Điều kiện: x>2

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

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

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

P= \(\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)

b) P= \(\dfrac{1}{4}\)

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

⇔\(4\sqrt{x}-8=3\sqrt{x}\)

⇔\(\sqrt{x}=8\)

⇔x=64 (TM) 

Vậy X=64(TMĐK) thì P=\(\dfrac{1}{4}\)

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

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

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

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

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

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

Để \(P=-\dfrac{2}{5}\) thì \(\dfrac{\sqrt{x}-1}{x+1}=\dfrac{-2}{5}\)

\(\Leftrightarrow-2x-2=5\sqrt{x}-5\)

\(\Leftrightarrow-2x-2-5\sqrt{x}+5=0\)

\(\Leftrightarrow-2x-5\sqrt{x}+3=0\)

\(\Leftrightarrow-2x-6\sqrt{x}+\sqrt{x}+3=0\)

\(\Leftrightarrow-2\sqrt{x}\left(\sqrt{x}+3\right)+\left(\sqrt{x}+3\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}+3\right)\left(-2\sqrt{x}+1\right)=0\)

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

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

\(\Leftrightarrow x=\dfrac{1}{4}\)(thỏa ĐK)

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)

\(a,P\) xác định \(\Leftrightarrow\left[{}\begin{matrix}x>0\\x\ne1\\x\ne4\end{matrix}\right.\)

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

\(c,P=\dfrac{1}{4}\Leftrightarrow\dfrac{\sqrt{x}-2}{3\sqrt{x}}=\dfrac{1}{4}\\ \Leftrightarrow\dfrac{4\left(\sqrt{x}-2\right)-3\sqrt{x}}{12\sqrt{x}}=0\\ \Leftrightarrow4\sqrt{x}-8-3\sqrt{x}=0\\ \Leftrightarrow\sqrt{x}=8\\ \Leftrightarrow x=64\left(tmdk\right)\)

Vậy \(x=64\) thì \(P=\dfrac{1}{4}\)

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

2/ \(y'=\dfrac{1-x^2-\left(1-x^2\right)'x}{\left(1-x^2\right)^2}=\dfrac{1+x^2}{\left(1-x^2\right)^2}\)

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

4/ \(y'=f'\left(x\right)=2x-\dfrac{2x}{x^4}=2x-\dfrac{2}{x^3}\)

\(y'=0\Leftrightarrow\dfrac{2x^4-2}{x^3}=0\Leftrightarrow x=\pm1\)

5/ \(y'=\dfrac{\dfrac{1}{2\sqrt{1+x}}}{2\sqrt{1+\sqrt{1+x}}}\Rightarrow f\left(x\right).f'\left(x\right)=\sqrt{1+\sqrt{1+x}}.\dfrac{1}{4\sqrt{1+x}.\sqrt{1+\sqrt{1+x}}}=\dfrac{1}{4\sqrt{1+x}}=\dfrac{1}{2\sqrt{2}}\)

\(\Leftrightarrow2\sqrt{1+x}=\sqrt{2}\Leftrightarrow1+x=\dfrac{1}{2}\Leftrightarrow x=-\dfrac{1}{2}\)

Hãy nhớ câu tính đạo hàm này, bởi nó liên quan đến nguyên hàm sau này sẽ học

ĐKXĐ: \(x\ge0;x\ne1\)

\(A=\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}+\dfrac{x+2}{x\sqrt{x}-1}\)

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

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

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

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

\(A=\dfrac{2}{7}\Rightarrow\dfrac{\sqrt{x}}{x+\sqrt{x}+1}=\dfrac{2}{7}\)

\(\Rightarrow2\left(x+\sqrt{x}+1\right)=7\sqrt{x}\)

\(\Leftrightarrow2x-5\sqrt{x}+2=0\Leftrightarrow\left(\sqrt{x}-2\right)\left(2\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{1}{4}\end{matrix}\right.\)

a) ĐKXĐ: \(x\ge0,x\ne1\)

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

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

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

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

\(=\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{1}{\sqrt{x}-1}=\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)\)

\(=\dfrac{4\sqrt{x}}{\sqrt{x}+1}\)

b) \(P=\sqrt{x}-1\Rightarrow\dfrac{4\sqrt{x}}{\sqrt{x}+1}=\sqrt{x}-1\Rightarrow4\sqrt{x}=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)

\(\Rightarrow4\sqrt{x}=x-1\Rightarrow x-4\sqrt{x}-1=0\)

\(\Delta=\left(-4\right)^2-4.\left(-1\right)=20\Rightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{4-2\sqrt{5}}{2}=2-\sqrt{5}\\\sqrt{x}=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{4+2\sqrt{5}}{2}=2+\sqrt{5}\end{matrix}\right.\)

mà \(\sqrt{x}\ge0\Rightarrow\sqrt{x}=2+\sqrt{5}\Rightarrow x=9+4\sqrt{5}\)

c) \(P=\dfrac{4\sqrt{x}}{\sqrt{x}+1}=\dfrac{4\left(\sqrt{x}+1\right)-4}{\sqrt{x}+1}=4-\dfrac{4}{\sqrt{x}+1}\)

Để \(P\in Z\Rightarrow4⋮\sqrt{x}+1\Rightarrow\sqrt{x}+1\in\left\{1;2;4\right\}\left(\sqrt{x}+1\ge1\right)\)

\(\Rightarrow x\in\left\{0;1;9\right\}\) mà \(x\ne1\Rightarrow x\in\left\{0;9\right\}\)

giúp mình với ạ:<

a, ĐK: \(x\ge0,x\ne1\)

\(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1}{x-1}\)

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

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

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

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

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

b, \(x=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\)

Khi đó: 

\(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)

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

\(=\dfrac{2\sqrt{3}-3}{\sqrt{3}}\)

\(=2-\sqrt{3}\)

c, \(A=\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{1}{2}\)

\(\Leftrightarrow4\sqrt{x}-2=\sqrt{x}+1\)

\(\Leftrightarrow3\sqrt{x}=3\)

\(\Leftrightarrow x=1\left(l\right)\)

Vậy không tồn tại giá trị x thỏa mãn \(A=\dfrac{1}{2}\).

ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

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

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

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

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

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

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

b) Để P=2 thì \(-x+1=2\sqrt{x}\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+1=\sqrt{2}\\\sqrt{x}+1=-\sqrt{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\sqrt{2}-1\\\sqrt{x}=-\sqrt{2}-1\left(loại\right)\end{matrix}\right.\Leftrightarrow x=3-2\sqrt{2}\)

Vậy: Để P=2 thì \(x=3-2\sqrt{2}\)