Question

calculati 3x-2√3,unde x=√432(1supra√27-1supra√48)si y=√3+1-√3 supra 3- √3

Answer 1

Răspuns:

Explicație pas cu pas:

$\displaystyle x=\sqrt{432} (\frac{1}{\sqrt{27} } -\frac{1}{\sqrt{48} } )\\x=\sqrt{4^{2}\times 3^{3} } (\frac{\sqrt{27} }{27} -\frac{\sqrt{48} }{48} )\\x=12\sqrt{3} (\frac{\sqrt{3^{3} } }{27} -\frac{\sqrt{4^{2}\times 3 } }{48} )\\x=12\sqrt{3} (\frac{\not3\sqrt{3} }{\not27} -\frac{\not4\sqrt{3} }{\not48} )\\x=12\sqrt{3} ( \frac{\sqrt{3} }{9} -\frac{\sqrt{3} }{12} )\\x=12\sqrt{3} (\frac{4\sqrt{3} }{36} -\frac{3\sqrt{3} }{36} )\\x=12\sqrt{3} \times \frac{\sqrt{3} }{36} \\x=\frac{12\times 3}{36} \\x=1$

$\displaystyle y=\sqrt{3} +1-\frac{\sqrt{3} }{3-\sqrt{3} } \\y=\frac{(\sqrt{3}+1)(3-\sqrt{3} )-\sqrt{3} }{3-\sqrt{3} } \\y=\frac{3\sqrt{3}-3+3-\sqrt{3} -\sqrt{3} }{3-\sqrt{3} } \\y=\frac{\sqrt{3} }{3-\sqrt{3} } \\y=\frac{\sqrt{3} (3+\sqrt{3} )}{9-3} \\y=\frac{3\sqrt{3}+3 }{6}$

3x+2√3=3·1+2√3=3+2√3