Câu 3 :

\(ĐKXĐ:x>0\)

 \(P=\left(\frac{2}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+2}\right):\frac{2\sqrt{x}}{x+2\sqrt{x}}\)

\(\Leftrightarrow P=\frac{2\sqrt{x}+4+x}{x+2\sqrt{x}}\cdot\frac{x+2\sqrt{x}}{2\sqrt{x}}\)

\(\Leftrightarrow P=\frac{2\sqrt{x}+4+x}{2\sqrt{x}}\)

b) Để P = 3

\(\Leftrightarrow\frac{2\sqrt{x}+4+x}{x+2\sqrt{x}}=3\)

\(\Leftrightarrow2\sqrt{x}+4+x=6\sqrt{x}\)

\(\Leftrightarrow x-4\sqrt{x}+4=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)^2=0\)

\(\Leftrightarrow\sqrt{x}-2=0\)

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

\(\Leftrightarrow x=4\)(tm)

Vậy để \(P=3\Leftrightarrow x=4\)

Câu 1 : Hình như sai đề !! Mik sửa :

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

\(A=\left(\frac{x}{x\sqrt{x}-4\sqrt{x}}-\frac{6}{3\sqrt{x}-6}+\frac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\frac{10-x}{\sqrt{x}+2}\right)\)

\(\Leftrightarrow A=\left(\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\right):\left(\frac{x-4+10-x}{\sqrt{x}+2}\right)\)

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

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

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

b) Để A < 2

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

\(\Leftrightarrow-1< 2\sqrt{x}-4\)

\(\Leftrightarrow2\sqrt{x}>3\)

\(\Leftrightarrow\sqrt{x}>1,5\)

\(\Leftrightarrow x>2,25\)

Vậy để \(A< 2\Leftrightarrow x>2,25\)

Câu 2 :

\(ĐKXĐ:\hept{\begin{cases}x\ne\frac{4}{25}\\x\ne9\\x\ne1\end{cases}}\)

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

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

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

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

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

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

b) Mik quên mất cách chứng minh rồi :((

Chỉ biết : Dấu " = " xảy ra : \(\Leftrightarrow x=0\)

\(\)

a.

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

\(=\frac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

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

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

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

b.Ta co:

\(A=\frac{2-5\sqrt{x}}{\sqrt{x}+3}=\frac{-5\left(\sqrt{x}+3\right)+17}{\sqrt{x}+3}=-5+\frac{17}{\sqrt{x}+3}\le-5+\frac{17}{3}=\frac{2}{3}\)

Dau '=' xay ra khi \(x=0\)

Vay \(A_{max}=\frac{2}{3}\)khi \(x=0\)

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 ~~~~~~~~~

\(A=\frac{\left(1+\sqrt{x}\right)^2-4\sqrt{x}}{\sqrt{x}-1}\)  \(\left(x\ge0;x\ne1\right)\)

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

và \(B=\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{2}}+\frac{2+\sqrt{2}}{\sqrt{x}+1}\)

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

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

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

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

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

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

để A = B thì \(\sqrt{x}-1\)= \(\frac{2}{\sqrt{3}-\sqrt{2}}+2\)

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

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

\(\sqrt{x}=2\sqrt{3}+2\sqrt{2}+3\)

tới bước này tui bí :(( mong các bạn giỏi khác giúp bạn :D

a/ Ta có: \(x+2\sqrt{x}+1=\left(\sqrt{x}+1\right)^2\)

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

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

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

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

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

b/ Thay \(x=\frac{\sqrt{3}}{2+\sqrt{3}}\)  => \(P=\frac{2}{\frac{\sqrt{3}}{2+\sqrt{3}}-1}=\frac{2\left(2+\sqrt{3}\right)}{\sqrt{3}-2-\sqrt{3}}\)

=> \(P=-\left(2+\sqrt{3}\right)\)

c/ \(P=\frac{2}{x-1}=-\frac{4}{\sqrt{x}+1}\) <=> \(\frac{1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=-\frac{2}{\sqrt{x}+1}\)

<=> \(\frac{1}{\sqrt{x}-1}=-2\)

<=> \(1=-2\sqrt{x}+2\)

<=> \(2\sqrt{x}=1=>\sqrt{x}=\frac{1}{2}=>x=\frac{1}{4}\)