plz la solution c'est pour demain x²−3+(x−√3)(2x+3√2)

Bonjour Yanyouness

[tex]x^2-3+(x-\sqrt{3})(2x+3\sqrt{2})=x^2-(\sqrt{3})^2+(x-\sqrt{3})(2x+3\sqrt{2})\\\\=(x-\sqrt{3})(x+\sqrt{3})+(x-\sqrt{3})(2x+3\sqrt{2})\\\\=(x-\sqrt{3})[(x+\sqrt{3})+(2x+3\sqrt{2})]\\\\=(x-\sqrt{3})(x+\sqrt{3}+2x+3\sqrt{2})\\\\=(x-\sqrt{3})(3x+\sqrt{3}+3\sqrt{2})[/tex]

Par conséquent,

[tex]\boxed{x^2-3+(x-\sqrt{3})(2x+3\sqrt{2})=(x-\sqrt{3})(3x+\sqrt{3}+3\sqrt{2})}[/tex]