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14 tháng 7 2016

a/ \(P=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}-\frac{3-11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

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

       \(=\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

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

b/ \(P< 1\Rightarrow\frac{3\sqrt{x}}{\sqrt{x}-3}< 1\Rightarrow\frac{2\sqrt{x}+3}{\sqrt{x}-3}< 0\)

   Xét 2 trường hợp:

  • \(\hept{\begin{cases}2\sqrt{x}+3>0\\\sqrt{x}-3< 0\end{cases}\Rightarrow\hept{\begin{cases}2\sqrt{x}>-3\\\sqrt{x}< 3\end{cases}\Rightarrow}\hept{\begin{cases}\sqrt{x}>-\frac{3}{2}\\\sqrt{x}< 3\end{cases}}\Rightarrow-\frac{3}{2}< \sqrt{x}< 3}\)

                                       \(\Rightarrow-\frac{9}{4}< x< 9\)

  •  \(\hept{\begin{cases}2\sqrt{x}+3< 0\\\sqrt{x}>3\end{cases}\Rightarrow\hept{\begin{cases}\sqrt{x}< -\frac{3}{2}\\\sqrt{x}>3\end{cases}}}\) (vô lí)

                                                   Vậy -9/4 < x < 9

a: 


Sửa đề: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{3x+3}{9-x}\right)\cdot\left(\dfrac{\sqrt{x}-7}{\sqrt{x}+1}+1\right)\)

\(P=\left(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\right)\cdot\dfrac{\sqrt{x}-7+\sqrt{x}+1}{\sqrt{x}+1}\)

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\cdot\dfrac{2\sqrt{x}-6}{\sqrt{x}+1}\)

\(=\dfrac{-3\sqrt{x}-3}{\sqrt{x}+3}\cdot\dfrac{2}{\sqrt{x}+1}=\dfrac{-6}{\sqrt{x}+3}\)

b: P>=1/2

=>P-1/2>=0

=>\(\dfrac{-6}{\sqrt{x}+3}-\dfrac{1}{2}>=0\)

=>\(\dfrac{-12-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>=0\)

=>\(-\sqrt{x}-15>=0\)

=>\(-\sqrt{x}>=15\)

=>căn x<=-15

=>\(x\in\varnothing\)

c: căn x+3>=3

=>6/căn x+3<=6/3=2

=>P>=-2

Dấu = xảy ra khi x=0

9 tháng 7 2019

a) \(A=\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}+1}+\frac{5+3\sqrt{5}}{\sqrt{5}}-\left(\sqrt{5}+3\right)\)

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

\(A=\frac{\sqrt{3}+1}{\sqrt{3}+1}+\frac{5+3\sqrt{5}}{\sqrt{5}}-\frac{5+3\sqrt{5}}{\sqrt{5}}\)

\(A=1\)

b) Ta có:

\(B=\frac{1}{3-\sqrt{x}}+\frac{\sqrt{x}}{3+\sqrt{x}}-\frac{x+9}{x-9}\) ( x >= 0, x khác 9 )

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

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

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

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

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

\(B=\frac{4}{3-\sqrt{x}}\)

Để B > A

\(\Rightarrow\frac{4}{3-\sqrt{x}}>1\)

\(\Rightarrow4>3-\sqrt{x}\)

\(\Rightarrow4-3+\sqrt{x}>0\)

\(\Rightarrow1+\sqrt{x}>0\)

\(\Rightarrow\sqrt{x}>-1\)

\(\Rightarrow x>1\)

9 tháng 7 2019

a) A=\(\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}+1}+\frac{5+3\sqrt{5}}{\sqrt{5}}-\left(\sqrt{5}+3\right)\)

\(=\frac{\sqrt{3+2\sqrt{3}+1}}{\sqrt{3}+1}+\frac{\sqrt{5}\cdot\left(\sqrt{5}+3\right)}{\sqrt{5}}\)

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

\(=\frac{\sqrt{3}+1}{\sqrt{3}+1}+0=1\)

b) B=\(\frac{1}{3-\sqrt{x}}+\frac{\sqrt{x}}{3+\sqrt{x}}-\frac{x+9}{x-9}\)

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

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

\(=\frac{4\text{​​}\sqrt{x}+12}{\left(3-\sqrt{x}\right)\cdot\left(3+\sqrt{x}\right)}\)

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

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

\(B>A \Leftrightarrow\frac{4}{3-\sqrt{x}}>1\)

các giá trị của x là \(\left\{x\in R\backslash0\le x\le9\right\}\)

18 tháng 7 2018

mk làm luôn.

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

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

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

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

mk làm phần rút gọn xong mk bận nên bn tự làm câu b nha ^^

5 tháng 7 2017

a)\(P=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{11\sqrt{x}-3}{x-9}\)

\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)+11\sqrt{x}}{x-9}\)

\(=\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}}{x-9}\)

\(=\frac{3x+9\sqrt{x}+3}{x-9}\)

\(=\)...

a) Ta có: \(A=\sqrt{3+2\sqrt{2}}-\frac{1}{1+\sqrt{2}}\)

\(=\sqrt{1+2\cdot1\cdot\sqrt{2}+2}-\frac{1}{1+\sqrt{2}}\)

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

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

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

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

\(=\frac{2\sqrt{2}+2}{1+\sqrt{2}}\)

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

b) Ta có: \(\left(\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{3}{\sqrt{x}-3}\right)\cdot\frac{\sqrt{x}+3}{x+9}\)

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

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

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

\(=\frac{1}{\sqrt{x}-3}\)(đpcm)

13 tháng 11 2021

\(a,P=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\\ P=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\\ P=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}=\dfrac{-3}{\sqrt{x}+3}\\ b,P=\dfrac{-3}{\sqrt{x}+3}\ge\dfrac{-3}{0+3}=-1\\ P_{min}=-1\Leftrightarrow x=0\)