Question

ajutor. rezolvare completa va rog​

Answer 1

Răspuns: $\bf \dfrac{\sqrt{2}}{20}$

Explicație pas cu pas:

✻ Salutare! ✻

$\bf \Big(\dfrac{3}{\sqrt{12}} - \dfrac{2}{\sqrt{75}}+\dfrac{3}{2\sqrt{48}}-\dfrac{21}{5\sqrt{27}} \Big):\dfrac{\sqrt{6}}{4}=$

$\bf \Big(\dfrac{3}{2\sqrt{3}} - \dfrac{2}{5\sqrt{3}}+\dfrac{3}{8\sqrt{3}}-\dfrac{21}{15\sqrt{3}} \Big)\cdot\dfrac{4}{\sqrt{6}}=$

$\bf \Big(\dfrac{3}{2\sqrt{3}} - \dfrac{2}{5\sqrt{3}}+\dfrac{3}{8\sqrt{3}}-\dfrac{\not21}{\not15\sqrt{3}} \Big)\cdot\dfrac{4}{\sqrt{6}}=$

$\bf \Big(\dfrac{3}{2\sqrt{3}} - \dfrac{2}{5\sqrt{3}}+\dfrac{3}{8\sqrt{3}}-\dfrac{7}{5\sqrt{3}} \Big)\cdot\dfrac{4}{\sqrt{6}}=$

$\bf \dfrac{1}{\sqrt{3}}\cdot \Big(\dfrac{3}{2} - \dfrac{2}{5}+\dfrac{3}{8}-\dfrac{7}{5} \Big)\cdot\dfrac{4}{\sqrt{6}}=$

$\bf \dfrac{1}{\sqrt{3}}\cdot \Big(\dfrac{3}{2} - \dfrac{9}{5}+\dfrac{3}{8} \Big)\cdot\dfrac{4}{\sqrt{6}}=$

$\bf \dfrac{1}{\sqrt{3}}\cdot \Big(\dfrac{3\cdot 20-9\cdot 8+3\cdot5 }{40} \Big)\cdot\dfrac{4}{\sqrt{6}}=$

$\bf \dfrac{1}{\sqrt{3}}\cdot \dfrac{3}{40}\cdot\dfrac{4}{\sqrt{6}}=$

$\bf \dfrac{1}{\sqrt{3}}\cdot \dfrac{3}{\not40}\cdot\dfrac{\not4}{\sqrt{6}}=$

$\bf \dfrac{1}{\sqrt{3}}\cdot \dfrac{3}{10}\cdot\dfrac{1}{\sqrt{6}}=$

$\bf \dfrac{3}{10\sqrt{18}}=$

$\bf \dfrac{3}{30\sqrt{2}}=$

$\bf \dfrac{1}{10\sqrt{2}}=$

$\bf \dfrac{\sqrt{2}}{10\cdot2}=$

$\boxed{\bf \dfrac{\sqrt{2}}{20}}$

⊱─────✧pav38✧─────⊰

Answer 2

Răspuns:

$\frac{\sqrt{2} }{20}$

Explicație pas cu pas:

$(\frac{3}{\sqrt{12} } -\frac{2}{\sqrt{75} } +\frac{3}{2\sqrt{48} } -\frac{21}{5\sqrt{27} } ):\frac{\sqrt{6} }{4} =\\(\frac{3}{2\sqrt{3} } -\frac{2}{5\sqrt{3} } +\frac{3}{2*4\sqrt{3} } -\frac{21}{5*3\sqrt{3} } ):\frac{\sqrt{6} }{4} =\\(\frac{3}{2\sqrt{3} } -\frac{2}{5\sqrt{3} } +\frac{3}{8\sqrt{3} } -\frac{21}{15\sqrt{3} } ):\frac{\sqrt{6} }{4} =\\(\frac{3}{2\sqrt{3} } -\frac{2}{5\sqrt{3} } +\frac{3}{8\sqrt{3} } -\frac{7}{5\sqrt{3} } ):\frac{\sqrt{6} }{4} =$

$(\frac{3}{2} *\frac{1}{\sqrt{3} }-\frac{2}{5} *\frac{1}{\sqrt{3} } +\frac{3}{8} *\frac{1}{\sqrt{3} } -\frac{7}{5} *\frac{1}{\sqrt{3} } ):\frac{\sqrt{6} }{4} =\\\frac{1}{\sqrt{3} } *(\frac{3}{2} -\frac{9}{5} +\frac{3}{8} ):\frac{\sqrt{6} }{4} =\\\frac{1}{\sqrt{3} } *(\frac{3*20}{40} -\frac{9*8}{40} +\frac{3*5}{40} ):\frac{\sqrt{6} }{4} =\\\frac{1}{\sqrt{3} } *(\frac{60-72+15}{40} ):\frac{\sqrt{6} }{4} =\\\frac{1}{\sqrt{3} } *\frac{3}{40} *\frac{4}{\sqrt{6} } =$

$\frac{1}{\sqrt{3} } *\frac{3}{10} *\frac{1}{\sqrt{6} } =\\\frac{1}{\sqrt{3} } *\frac{1}{\sqrt{6} } *\frac{3}{10} =\\\frac{1}{\sqrt{18} } *\frac{3}{10} =\\\frac{1}{3\sqrt{2} } *\frac{3}{10} =\\\frac{3}{10*3\sqrt{2} } =\\\frac{1}{10\sqrt{2} } =\\\frac{\sqrt{2} }{10*2} =\frac{\sqrt{2} }{20}$

↔ΔTriunghiul1Δ↔