Question

Cine mă poate ajuta la matematică ​

Answer 1

1)

$\bf -\dfrac{\sqrt{2} }{7}\cdot\sqrt{3\dfrac{1}{2} }=$

$\bf -\dfrac{\sqrt{2} }{7}\cdot\sqrt{\dfrac{7}{2} }=$

$\bf -\dfrac{\sqrt{2} }{7}\cdot\dfrac{\sqrt{7} }{\sqrt{2}} =$

$\bf -\dfrac{\not\sqrt{2} }{7}\cdot\dfrac{\sqrt{7} }{\not\sqrt{2}} =$

$\bf -\dfrac{1}{7}\cdot\dfrac{\sqrt{7} }{1} =$

$\boxed{\bf -\dfrac{\sqrt{7}}{7}}$

2)

$\bf -388:\sqrt{98} =$

$\bf -\dfrac{338}{7\sqrt{2}} =$

$\bf -\dfrac{338}{7\sqrt{2}}\cdot \dfrac{\sqrt{2} }{\sqrt{2}} =$

$\bf -\dfrac{338\cdot{\sqrt{2}} }{7\sqrt{2}\cdot \sqrt{2}}} =$

$\bf -\dfrac{338{\sqrt{2}} }{14}} =$

$\bf -\dfrac{\not338{\sqrt{2}}}{\not14}} =$

$\boxed{\bf -\dfrac{169{\sqrt{2}} }{7}}}$

3)

$\bf \dfrac{{\sqrt{5,4}} }{\sqrt{9,4} }} =$

$\bf \dfrac{{\dfrac{\sqrt{27}}{\sqrt{5}}}}{\dfrac{\sqrt{47}}{\sqrt{5}}}} =$

$\bf \dfrac{\sqrt{27}}{\sqrt{5}}}}\cdot{\dfrac{\sqrt{5}}{\sqrt{47}}}} =$

$\bf \dfrac{\sqrt{27}}{\not\sqrt{5}}}}\cdot{\dfrac{\not\sqrt{5}}{\sqrt{47}}}} =$

$\bf \dfrac{\sqrt{27}}{\sqrt{47}}}} =$

$\bf \dfrac{3\sqrt{3}}{\sqrt{47}}}}\cdot{\dfrac{\sqrt{47}}{\sqrt{47}}}} =$

$\boxed{\bf \dfrac{3\sqrt{141}}{\sqrt{47}}}}$