E.199. Determinați soluția ecuației: 2(x+2−5)+3=x(5−2).2(x+\sqrt{2}-\sqrt{5})+3 = x(\sqrt{5}-\sqrt{2}).2(x+2−5)+3=x(5−2).
Răspuns: x=−2.x=-2.x=−2.
2x+22−25+4=x5−x2.2x+2\sqrt{2}-2\sqrt{5}+4= x\sqrt{5}-x\sqrt{2}.2x+22−25+4=x5−x2. 2x−x5+x2=25−22−4.2x-x\sqrt{5} + x\sqrt{2} = 2\sqrt{5}-2\sqrt{2}-4.2x−x5+x2=25−22−4. x(2−5+2)=2(5−2−2).x(2-\sqrt{5} + \sqrt{2}) = 2(\sqrt{5}-\sqrt{2}-2).x(2−5+2)=2(5−2−2).
x=2(5−2−2)2−5+2=−2(2−5+2)2−5+2⇒x=−2.x=\dfrac{2(\sqrt{5}-\sqrt{2}-2)}{2-\sqrt{5} + \sqrt{2}} = \dfrac{-2\cancel{(2-\sqrt{5}+\sqrt{2})}}{\cancel{2-\sqrt{5} + \sqrt{2}}} \Rightarrow \boxed{x=-2}.x=2−5+22(5−2−2)=2−5+2−2(2−5+2)⇒x=−2.