Réponses

2014-06-22T23:16:54+02:00
Bonsoir,

\ln(x^2-1)=2\ln(x+1)+\ln(2x-5)\\\\Conditions : \left\{\begin{matrix}x^2-1>0\\x+1>0\\2x-5>0\end{matrix}\right.\ \ \ \Longrightarrow \left\{\begin{matrix}x\in]-\infty;-1[\cup]1;+\infty[\\x\in]-1;+\infty[\\x\in]\dfrac{5}{2};+\infty[\end{matrix}\right.\\\\\\\ \ \ \Longrightarrow \boxed{x\in]\dfrac{5}{2};+\infty[}

\ln[(x+1)(x-1)]=2\ln(x+1)+\ln(2x-5)\\\ln(x+1)+\ln(x-1)=2\ln(x+1)+\ln(2x-5)\\\ln(x-1)=2\ln(x+1)+\ln(2x-5)-\ln(x+1)\\\ln(x-1)=\ln(x+1)+\ln(2x-5)\\\ln(x-1)=\ln[(x+1)(2x-5)]\\\\x-1=(x+1)(2x-5)\\x-1=2x^2-5x+2x-5\\2x^2-5x+2x-5=x-1

2x^2-5x+2x-5-x+1=0\\2x^2-4x-4=0\\x^2-2x-2=0\\\\\Delta=(-2)^2-4\times1\times(-2)=4+8=12\\\\x_1=\dfrac{2-\sqrt{12}}{2}=\dfrac{2-2\sqrt{3}}{2}\\=\dfrac{2(1-\sqrt{3})}{2}=1-\sqrt{3}\approx-0,73\ \ \ (\grave{a}\ rejeter\ car\ x\in]\dfrac{5}{2};+\infty[)

\\\\x_2=\dfrac{2+\sqrt{12}}{2}=\dfrac{2+2\sqrt{3}}{2}\\=\dfrac{2(1+\sqrt{3})}{2}=1+\sqrt{3}\approx2,73

L'ensemble des solutions est  \boxed{ S={1+\sqrt{3}}}