Question

amplificați cu numărul natural nenul n fracțiile:3/4;7/2;3a/7b;a+1/b+3;a+b/x+y.​

Answer 1

Răspuns:

         

Explicație pas cu pas:

$\bf \!{^{^{^{^{^{^{\displaystyle n)}}}}}}}\!\!\!\dfrac{3}{4} = \dfrac{3n}{4n}$

$\bf \!{^{^{^{^{^{^{\displaystyle n)}}}}}}}\!\!\!\dfrac{7}{2} =\dfrac{7 \cdot n}{2 \cdot n} = \dfrac{7n}{2n}$

$\bf \!{^{^{^{^{^{^{\displaystyle n)}}}}}}}\!\!\!\dfrac{3a}{7b} =\dfrac{n\cdot 3a}{n \cdot 7b} = \dfrac{3an}{7bn}$

$\bf \!{^{^{^{^{^{^{\displaystyle n)}}}}}}}\!\!\!\dfrac{3a}{7b} =\dfrac{n \cdot 3a}{n \cdot 7b} = \dfrac{3an}{7bn}$

$\bf \!{^{^{^{^{^{^{\displaystyle n)}}}}}}}\!\!\!\dfrac{(a+1)}{(b+3)} =\dfrac{n \cdot (a+1)}{n \cdot (b+3)} = \dfrac{an+n}{bn+3n}$

$\bf \!{^{^{^{^{^{^{\displaystyle n)}}}}}}}\!\!\!\dfrac{(a+b)}{(x+y)} =\dfrac{n \cdot (a+b)}{n \cdot (x+y)} = \dfrac{an+bn}{xn+yn}$