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11 tháng 11 2017

giả sử:

\(\sqrt{x}=\sqrt{a^2}=a\)     ( a thuộc z)

 ta có :

\(\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{a+1}{a-3}=\frac{a-3+3+1}{a-3}=\frac{a-3+4}{a-3}\)

                                                                              \(=\frac{a-3}{a-3}+\frac{4}{a-3}=\frac{4}{a-3}\)

từ đó ta suy ra :

\(4=\left\{-4;-2;-1;0;1;2;4\right\}\)

\(a-3=-4\Rightarrow a=-1\)

\(a-3=-4\Rightarrow a=-1\)

\(a-3=4\Rightarrow a=7\)

\(a-3=-2\Rightarrow a=1\)

\(a-3=2\Rightarrow a=5\)

\(a-3=-1\Rightarrow a=2\)

\(a-3=1\Rightarrow a=4\)

\(a-3=0\Rightarrow a=3\)

\(\Rightarrow a=\left\{-1;1;2;3;4;5;7\right\}\)

mà \(x=a^2\) lấy các phần tử mũ hai lên là được

chúc bạn học tôt 

15 tháng 8 2017

Em không biết làm

15 tháng 8 2017

a. x khac 1

b. 5-2√5 / 5

1 tháng 7 2019

Thuy Duong Nguyen đánh đề cẩn thận hơn bạn nhé

Lời giải :

a) ĐKXĐ : \(x\ne1\)

 \(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(A=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\frac{\left(\sqrt{x}+3\right)\left(2-3\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(A=\frac{15\sqrt{x}-11-3x+6-7\sqrt{x}-2x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(A=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

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

\(A=\frac{2-5\sqrt{x}}{\sqrt{x}+3}\)

b) \(x=3-2\sqrt{2}=2-2\sqrt{2}+1=\left(\sqrt{2}-1\right)^2\)

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

Khi đó \(A=\frac{2-5\left(\sqrt{2}-1\right)}{\sqrt{2}-1+3}\)

\(A=\frac{2-5\sqrt{2}+5}{\sqrt{2}+2}=\frac{7-5\sqrt{2}}{\sqrt{2}+2}\)

c) \(A=\frac{1}{2}\)

\(\Leftrightarrow\frac{2-5\sqrt{x}}{\sqrt{x}+3}=\frac{1}{2}\)

\(\Leftrightarrow2\left(2-5\sqrt{x}\right)=\sqrt{x}+3\)

\(\Leftrightarrow4-10\sqrt{x}-\sqrt{x}-3=0\)

\(\Leftrightarrow1-11\sqrt{x}=0\)

\(\Leftrightarrow11\sqrt{x}=1\)

\(\Leftrightarrow\sqrt{x}=\frac{1}{11}\)

\(\Leftrightarrow x=\frac{1}{121}\)( thỏa )

d) A nguyên \(\Leftrightarrow2-5\sqrt{x}⋮\sqrt{x}+3\)

\(\Leftrightarrow-5\left(\sqrt{x}+3\right)+17⋮\sqrt{x}+3\)

Vì \(-5\left(\sqrt{x}+3\right)⋮\sqrt{x}+3\)

\(\Rightarrow17⋮\sqrt{x}+3\)

\(\Rightarrow\sqrt{x}+3\inƯ\left(17\right)=\left\{17\right\}\)( vì \(\sqrt{x}+3\ge3\))

\(\Leftrightarrow\sqrt{x}=14\)

\(\Leftrightarrow x=196\)( thỏa )

Vậy....

\(a,ĐKXĐ:\orbr{\begin{cases}x+2\sqrt{x}+3\ne0\\\sqrt{x}+3\ne0\end{cases}}\)

\(\Leftrightarrow\orbr{ }\sqrt{x}\ne-3\)

Rút gọn: p/s: sau phân số thứ 2 ở mẫu ko có x à? Bạn chép đề sai?

22 tháng 8 2019

\(Q=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}\)

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

\(=\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

b.\(Q< 1\)

\(\Leftrightarrow x-\sqrt{x}-2< x-5\sqrt{x}+6\)

\(\Leftrightarrow4\sqrt{x}-8< 0\)

\(\Leftrightarrow0\le x< 4\)

Vay de Q<1 thi \(0\le0< 4\)

26 tháng 5 2018

tích đi rồi t làm 

27 tháng 5 2018

9 T I C H  sai buồn

\(A=\frac{\sqrt{x^3}}{\sqrt{xy}-2y}-\frac{2x}{x+\sqrt{x}-2\sqrt{xy}-2\sqrt{y}}.\frac{1-x}{1-\sqrt{x}}..\)

nhờ vào năng lực rinegan tối hậu của ta , ta có thể dễ dàng nhìn thấy mẫu chung 

\(x+\sqrt{x}-2\sqrt{xy}-2\sqrt{y}=\sqrt{x}\left(\sqrt{x}-2\sqrt{xy}\right)+\left(\sqrt{x}-2\sqrt{y}\right)=\left(\sqrt{x}-2\sqrt{y}\right)\left(\sqrt{x}+1\right)\)

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

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

\(A=\frac{\sqrt{x^3}-2x\sqrt{y}}{\sqrt{y}\left(\sqrt{x}-2\sqrt{y}\right)}=\frac{x\sqrt{x}-2x\sqrt{y}}{\sqrt{y}\left(\sqrt{x}-2\sqrt{y}\right)}=\frac{x\left(\sqrt{x}-2\sqrt{y}\right)}{\sqrt{y}\left(\sqrt{x}-2\sqrt{y}\right)}=\frac{x}{\sqrt{y}}\)

b) thay y=625 vào ta được

\(\frac{x}{\sqrt{625}}=\frac{x}{25}< 0.2\Leftrightarrow x< 5\)

vậy   \(0< x< 5\)

7 tháng 7 2019

\(dkxd\Leftrightarrow\hept{\begin{cases}x\ge0\\\sqrt{x}-2\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}}\)

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

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

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

\(=\frac{-6\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=-\frac{6}{\sqrt{x}-2}\)

7 tháng 7 2019

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

a,ĐKXĐ:\(\hept{\begin{cases}x\ge0\\2-\sqrt{x}\\x-4\ne0\end{cases}\ne0}\)\(\Rightarrow\)\(\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)

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

\(A=\)\(\left(\frac{\sqrt{x}}{x-4}-\frac{2}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\right)\)\(.\left(\sqrt{x}+2\right)\)

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

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

\(A=\)\(\left(\frac{-6}{x-4}\right)\)\(.\left(\sqrt{x}+2\right)\)

\(A=\)\(\frac{-6}{\sqrt{x}-2}\)

b,\(x=9-4\sqrt{5}\)\(\Rightarrow\)\(A=\)\(\frac{-6}{\sqrt{9-4\sqrt{5}}-2}\)\(=\frac{-6}{\sqrt{5-2.2\sqrt{5}+4}-2}\)

\(A=\)\(\frac{-6}{\sqrt{\left(\sqrt{5}-2\right)^2}-2}\)\(=\frac{-6}{\sqrt{5}-2-2}\)\(=\frac{-6}{\sqrt{5}-4}\)

c,\(A>-1\)\(\Rightarrow\)\(\frac{-6}{\sqrt{x}-2}\)\(>-1\)\(\Rightarrow\)\(\frac{-6}{\sqrt{x}-2}+1>0\)

\(\Leftrightarrow\)\(\frac{-6+\sqrt{x}-2}{\sqrt{x}-2}>0\)

\(\Leftrightarrow\)\(\frac{\sqrt{x}-8}{\sqrt{x}-2}>0\)

4 tháng 10 2020

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

ĐKXĐ : x > 1

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

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

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

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

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

Để A = 9/2

=> \(\frac{x}{\sqrt{x}-1}=\frac{9}{2}\)( ĐK : x > 1 )

<=> 2x = 9( √x - 1 )

<=> 2x = 9√x - 9

<=> 2x + 9 = 9√x (1)

Bình phương hai vế

(1) <=> 4x2 + 36x + 81 = 81x

     <=> 4x2 + 36x + 81 - 81x = 0

     <=> 4x2 - 45x + 81 = 0

     <=> 4x2 - 36x - 9x + 81 = 0

     <=> 4x( x - 9 ) - 9( x - 9 ) = 0

     <=> ( x - 9 )( 4x - 9 ) = 0

     <=> \(\orbr{\begin{cases}x-9=0\\4x-9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=9\\x=\frac{9}{4}\end{cases}}\)( tm )