Question

Radical din 3-x- radical din2-x =1

Answer 1

[tex]\sqrt{3-x}-\sqrt{2-x}=1\\ C.E. \left \{ {{3-x \geq 0} \atop {2-x \geq 0}} \right. \\ \sqrt{3-x}-\sqrt{2-x}=1\\ ridicam\:la\:patrat\\ (\sqrt{3-x}-\sqrt{2-x})^2=1\\ (\sqrt{3-x})^2-2\sqrt{3-x}\cdot\sqrt{2-x}+(\sqrt{2-x})^2=1\\ 3-x-2\sqrt{(3-x)(2-x)}+2-x=1\\ 5-2x-2\sqrt{(3-x)(2-x)}=1\\ -2\sqrt{(3-x)(2-x)}=1-5+2x\\ -2\sqrt{(3-x)(2-x)}=-4+2x\\ \sqrt{(3-x)(2-x)}=2-x\\ ridicam\:la\:patrat\\ (3-x)(2-x)=(2-x)^2\\ 6-3x-2x+x^2=4-4x+x^2\\ 6-5x+x^2=4-4x+x^2\\ 6-5x=4-4x\\ x=2\:verifica\:C.E. [/tex]