K
Khách

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

29 tháng 1 2020

\(ĐKXĐ:x\ne2;x\ne4\)

\(\frac{x-3}{x-2}-\frac{x-2}{x-4}=3\frac{1}{5}\)

\(\Rightarrow\frac{\left(x-3\right)\left(x-4\right)-\left(x-2\right)^2}{\left(x-2\right)\left(x-4\right)}=\frac{16}{5}\)

\(\Rightarrow\frac{x^2-7x+12-x^2+4x-4}{x^2-6x+8}=\frac{16}{5}\)

\(\Rightarrow\frac{-3x+8}{x^2-6x+8}=\frac{16}{5}\)

\(\Rightarrow-3x+8=\frac{16}{5}\left(x^2-6x+8\right)\)

\(\Rightarrow-3x+8=\frac{16}{5}x^2-\frac{96}{5}x+\frac{128}{5}\)

\(\Rightarrow\frac{16}{5}x^2-\frac{81}{5}x+\frac{88}{5}=0\)

Ta có \(\Delta=\frac{81^2}{5^2}-4.\frac{16}{5}.\frac{88}{5}=\frac{929}{25},\sqrt{\Delta}=\frac{\sqrt{929}}{5}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{81+\sqrt{929}}{32}\\x=\frac{81-\sqrt{929}}{32}\end{cases}}\)

\(\left(\frac{1}{x-1}+\frac{1}{x-4}\right)-\left(\frac{1}{x-2}+\frac{1}{x-3}\right)=0\)

\(\Leftrightarrow\frac{x-4+x-1}{\left(x-1\right).\left(x-4\right)}-\frac{x-3-x-2}{\left(x-2\right).\left(x-3\right)}=0\)

\(\Leftrightarrow\frac{2x-5}{x^2-5x+4}-\frac{2x-5}{x^2-5x+6}=0\)

\(\Leftrightarrow\left(2x-5\right).\left(\frac{1}{x^2-5x+4}-\frac{1}{x^2-5x+6}=0\right)\)

\(\Leftrightarrow\orbr{\begin{cases}2x-5=0\\\frac{1}{x^2-5x+4}-\frac{1}{x^2-5x+6}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x^2-5x+4=x^2-5x+6\left(loai\right)\end{cases}}}\)

Vậy..

14 tháng 2 2019

\(\frac{x-4}{2000}+\frac{x-3}{2001}+\frac{x-2}{2002}=\frac{x-2002}{2}+\frac{x-2001}{3}+\frac{x-2000}{4}\)

\(\Rightarrow\left(\frac{x-4}{2000}-1\right)+\left(\frac{x-3}{2001}-1\right)+\left(\frac{x-2}{2002}-1\right)=\left(\frac{x-2002}{2}-1\right)+\left(\frac{x-2001}{3}-1\right)+\left(\frac{x-2000}{4}-1\right)\)\(\Rightarrow\frac{x-2004}{2000}+\frac{x-2004}{2001}+\frac{x-2004}{2002}=\frac{x-2004}{2}+\frac{x-2004}{3}+\frac{x-2004}{4}\)

\(\Rightarrow\left(x-2004\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}\right)=\left(x-2004\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)\)

Với \(x-2004\ne0\)

\(\Rightarrow\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\left(KTM\right)\)

Với \(x-2004=0\)

\(\Rightarrow x=2004\)

7 tháng 1 2016

(x+1)/2011+1+(x+2)/2010+1+(x+3)/2009+1-((x+4)/2008+1+(x+5)/2007+1+(x+6)/2006+1)=0

(x+2012)/2011+(x+2012)/2010+(x+2012/2009-(x+2012)/2008-(x+2012)/2007-(x+2012)/2006=0

(x+2012)(1/2011+1/2010+1/2009-1/2008-1/2007-1/2006)=0

x+2012=0

x=-2012

9 tháng 8 2015

\(-2=\frac{2}{\left(x^2+5\right)\left(x^2+4\right)}+\frac{2}{\left(x^2+4\right)\left(x^2+3\right)}+\frac{2}{\left(x^2+3\right)\left(x^2+2\right)}+\frac{2}{\left(x^2+2\right)\left(x^2+1\right)}\)

<=>\(\frac{1}{\left(x^2+5\right)\left(x^2+4\right)}+\frac{1}{\left(x^2+4\right)\left(x^2+3\right)}+\frac{1}{\left(x^2+3\right)\left(x^2+2\right)}+\frac{1}{\left(x^2+2\right)\left(x^2+1\right)}=-1\)

<=>\(\frac{1}{x^2+1}-\frac{1}{x^2+2}+\frac{1}{x^2+2}-\frac{1}{x^2+3}+...+\frac{1}{x^2+4}-\frac{1}{x^2+5}=-1\)

<=>\(\frac{1}{x^2+1}-\frac{1}{x^2+5}=-1\)

<=>(x2+5)-(x2+1)=-(x2+1)(x2+5)

<=>4=-x4-6x2-5

<=>x4+6x2+9=0

<=>(x2+3)2=0

<=>x2+3=0

Do x2>0

=>x2+3>0 nên PT vô nghiệm

11 tháng 3 2017

<=> \(\frac{3\left(x+2\right)-5x}{5.3}=\frac{2x-5}{2}< =>\frac{3x+6-5x}{15}=\frac{2x-5}{2}\) <=> 2(6-2x)=15(2x-5)

<=> 12-4x=30x-75 => 34x=87 => x=\(\frac{87}{34}\)

11 tháng 3 2017

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

\(\Leftrightarrow\frac{6\left(x+2\right)-10x-15\left(2x-5\right)}{30}\)=0

\(\Leftrightarrow6x+12-10x-30x+75\)=0

\(\Leftrightarrow-34x=-87\)

\(\Leftrightarrow x=\frac{87}{34}\)

Vay S={\(\frac{87}{34}\)}