\(\sqrt{X}+3\)

">
K
Khách

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

9 tháng 11 2019

\(A=\frac{15\sqrt{x}-11}{x-\sqrt{x}+3\sqrt{x}-3}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\frac{45\sqrt{x}-11}{\left(\sqrt{x}+3\right)(\sqrt{x}-1)}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\frac{45\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{37\sqrt{x}-5x-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

6 tháng 8 2016

a) \(P=\left[\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-\left(3x+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]:\left[\frac{\left(2\sqrt{x}-2\right)-\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\right]\left(ĐK:x\ge0;x\ne9\right)\) 

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

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

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

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

 

 

 

 

10 tháng 8 2017

a/ \(P=\left(\frac{x-7\sqrt{x}+12}{x-4\sqrt{x}+3}+\frac{1}{\sqrt{x}-1}\right).\frac{\sqrt{x}+3}{\sqrt{x}-3}.\)

\(P=\left(\frac{x-7\sqrt{x}+12}{\left(x-4\sqrt{x}+4\right)-1}+\frac{1}{\sqrt{x}-1}\right).\frac{\sqrt{x}+3}{\sqrt{x}-3}\)

\(P=\left(\frac{x-7\sqrt{x}+12}{\left(\sqrt{x}-2\right)^2-1}+\frac{1}{\sqrt{x}-1}\right).\frac{\sqrt{x}+3}{\sqrt{x}-3}\)

\(P=\left(\frac{x-7\sqrt{x}+12}{\left(\sqrt{x}-2-1\right)\left(\sqrt{x}-2+1\right)}+\frac{1}{\sqrt{x}-1}\right).\frac{\sqrt{x}+3}{\sqrt{x}-3}\)

\(P=\left(\frac{x-7\sqrt{x}+12}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right).\frac{\sqrt{x}+3}{\sqrt{x}-3}\)

\(P=\frac{x-7\sqrt{x}+12+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}+3}{\sqrt{x}-3}\)

\(P=\frac{x-6\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}+3}{\sqrt{x}-3}\)

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

b/ Để P>3/4 => \(P=\frac{\sqrt{x}+3}{\sqrt{x}-1}>\frac{3}{4}\)

+/ TH1: x>1 => \(4\left(\sqrt{x}+3\right)>3\left(\sqrt{x}-1\right)\)

  <=> \(\sqrt{x}>-16\)  => x>1

+/ TH2: 0<x<1 => \(4\left(\sqrt{x}+3\right)< 3\left(\sqrt{x}-1\right)\)  => \(\sqrt{x}< -16\)=> Loại

       ĐS: x>1

c/ P=2  <=> \(P=\frac{\sqrt{x}+3}{\sqrt{x}-1}=2\)

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

<=> \(\sqrt{x}=5=>x=25\)

9 tháng 2 2018

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

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

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

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

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

\(M=\frac{3\left(x+\sqrt{x}-2\right)}{x+\sqrt{x}-2}\)

\(M=3\)

9 tháng 2 2018

b) \(\sqrt{x}=M\)

\(\Leftrightarrow x=M^2\)

thay vào ta có: 

\(x=3^2\)

\(x=9\)

c) \(M=3\in N\)

\(\Rightarrow x=3\)

d) \(M>1\Leftrightarrow x>1\)

1 tháng 8 2017

ĐK  \(\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)

a. Ta có \(P=\frac{\sqrt{x}-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{\sqrt{x}-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\frac{3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{3}=\frac{\sqrt{x}}{\sqrt{x}+3}\)

b.Để \(P< 0,5\Rightarrow\frac{\sqrt{x}}{\sqrt{x}+3}-0,5< 0\Leftrightarrow\frac{2\sqrt{x}-\sqrt{x}-3}{2\cdot\left(\sqrt{x}+3\right)}< 0\)

\(\Leftrightarrow\frac{\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}< 0\Leftrightarrow\sqrt{x}-3< 0\Leftrightarrow0\le x< 9\)

Vậy \(0\le x< 9\)thì \(P< 0,5\)

c. Để \(P=\frac{1}{2\sqrt{x}}\Rightarrow\frac{\sqrt{x}}{\sqrt{x}+3}=\frac{1}{2\sqrt{x}}\Leftrightarrow2x-\sqrt{x}-3=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=\frac{3}{2}\\\sqrt{x}=-1\left(l\right)\end{cases}\Leftrightarrow x=\frac{9}{4}\left(tm\right)}\)

Vậy \(x=\frac{9}{4}\)  

1 tháng 8 2017

các bạn sửa lại giúp mình đề bài  ở  đoạn P=.........-(1/căn x) thành P=.......+(1/căn x) với nha cảm ơn nhiều XD