Răspunsuri

  • Utilizator Brainly
2014-02-16T22:13:26+02:00
(6/rad2*x-9/rad3*x)+(3/rad18*x+10/rad75*x)-(24/2rad48*x-12/rad108*x)=
=(6rad3x-9rad2x)/rad6+(3/3rad2*x+10/5rad3*x)-(12/4rad3*x-12/6rad3*x)=
=(6rad3x-9rad2x)/rad6 +(xrad3+2xrad2)/rad6-x/rad3=
=(6xrad3-9xrad2+xrad3+2xrad2-xrad2)/rad6=(7xrad3-8xrad2)/rad6
Cel mai inteligent răspuns!
2014-02-16T23:17:09+02:00
După ce scoţi întregii şi desfaci parantezele, se obţine:
\dfrac{6x}{\sqrt2}-\dfrac{9x}{\sqrt3}+\dfrac{3x}{3\sqrt2}+\dfrac{10x}{5\sqrt3}-\dfrac{12x}{4\sqrt3}+\dfrac{12x}{6\sqrt3}= acum simplificam si adunam fractiile cu acelasi numitor, si obtinem
=\dfrac{7x}{\sqrt2}-\dfrac{8x}{\sqrt3}=\dfrac{7x\sqrt2}{2}-\dfrac{8x\sqrt3}{3}=\dfrac{x(21\sqrt2-16\sqrt3)}{6}
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