Ta có:

\(B=\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\) (ĐK: \(x\ne4;x\ge0\)) 

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

\(B=\dfrac{x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(B=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(B=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(B=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(B=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

\(\Rightarrow P=\dfrac{A}{B}=\dfrac{\dfrac{\sqrt{x}+2}{\sqrt{x}}}{\dfrac{\sqrt{x}}{\sqrt{x}-2}}=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\sqrt{x}\cdot\sqrt{x}}=\dfrac{x-4}{x}\) (ĐK: \(x\ne0\)) 

Theo đề ta có:

\(P\cdot x\le10\sqrt{x}-29-\sqrt{x}+25\) (ĐK: \(x\ge0\)) 

\(\Leftrightarrow\dfrac{x-4}{x}\cdot x\le9\sqrt{x}-4\)

\(\Leftrightarrow x-4\le9\sqrt{x}-4\)

\(\Leftrightarrow x-9\sqrt{x}\le0\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-9\right)\le0\)

Mà: \(\sqrt{x}\ge0\)

\(\Leftrightarrow\sqrt{x}-9\le0\)

\(\Leftrightarrow\sqrt{x}\le9\)

\(\Leftrightarrow x\le81\)

Kết hợp với đk:

\(0\le x\le81\)

Ta có: \(\Delta=4m^2+4m-11\)

Để phương trình có 2 nghiệm phân biệt \(\Leftrightarrow4m^2+4m-11>0\)

Theo Vi-ét, ta có: \(\left\{{}\begin{matrix}x_1+x_2=2m+3\\x_1x_2=2m+5\end{matrix}\right.\)

Để phương trình có 2 nghiệm dương phân biệt

\(\Leftrightarrow\left\{{}\begin{matrix}4m^2+4m-11>0\\2m+3>0\\2m+5>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m< \dfrac{-1-2\sqrt{3}}{2}\\m>\dfrac{-1+2\sqrt{3}}{2}\end{matrix}\right.\\m>-\dfrac{3}{2}\\m>-\dfrac{5}{2}\end{matrix}\right.\) \(\Leftrightarrow m>\dfrac{-1+2\sqrt{3}}{2}\)

 Mặt khác: \(\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}=\dfrac{4}{3}\)

\(\Rightarrow\dfrac{x_1+x_2+2\sqrt{x_1x_2}}{x_1x_2}=\dfrac{16}{9}\) \(\Rightarrow\dfrac{2m+3+2\sqrt{2m+5}}{2m+5}=\dfrac{16}{9}\)

\(\Rightarrow18m+27+18\sqrt{2m+5}=32m+80\)

\(\Leftrightarrow14m-53=18\sqrt{2m+5}\)

\(\Rightarrow\) ...

giúp mình với ạ ! Mình cảm ơn ạ 

a) Ta có: \(P=\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\)

\(=\dfrac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

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

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

a) Ta có: \(M=\left(\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\right)\cdot\dfrac{x+3\sqrt{x}}{7-\sqrt{x}}\)

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

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

\(=\dfrac{x-9-x+\sqrt{x}+2}{\sqrt{x}-2}\cdot\dfrac{-\sqrt{x}}{\sqrt{x}-7}\)

\(=\dfrac{\sqrt{x}-7}{\sqrt{x}-2}\cdot\dfrac{-\sqrt{x}}{\sqrt{x}-7}\)

\(=\dfrac{-\sqrt{x}}{\sqrt{x}-2}\)

b) Ta có: \(x^2-4x=0\)

\(\Leftrightarrow x\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=4\left(loại\right)\end{matrix}\right.\)

Thay x=0 vào biểu thức \(M=\dfrac{-\sqrt{x}}{\sqrt{x}-2}\), ta được:

\(M=\dfrac{-\sqrt{0}}{\sqrt{0}-2}=-\dfrac{0}{-2}=0\)

Vậy: Khi \(x^2-4x=0\) thì M=0

a: \(P=\dfrac{x-1}{\sqrt{x}}:\dfrac{x-1+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

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

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

b: \(x=\dfrac{2}{2+\sqrt{3}}=2\left(2-\sqrt{3}\right)=4-2\sqrt{3}\)

Khi x=4-2căn 3 thì \(P=\dfrac{\left(\sqrt{3}-1+1\right)^2}{\sqrt{3}-1}=\dfrac{3}{\sqrt{3}-1}=\dfrac{3\sqrt{3}+3}{2}\)

a: \(B=\dfrac{1}{\sqrt{x}+1}\)

\(B-1=\dfrac{\sqrt{x}+1-1}{\sqrt{x}+1}=\dfrac{\sqrt{x}}{\sqrt{x}+1}>=0\)

=>B>=1

b: \(P=\dfrac{\sqrt{x}+1+x}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)

\(P\cdot\sqrt{x}+2x-\sqrt{x}=3x-2\sqrt{x-4}+3\)

=>\(x+\sqrt{x}+1+2x-\sqrt{x}=3x+3-2\sqrt{x-4}\)

=>\(-2\sqrt{x-4}+3=1\)

=>x-4=1

=>x=5

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

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

2: P<1/2
=>P-1/2<0

=>\(2\sqrt{x}-2-x-1< 0\)

=>-x+2căn x-1<0

=>(căn x-1)^2>0(luôn đúng)