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9 tháng 8 2018

Mình cần gấp nha mn 😭😭 

22 tháng 7 2020

1) Ta có: \(\frac{x+6\sqrt{x}+9}{x-9}=\frac{\left(\sqrt{x}+3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+3}{\sqrt{x}-3}\)

31 tháng 12 2021

a: \(P=\dfrac{x}{x+3}-\dfrac{x^2-5x-6}{\left(x-3\right)\left(x+3\right)}+\dfrac{3}{x-3}\)

\(=\dfrac{x^2-3x-x^2+5x+6+3x+9}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{5}{x-3}\)

13 tháng 5 2023

\(\left(\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{2\sqrt{x}+6}{x-9}\right):\dfrac{x-2\sqrt{x}}{\sqrt{x}-3}\left(x>3;x\ne9\right)\)

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

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

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

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

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

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

 

 

29 tháng 8 2020

Bài làm:

Ta có: 

\(P=\left(1-\frac{x-3\sqrt{x}}{x-9}\right)\div\left(\frac{\sqrt{x}-9}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\right)\)

\(P=\frac{x-9-x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\div\left[\frac{\left(9-\sqrt{x}\right)\left(3+\sqrt{x}\right)+\left(\sqrt{x}-2\right)^2-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]\)

\(P=\frac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\div\frac{-x+6\sqrt{x}+27+x-4\sqrt{x}+2-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(P=\frac{3}{\sqrt{x}+3}\div\frac{x+2\sqrt{x}+20}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(P=\frac{3}{\sqrt{x}+3}\cdot\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{x+2\sqrt{x}+20}\)

\(P=\frac{3\left(\sqrt{x}-2\right)}{x+2\sqrt{x}+20}=\frac{3\sqrt{x}-6}{x+2\sqrt{x}+20}\)

12 tháng 7 2018

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

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

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

\(B=\frac{\sqrt{\left(3x+2\right)^2}}{3x+2}=\frac{|3x+2|}{3x+2}\)

\(TH1:3x+2>0\Rightarrow B=1\)

\(TH2:3x+2< 0\Rightarrow B=-1\)

12 tháng 7 2018

A <=> 4√x - [ ( (√x )^2 + 2√x3+ 3^2)*( √x -3)]/ (x-9)

<=> 4√x - [(√x+3)^2×(√x-3)]/( x-9)

<=> 4√x - [(√x+3)*(x-9)]/(x-9)

<=> 4√x - √x -3

<=> 3√x -3

b, <=> √[(3*x) ^2+2*3x*2+2^2]/(3x+2)

<=> √[( 3x+2)^2] /(3x+2) 

<=> (3x+2)/(3x+2) = 1

26 tháng 8 2023

Bài 1:

a) \(x.\dfrac{3}{4}=\dfrac{9}{14}\)

\(\Rightarrow x=\dfrac{9}{14}:\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{6}{7}\)

b) \(x:\dfrac{5}{9}=\dfrac{3}{10}\)

\(\Rightarrow x=\dfrac{3}{10}.\dfrac{5}{9}\)

\(\Rightarrow x=\dfrac{1}{6}\)

26 tháng 8 2023

help me

\(=\dfrac{2x-9}{\left(x-2\right)\left(x-3\right)}-\dfrac{x+3}{x-2}+\dfrac{2x+1}{x-3}\)

\(=\dfrac{2x-9-x^2+9+\left(2x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{-x^2+2x+2x^2-4x+x-2}{\left(x-2\right)\left(x-3\right)}\)

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

20 tháng 2 2022

\(\dfrac{2x-9}{x^2-5x+6}-\dfrac{x+3}{x-2}-\dfrac{2x+1}{3-x}\left(ĐKXĐ:x\ne2,x\ne3\right)\)

\(=\dfrac{2x-9}{x^2-3x-2x+6}-\dfrac{x+3}{x-2}+\dfrac{2x+1}{x-3}\)

\(=\dfrac{2x-9}{x\left(x-3\right)-2\left(x-3\right)}-\dfrac{x+3}{x-2}+\dfrac{2x+1}{x-3}\)

\(=\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{x+3}{x-2}+\dfrac{2x+1}{x-3}\)

\(=\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{\left(x+3\right)\left(x-3\right)}{x-2}+\dfrac{\left(2x+1\right)\left(x-2\right)}{x-3}\)

\(=\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{x^2-9}{x-2}+\dfrac{2x^2-4x+x-2}{x-3}\)

\(=\dfrac{2x-9-x^2+9+2x^2-4x+x-2}{\left(x-3\right)\left(x-2\right)}\)

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

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

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

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

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