a) \(A=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{3-11\sqrt{x}}{9-x}=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{11\sqrt{x}-3}{x-9}=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{2x-6\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{x+4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}}{\sqrt{x}-3}\)

Bài 1:

a)  \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)

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

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

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

b)   \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)

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

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

\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)

c)  ĐK:  \(a\ge0;a\ne1\)

  \(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)

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

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

\(=1-a+a=1\)

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

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

\(\Leftrightarrow\frac{-15\sqrt{x}+11-3x-9\sqrt{x}+2\sqrt{x}+6-3+3\sqrt{x}}{-x-2\sqrt{x}+3}\)

\(\Leftrightarrow\frac{19\sqrt{x}+3x-14}{x+2\sqrt{x}-3}\)

\(b,\) Xét \(\frac{19\sqrt{x}+3x-14}{x+2\sqrt{x}-3}\) phân tử \(13\sqrt{x}-5\)

Vậy để biểu thức trên nguyên thì \(13\sqrt{x}-5=0\)

\(\Leftrightarrow x=\left(\frac{5}{13}\right)^2\)

Vậy .......................

~~~~~~~~~ Học Tốt ~~~~~~~~~

a/ \(P=\left[1-\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\left[\frac{3-\sqrt{x}}{\sqrt{x}-2}+\frac{\sqrt{x}-2}{\sqrt{x}+3}-\frac{9x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]\)

\(=\left(1-\frac{\sqrt{x}}{\sqrt{x}+3}\right):\left[\frac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)+\left(\sqrt{x}-2\right)^2-9x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]\)

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

\(=\frac{3}{\sqrt{x}+3}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{13-4\sqrt{x}-9x}\)

\(=\frac{3\sqrt{x}-6}{13-4\sqrt{x}-9x}\)

b/ \(P=1\Rightarrow\frac{3\sqrt{x}-6}{13-4\sqrt{x}-9x}=1\Rightarrow3\sqrt{x}-6=13-4\sqrt{x}-9x\)

\(\Rightarrow9x+7\sqrt{x}-19=0\)

Mình k biết mình sai chỗ nào nữa, bạn xem giúp mình với

Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé

a, - Thay x = 25 vào biểu thức B ta được :

\(B=\frac{\sqrt{25}-3}{\sqrt{25}+1}=\frac{1}{3}\)

b, Ta có : \(A=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{3-11\sqrt{x}}{9-x}\)

=> ​​​\(A=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{11\sqrt{x}-3}{x-9}\)​

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

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

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

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

c, Ta có : \(P=AB+1=\frac{3\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+1\)

=> \(P=\frac{3\sqrt{x}}{\sqrt{x}+1}+1\)

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

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

Ta thấy : \(\sqrt{x}\ge0\)

=> \(4-\frac{3}{\sqrt{x}+1}\ge1\)

Vậy Min_P = 1 khi x = 0

a) biểu thức có nghĩa khi và chỉ khi: \(\Leftrightarrow\hept{\begin{cases}\sqrt{x}+3\ne0\\\sqrt{x}-3\ne0\\x-9\ne0\end{cases}\Leftrightarrow x\ne9}\)    và     \(x\ge0\)

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

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

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

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

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

c) để Q < 1 thì:

\(\frac{3\sqrt{x}}{\sqrt{x}+3}< 1\)đkxđ:  \(x\ge0\)

\(\Leftrightarrow\frac{3\sqrt{x}-\sqrt{x}-3}{\sqrt{x}+3}< 0\)

\(\Leftrightarrow\frac{2\sqrt{x}-3}{\sqrt{x}+3}< 0\)(1)

do \(\sqrt{x}+3>0\forall x\)

\(\Rightarrow\left(1\right)< 0\)khi và chỉ khi \(2\sqrt{x}-3< 0\)

                                              \(\Leftrightarrow2\sqrt{x}< 3\Leftrightarrow\sqrt{x}< \frac{3}{2}\Leftrightarrow x< \frac{9}{4}\)

kết hợp với điều kiện ban đầu \(\Rightarrow Q< 1khi0\le x< \frac{9}{4}\)

Lời giải:

a)

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

b)

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

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