Question

cum se rezolva ecuația de la ex 22b)​

Answer 1

Răspuns:

Explicație pas cu pas:

x-1≥0, ⇒x≥1. Este multimea de valori admisibile.

$\sqrt{x+8-6\sqrt{x-1}}=\sqrt{x-1+1+8-6\sqrt{x-1}} = \sqrt{x-1-2*\sqrt{x-1}*3+9} =  \sqrt{(\sqrt{x-1} ^{2}-2*\sqrt{x-1}*3+3^{2}} =\sqrt{(\sqrt{x-1}-3)^{2} }=| \sqrt{x-1}-3|\\\sqrt{4x-3-4\sqrt{x-1} } =\sqrt{4(x-1)-4\sqrt{x-1}+1} =\sqrt{(2\sqrt{x-1} )^{2}-2*2\sqrt{x-1}*1+1^{2}}=\sqrt{(2\sqrt{x-1}-1)^{2}} =|2\sqrt{x-1}-1|.~Obtinem~ecuatia:\\| \sqrt{x-1}-3|+|2\sqrt{x-1}-1|=3.~Aflam~zerourile~modulelor:~\sqrt{x-1}-3=0,~\sqrt{x-1}=3,~x-1=3^{2},~x=10.\\2\sqrt{x-1}-1=0,~2\sqrt{x-1}=1,~\sqrt{x-1}=\frac{1}{2}$

$x=1+\frac{1}{4} =\frac{5}{4}$

Vom diviza multimea valorilor admisibile in 3 intervale: [1; 5/4), [5/4; 10) si [10; +∞)

$caz~1:~pe~[1; 5/4),~|\sqrt{x-1}-3|=3-\sqrt{x-1},~|2\sqrt{x-1}-1|=1-2\sqrt{x-1}.~obtinem~ecuatia~3-\sqrt{x-1}+1-2\sqrt{x-1}=3,~3\sqrt{x-1}=1,~\sqrt{x-1}=\frac{1}{3},~x-1=\frac{1}{9},~x=\frac{10}{9}~apartine~[1; 5/4).\\caz~2:~pe[5/4;10),~|\sqrt{x-1}-3|=3-\sqrt{x-1},~|2\sqrt{x-1}-1|=2\sqrt{x-1}-1.~obtinem~ecuatia~3-\sqrt{x-1}+2\sqrt{x-1}-1=3,~\sqrt{x-1}=1,~x=2~apartine~[5/4;10).\\caz~3.~pe~[10;+infinit)~|\sqrt{x-1}-3|=\sqrt{x-1}-3,~|2\sqrt{x-1}-1|=2\sqrt{x-1}-1.~obtinem~ecuatia~\sqrt{x-1}-3+2\sqrt{x-1}-1=3,$

$3\sqrt{x-1}=7,~\sqrt{x-1}=\frac{7}{3},~x-1=\frac{49}{9},~x=\frac{58}{9}~nu~apartine~[10;+infinit).$

deci am obtinut dou[ solutii: x∈{10/9; 2}