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

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

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

b: P=1/4

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

=>\(4\left(\sqrt{x}-2\right)=3\sqrt{x}\)

=>\(4\sqrt{x}-8-3\sqrt{x}=0\)

=>\(\sqrt{x}=8\)

=>x=64

c: Khi \(x=4+2\sqrt{3}\) thì \(P=\dfrac{\sqrt{4+2\sqrt{3}}-2}{3\cdot\sqrt{4+2\sqrt{3}}}\)

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

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

b) để P>0\(\Rightarrow\)\(\frac{\left(x-1\right)}{\sqrt{x}}>0\)

do \(\sqrt{x}>0\Rightarrow x-1>0\)

\(\Leftrightarrow x>1\)

c)P=\(\frac{8}{3}\)

Giúp mình với

1: Khi x=9 thì \(A=\dfrac{3+1}{3-1}=\dfrac{4}{2}=2\)

2:

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

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

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

b: \(2P=2\sqrt{x}+5\)

=>\(P=\sqrt{x}+\dfrac{5}{2}\)

=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}+\dfrac{5}{2}=\dfrac{2\sqrt{x}+5}{2}\)

=>\(\sqrt{x}\left(2\sqrt{x}+5\right)=2\sqrt{x}+2\)

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

=>\(\left(\sqrt{x}+2\right)\left(2\sqrt{x}-1\right)=0\)

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

=>x=1/4

Bạn có thể làm hộ mình câu c được không?Nếu được thì mình cảm ơn bạn nhiều!

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

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

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

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

Tại \(x=4+2\sqrt{3}\): \(\sqrt{x}=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)

\(A=\frac{\sqrt{3}}{4+2\sqrt{3}+\sqrt{3}+1}=\frac{\sqrt{3}}{5+3\sqrt{3}}\)

\(A-1=\frac{\sqrt{x}-1}{x+\sqrt{x}}-1=\frac{-1-x}{x+\sqrt{x}}< 0\)do \(x>0\).

Vậy \(A< 1\).

a, ĐK \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)

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

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

Ta thấy \(P=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}-1}>0\forall x>0,x\ne1\)

b, P=\(\frac{x+2\sqrt{x}+1}{\sqrt{x}-1}=\frac{\frac{2}{2+\sqrt{3}}+2\sqrt{\frac{2}{2+\sqrt{3}}}+1}{\sqrt{\frac{2}{2+\sqrt{3}}}-1}\)

=\(\frac{\frac{4}{\left(\sqrt{3}+1\right)^2}+2.\sqrt{\left(\frac{2}{\left(\sqrt{3}+1\right)^2}\right)}+1}{\sqrt{\left(\frac{2}{2+\sqrt{3}}\right)^2}-1}=\frac{\frac{4}{\left(\sqrt{3}+1\right)^2}+2.\frac{2}{\sqrt{3}+1}+1}{\frac{2}{\sqrt{3}+1}-1}\)

\(=\frac{12+6\sqrt{3}}{1-3}=-6-3\sqrt{3}\)

cậu ơi câu c đâu ạ??

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

a) \(ĐKXĐ:\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)

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

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

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

b) Vì \(\sqrt{x}>0\)

Nên để \(P< 0\)

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

\(\Leftrightarrow\sqrt{x}< 1\)

\(\Leftrightarrow x< 1\)

Vậy để \(P< 0\Leftrightarrow0< x< 1\)