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20 tháng 11 2023

Ta có: \(\sqrt{P}< \dfrac{1}{2}\Rightarrow P< \left(\dfrac{1}{2}\right)^2\Leftrightarrow P< \dfrac{1}{4}\) (1) 

Với đk: \(P\ge0\)

\(\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\ge0\)

\(\Leftrightarrow\sqrt{x}-2\ge0\) (vì \(\sqrt{x}+1>0\forall x\ge0\))

\(\Leftrightarrow\sqrt{x}\ge2\)

\(\Leftrightarrow x\ge4\) 

\(\left(1\right)\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< \dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}-\dfrac{1}{4}< 0\)

\(\Leftrightarrow\dfrac{4\sqrt{x}-8}{4\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+1}{4\left(\sqrt{x}+1\right)}< 0\)

\(\Leftrightarrow\dfrac{4\sqrt{x}-8-\sqrt{x}-1}{4\left(\sqrt{x}+1\right)}< 0\)

\(\Leftrightarrow\dfrac{3\sqrt{x}-9}{4\left(\sqrt{x}+1\right)}< 0\)

\(\Leftrightarrow3\sqrt{x}-9< 0\)

\(\Leftrightarrow\sqrt{x}< 3\)

\(\Leftrightarrow x< 9\)

Kết hợp với đk: \(4\le x< 9\)

20 tháng 11 2023

sai dấu r bn ơi \(\sqrt{x}+1\\ \) mà bn bn nhìn lại đi

 

20 tháng 11 2023

Ta biết: \(\sqrt{P}=\dfrac{1}{2}\Rightarrow P=\left(\dfrac{1}{2}\right)^2=\dfrac{1}{4}\) (1)

Với đk: \(P\ge0\)

\(\Rightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\ge0\) 

\(\Leftrightarrow\sqrt{x}-2\ge0\) (vì \(\sqrt{x}+1\ge1>0\forall x\ge0\))

\(\Leftrightarrow\sqrt{x}\ge2\)

\(\Leftrightarrow x\ge4\) 

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

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

\(\Leftrightarrow4\sqrt{x}-8=\sqrt{x}+1\)

\(\Leftrightarrow4\sqrt{x}-\sqrt{x}=1+8\)

\(\Leftrightarrow3\sqrt{x}=9\)

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

\(\Leftrightarrow x=3^2\)

\(\Leftrightarrow x=9\left(tm\right)\)

Vậy: ... 

3 tháng 2 2021

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=\(\dfrac{1}{4}\)

 

 

10 tháng 12 2023

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

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

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

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

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

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

\(ĐặtP=\dfrac{A}{B}\)

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

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

Để P<1 thì P-1<0

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

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

=>\(\sqrt{x}< 2\)

=>0<=x<4

mà x nguyên

nên \(x\in\left\{0;1;2;3\right\}\)

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 \(x=3-2\sqrt{2}\)

13 tháng 7 2021

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

 

13 tháng 7 2021

Từ khúc có \(x-4\sqrt{x}-1=0\)

Ta có: \(\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)=4-5=-1\)

Thế vào \(\Rightarrow x-4\sqrt{x}+\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)=0\)

\(\Rightarrow x-\sqrt{x}\left(2-\sqrt{5}+2+\sqrt{5}\right)+\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)=0\)

\(\Rightarrow x-\left(2-\sqrt{5}\right)\sqrt{x}-\left(2+\sqrt{5}\right)\sqrt{x}+\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)=0\)

\(\Rightarrow\sqrt{x}\left(\sqrt{x}-\left(2-\sqrt{5}\right)\right)-\left(2+\sqrt{5}\right)\left(\sqrt{x}-\left(2-\sqrt{5}\right)\right)=0\)

\(\Rightarrow\left(\sqrt{x}-\left(2-\sqrt{5}\right)\right)\left(\sqrt{x}-\left(2+\sqrt{5}\right)\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=2-\sqrt{5}\\\sqrt{x}=2+\sqrt{5}\end{matrix}\right.\) rồi khúc sau như trên

8 tháng 7 2023

\(a,P\) 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 \(P=\dfrac{1}{4}\)

a: Ta có: \(P=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)

\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

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

b: Thay \(x=\dfrac{1}{4}\) vào P, ta được:

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

c: Ta có: \(P< \dfrac{1}{2}\)

\(\Leftrightarrow P-\dfrac{1}{2}< 0\)

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

\(\Leftrightarrow\dfrac{2\sqrt{x}-2-\sqrt{x}-1}{2\left(\sqrt{x}+1\right)}< 0\)

\(\Leftrightarrow\sqrt{x}< 3\)

hay x<9

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne1\end{matrix}\right.\)

3 tháng 7 2023

Với \(x\ge0;x\ne4\) có:

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

a

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

b

\(P^2=P+2\\ \Leftrightarrow P^2-P-2=0\\ \Leftrightarrow P^2-2P+P-2=0\\ \Leftrightarrow P\left(P-2\right)+\left(P-2\right)=0\\ \Leftrightarrow\left(P-2\right)\left(P+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}P=2\\P=-1\end{matrix}\right.\)

Với P = 2 có:

\(\dfrac{4\sqrt{x}+1}{\sqrt{x}+1}=2\\ \Leftrightarrow2\left(\sqrt{x}+1\right)=4\sqrt{x}+1\\ \Leftrightarrow2\sqrt{x}+2-4\sqrt{x}-1=0\\\Leftrightarrow -2\sqrt{x}+1=0\\\Leftrightarrow-2\sqrt{x}=-1\\\Leftrightarrow \sqrt{x}=\dfrac{1}{2}\\ \Leftrightarrow x=\dfrac{1}{4} \)

Với P = -1 có:

\(\dfrac{4\sqrt{x}+1}{\sqrt{x}+1}=-1\\ \Leftrightarrow-\sqrt{x}-1-4\sqrt{x}-1=0\\ \Leftrightarrow-5\sqrt{x}=2\\ \Leftrightarrow\sqrt{x}=-\dfrac{2}{5}\left(loại\right)\)

Vậy để \(P^2=P+2\) thì \(x=\dfrac{1}{4}\)

a: P=A:B

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

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

b: P^2=P+2

=>P^2-P-2=0

=>(P-2)(P+1)=0

=>P=2(nhận) hoặc P=-1(loại)

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

=>2căn x=1

=>x=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)