1) ĐKXĐ: \(x>0;x\ne4;x\ne9\)

(*lười lắm, ko chép lại đề nha :V*)

\(P=\frac{\left(2+\sqrt{x}\right)^2+\sqrt{x}\left(2-\sqrt{x}\right)+4x+2\sqrt{x}-4}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\frac{2\sqrt{x}-\left(\sqrt{x}+3\right)}{\sqrt{x}\left(2-\sqrt{x}\right)}\\ =\frac{4+4\sqrt{x}+x+2\sqrt{x}-x+4x+2\sqrt{x}-4}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\cdot\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}\\ =\frac{4x+8\sqrt{x}}{2+\sqrt{x}}\cdot\frac{\sqrt{x}}{\sqrt{x}-3}\\ =\frac{4\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\cdot\frac{\sqrt{x}}{\sqrt{x}-3}=\frac{4x}{\sqrt{x}-3}\)

2) Để P>0 thì

\(\frac{4x}{\sqrt{x}-3}>0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4x>0\\\sqrt{x}-3>0\end{matrix}\right.\\\left\{{}\begin{matrix}4x< 0\\\sqrt{x}-3< 0\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\\sqrt{x}>3\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\\sqrt{x}< 3\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x>9\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x< 9\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>9\\x< 0\left(ktm\right)\end{matrix}\right.\)

Vậy với \(x>9\) thì \(P>0\).

Chúc bạn học tốt nha.

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

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

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

Dễ thấy \(\sqrt{x}+2\ge2>0\forall x\ge0\)

Nên để \(P< 0\Leftrightarrow\sqrt{x}-1< 0\Leftrightarrow\sqrt{x}< 1\Leftrightarrow x< 1\)

Vậy với \(0\le x< 1\)thì P<0

(Câu trả lời bằng hình ảnh)

Có vẻ hum nào cũng có bào nhề

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

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

=1-x

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

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

=\(1-x\)

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

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

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

Để \(A=\frac{B}{6}\Leftrightarrow B=6A\Rightarrow\frac{\sqrt{x}-1}{\sqrt{x}}=-18\)

\(\Rightarrow\sqrt{x}-1=-18\sqrt{x}\Rightarrow\sqrt{x}=\frac{1}{19}\Rightarrow x=\frac{1}{361}\)

\( 1)P = \left( {\dfrac{{2x + 1}}{{\sqrt {{x^3}} - 1}} - \dfrac{1}{{\sqrt x - 1}}} \right):\left( {1 - \dfrac{{x + 4}}{{x + \sqrt x + 1}}} \right)\\ = \left( {\dfrac{{2x + 1}}{{x\sqrt x - 1}} - \dfrac{1}{{\sqrt x - 1}}} \right):\dfrac{{x + \sqrt x + 1 - \left( {x + 4} \right)}}{{x + \sqrt x + 1}}\\ = \left[ {\dfrac{{2x + 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} - \dfrac{1}{{\sqrt x - 1}}} \right]:\dfrac{{\sqrt x - 3}}{{x + \sqrt x + 1}}\\ = \dfrac{{2x + 1 - \left( {x + \sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{{x + \sqrt x + 1}}{{\sqrt x - 3}}\\ = \dfrac{{x - \sqrt x }}{{\sqrt x - 1}}.\dfrac{1}{{\sqrt x - 3}}\\ = \dfrac{{\sqrt x \left( {\sqrt x - 1} \right)}}{{\sqrt x - 1}}.\dfrac{1}{{\sqrt x - 3}}\\ = \dfrac{{\sqrt x }}{{\sqrt x - 3}} \)

a) \(P=\frac{x^2-9}{x-3}+\frac{4-4\sqrt{x}+x}{2-\sqrt{x}}+\frac{4-x}{2+\sqrt{x}}\)

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

\(x+3+2-\sqrt{x}+2-\sqrt{x}\) = \(x+7-2\sqrt{x}\)

b) Tại x = 9, ta có:

P = \(x+7-2\sqrt{x}\) = 9 + 7 - 2\(\sqrt{9}\) = 10

A=\(\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{1}{x+\sqrt{x}}\right)\):\(\left(\frac{1}{\sqrt{x}+1}+\frac{2}{x-1}\right)\)Đk x>0 x#0 x#1

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

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

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

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

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

Ta có 3+\(2\sqrt{2}=\left(\sqrt{2}+1\right)^2\)(thay và A ta dc

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

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

=2

k minh minh giai cho

\(\frac{2.\left(x+4\right)}{\left(\sqrt{x}+1\right).\left(\sqrt{x}-4\right)}+\frac{\sqrt{x}.\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+1\right).\left(\sqrt{x}-4\right)}-\frac{8.\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}\)      

=\(\frac{2x+8+x-4\sqrt{x}-8\sqrt{x}-8}{\left(\sqrt{x}+1\right).\left(\sqrt{x}-4\right)}\)

=\(\frac{3x-12\sqrt{x}}{mc}\)  

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

k tk mk cung lam cho

1) ĐKXĐ \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)

\(P=\left(\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{5\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right):\left(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{\left(\sqrt{x}\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\\ =\left(\frac{\sqrt{x}-5\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right):\left(\frac{x-4-x}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\\ =\frac{-4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}:\frac{-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ =\frac{-4\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{-4}\\ =\sqrt{x}+1\)

2)

\(P=\sqrt{x}+1=\sqrt{\frac{3-\sqrt{5}}{2}}+1\\ \sqrt{\frac{6-2\sqrt{5}}{4}}+1\\ =\sqrt{\frac{5-2\cdot\sqrt{5}\cdot1+1}{4}}+1\\ =\sqrt{\frac{\left(\sqrt{5}-1\right)^2}{4}}+1\\ =\frac{\sqrt{5}-1}{2}+1\\ \frac{\sqrt{5}-1+2}{2}\\ =\frac{\sqrt{5}+1}{2}\)