfrac{5-2\sqrt{5}}{5}\)

b: Để \(A< \dfrac{1}{2}\) thì \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{1}{2}< 0\)

\(\Leftrightarrow2\sqrt{x}-2-\sqrt{x}-1< 0\)

\(\Leftrightarrow x< 9\)

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne1\end{matrix}\right.\)

Bài 1: ĐKXĐ:`x + 3 ne 0` và `x^2+ x-6 ne 0 ; 2-x ne 0`

`<=> x ne -3 ; (x-2)(x+3) ne 0 ; x ne2`

`<=>x ne -3 ; x ne 2`

b) Với `x ne - 3 ; x ne 2` ta có:

`P= (x+2)/(x+3)  - 5/(x^2 +x -6) + 1/(2-x)`

`P = (x+2)/(x+3) - 5/[(x-2)(x+3)] + 1/(2-x)`

`= [(x+2)(x-2)]/[(x-2)(x+3)] - 5/[(x-2)(x+3)] - (x+3)/[(x-2)(x+3)]`

`= (x^2 -4)/[(x-2)(x+3)] - 5/[(x-2)(x+3)] - (x+3)/[(x-2)(x+3)]`

`=(x^2 - 4 - 5 - x-3)/[(x-2)(x+3)]`

`= (x^2 - x-12)/[(x-2)(x+3)]`

`= [(x-4)(x+3)]/[(x-2)(x+3)]`

`= (x-4)/(x-2)`

Vậy `P= (x-4)/(x-2)` với `x ne -3 ; x ne 2`

c) Để `P = -3/4`

`=> (x-4)/(x-2) = -3/4`

`=> 4(x-4) = -3(x-2)`

`<=>4x -16 = -3x + 6`

`<=> 4x + 3x = 6 + 16`

`<=> 7x = 22`

`<=> x= 22/7` (thỏa mãn ĐKXĐ)

Vậy `x = 22/7` thì `P = -3/4`

d) Ta có: `P= (x-4)/(x-2)`

`P= (x-2-2)/(x-2)`

`P= 1 - 2/(x-2)`

Để P nguyên thì `2/(x-2)` nguyên

`=> 2 vdots x-2`

`=> x -2 in Ư(2) ={ 1 ;2 ;-1;-2}`

+) Với `x -2 =1 => x= 3` (thỏa mãn ĐKXĐ)

+) Với `x -2 =2 => x= 4`  (thỏa mãn ĐKXĐ)

+) Với `x -2 = -1=> x= 1` (thỏa mãn ĐKXĐ)

+) Với `x -2 = -2 => x= 0`(thỏa mãn ĐKXĐ)

Vậy `x in{ 3 ;4; 1; 0}` thì `P` nguyên

e) Từ `x^2 -9 =0`

`<=> (x-3)(x+3)=0`

`<=> x= 3` hoặc `x= -3`

+) Với `x=3` (thỏa mãn ĐKXĐ) thì:

`P  = (3-4)/(3-2)`

`P= -1/1`

`P=-1`

+) Với `x= -3` thì không thỏa mãn ĐKXĐ

Vậy với x= 3 thì `P= -1`

a)

\(P=\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-4\right)+5\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}.\dfrac{3\sqrt{x}+1}{3}\)

\(P=\dfrac{3x-2\sqrt{x}-1-3\sqrt{x}+4+5\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}.\dfrac{3\sqrt{x}+1}{3}\)

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

\(P=\dfrac{x+1}{3\sqrt{x}-1}\)

b) Từ phương trình suy ra \(\left[{}\begin{matrix}x=1\\x=\dfrac{1}{9}\end{matrix}\right.\)

Vói x=1 

\(P=\dfrac{1}{3\sqrt{1}-1}=\dfrac{1}{2}\)

Với x= 1/9

\(P=\dfrac{\dfrac{1}{9}}{3\sqrt{\dfrac{1}{9}}-1}\) không có nghiệm

`A)đk:x>=0,x ne 25`

`A=9=>A=(3+2)/(3-5)=-5/2`

`B)B=(3sqrtx-15+20-2sqrtx)/(x-25)`

`=(sqrtx+5)/(x-25)`

`=1/(sqrtx-5)`

`A=B.|x-4|`

`<=>A/B=|x-4|`

`<=>\sqrtx+2=|x-4|`

`<=>\sqrtx+2=(sqrtx+2)|sqrtx-2|`

`<=>|sqrtx-2|=1`

`+)sqrtx-2=1<=>x=9(tm)`

`+)sqrtx-2=-1<=>x=1(tm)`

Vậy `S={1,9}`

a, Thay x=9 vào biểu thức A ta có

\(A=\dfrac{\sqrt{9}+2}{\sqrt{9}-5}\)

\(A=\dfrac{3+2}{3-5}=\dfrac{5}{-2}=-2,5\)

Vậy A =-2,5 khi x=9

a. A=\(\dfrac{\sqrt{x}+2}{\sqrt{x}-5}\)

=\(\dfrac{\sqrt{9}+2}{\sqrt{9}-5}=\dfrac{-5}{2}\)

ĐKXĐ: \(x\notin\left\{-1;-\dfrac{1}{2}\right\}\)

a) Ta có: \(P=\left(\dfrac{2x}{x^3+x^2+x+1}+\dfrac{1}{x+1}\right):\left(1+\dfrac{x}{x+1}\right)\)

\(=\left(\dfrac{2x}{\left(x+1\right)\left(x^2+1\right)}+\dfrac{x^2+1}{\left(x^2+1\right)\left(x+1\right)}\right):\left(\dfrac{x+1+x}{x+1}\right)\)

\(=\dfrac{x^2+2x+1}{\left(x+1\right)\left(x^2+1\right)}:\dfrac{2x+1}{x+1}\)

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

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

b) Vì \(x=\dfrac{1}{4}\) thỏa mãn ĐKXĐ

nên Thay \(x=\dfrac{1}{4}\) vào biểu thức \(P=\dfrac{x^2+2x+1}{\left(2x+1\right)\left(x^2+1\right)}\), ta được:

\(P=\left[\left(\dfrac{1}{4}\right)^2+2\cdot\dfrac{1}{4}+1\right]:\left[\left(2\cdot\dfrac{1}{4}+1\right)\left(\dfrac{1}{16}+1\right)\right]\)

\(=\left(\dfrac{1}{16}+\dfrac{1}{2}+1\right):\left[\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{16}+1\right)\right]\)

\(=\dfrac{25}{16}:\dfrac{51}{32}=\dfrac{25}{16}\cdot\dfrac{32}{51}=\dfrac{50}{51}\)

Vậy: Khi \(x=\dfrac{1}{4}\) thì \(P=\dfrac{50}{51}\)

a) \(B=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\left(x\ge0,x\ne1\right)\)

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

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

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

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

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

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

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