Question

Rezolvati in multimea nr reale
Doar d va rog

Answer 1

Răspuns:

Explicație pas cu pas:

$notam~ t=2x^{2}+x+1,~atunci~2x^{2}+x-1=2x^{2}+x+1-2=t-2.~Obtinem~\sqrt{t} =t-2\\pentru~t<2~n-are~solutii;~pentru~t\geq 0~si~t-2\geq 0,~are~solutii.\\\sqrt{t} =t-2 |^2,~t=(t-2)^{2},~t^{2}-5x+4=0,~delta=(-5)^{2}-4*1*4=25-16=9>0\\t_{1}=\frac{5-3}{2}=1<2,~deci~nevalabila;~ t_{2}=\frac{5+3}{2}=4>2,\\deci~2x^{2}+x+1=4,~2x^{2}+x-3=0,~delta=1^{2}-4*2*(-3)=1+24=25>0\\x_{1}=\frac{-1-5}{2*2}=-\frac{3}{2};~  x_{2}=\frac{-1+5}{2*2}=1$

S={-3/2; 1}

Answer 2

$\it \sqrt{2x^2+x+1}=2x^2+x-1\\ \\ O\ condi\c{\it t}ie\ de\ existen\c{\it t}\breve{a}\ este:\\ \\ 2x^2+x-1>0 \Rightarrow\ 2x^2+x>1\ \ \ \ \ (*) \\ \\ Vom\ nota\ 2x^2+x=t,\ t>1\ \ \ (*)\ iar\ ecua\c{\it t}ia\ devine:\\ \\ \sqrt{t+1}=t-1\ \Rightarrow\ (\sqrt{t+1})^2=(t-1)^2\ \Rightarrow t+1 =t^2-2t+1 \Rightarrow \\ \\ \\ \Rightarrow 0=t^2-2t+1-t-1\ \Rightarrow\ t^2-3t=0\ \Rightarrow\ t(t-3)=0\ \Rightarrow$

$\it \Rightarrow \begin{cases}\it t=0\ (nu\ convine,\ deoarece\ t>1\ \ \ \ (*)\ )\\ \\ t-3=0\ \Rightarrow\ t=3\end{cases}$

Revenim asupra notației:

$\it\ t=3 \Rightarrow 2x^2+x=3 \Rightarrow 2x^2+x-3=0 \Rightarrow\ x=1\ sau\ x=-\dfrac{3}{2}\\ \\ S=\{-\dfrac{3}{2},\ \ 1\}$