K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

1 tháng 8 2018

1 , ĐKXĐ : \(x\ge0,x\ne1\)

Với điều kiện xác định trên phương trình đã cho thánh :

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

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

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

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

10 tháng 7 2021

a) \(B=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\left(x\ge0,x\ne1\right)\)

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

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

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

 

10 tháng 7 2021

undefinedundefined

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

16 tháng 11 2021

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

25 tháng 5 2023

\(a,\) Rút gọn 

\(A=\dfrac{3}{\sqrt{7}-2}+\sqrt{\left(\sqrt{7}-3\right)^2}\)

\(=\dfrac{3}{\sqrt{7}-2}+\left|\sqrt{7}-3\right|\)

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

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

\(=\dfrac{3+3\sqrt{7}-6-7+2\sqrt{7}}{\sqrt{7}-2}\)

\(=\dfrac{5\sqrt{7}-10}{\sqrt{7}-2}\)

\(=\dfrac{5\left(\sqrt{7}-2\right)}{\sqrt{7}-2}\)

\(=5\)

Vậy \(A=5\)

\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}}{x-\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{x-1}\left(dkxd:x\ge0,x\ne1\right)\)

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

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

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

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

Vậy \(B=\sqrt{x}-1\)

\(b,\) Để \(B< A\) thì \(\sqrt{x}-1< 5\)

\(\Leftrightarrow\sqrt{x}< 6\)

\(\Leftrightarrow x< 36\)

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

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

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

b) Để \(P=\dfrac{3}{2}\) thì \(4\sqrt{x}+2=3\sqrt{x}+3\)

\(\Leftrightarrow x=1\)(Vô lý)

21 tháng 7 2021

em cảm ơn ạ vui

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)

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

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

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

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

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

c) Để \(P< -\dfrac{1}{2}\) thì \(P+\dfrac{1}{2}< 0\)

\(\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}+\dfrac{1}{2}< 0\)

\(\Leftrightarrow\dfrac{-6+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}< 0\)

\(\Leftrightarrow\sqrt{x}-3< 0\)

\(\Leftrightarrow x< 9\)

Kết hợp ĐKXĐ, ta được: \(0\le x< 9\)

31 tháng 10 2021

\(a,A=\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\dfrac{x-2-x+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\\ A=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\\ A=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\notin\left\{9;4\right\}\end{matrix}\right.\)

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

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

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

18 tháng 10 2021

a. B = \(\dfrac{\sqrt{36}}{\sqrt{36}-3}=\dfrac{6}{6-3}=2\)

 

18 tháng 10 2021

a: Thay x=36 vào B, ta được:

\(B=\dfrac{6}{6-3}=\dfrac{6}{3}=2\)