Question

Am nevoie pentru nota,urgent!Dau coroana!!!

Answer 1

Răspuns

Răspunsul atașat. .....

Explicație pas cu pas:

Answer 2

$\it \{a,b\}\ d.\ p.\ \{6,3\}\Rightarrow \dfrac{a}{6}=\dfrac{b}{3}=k\Rightarrow\begin{cases}\it a=6k\\ \\\it b=3k\end{cases},\ \ \ k\in\mathbb{N}^*\\ \\ \\ \dfrac{a}{a+b} =\dfrac{6k}{6k+3k}=\dfrac{6k}{9k}=\dfrac{\ 6^{(3}}{9}=\dfrac{2}{3}\\ \\ \\ \dfrac{a^2}{a^2+b^2} =\dfrac{(6k)^2}{(6k)^2+(3k)^2}=\dfrac{36k^2}{36k^2+9k^2}=\dfrac{36k^2}{45k^2}=\dfrac{\ 36^{(9}}{45}=\dfrac{4}{5}$

$\it \dfrac{a^3}{a^3+b^3} =\dfrac{(6k)^3}{(6k)^3+(3k)^3}=\dfrac{216k^3}{216k^3+27k^3}=\dfrac{216k^3}{243k^3}=\dfrac{\ 216^{(9}}{243}=\dfrac{\ 24^{(3}}{27}=\dfrac{8}{9}$

$\it \dfrac{^{15)}2}{\ \ 3}<\dfrac{^{9)}4}{\ 5}<\dfrac{^{5)}8}{\ 9}\Rightarrow \dfrac{30}{45}<\dfrac{36}{45}<\dfrac{40}{45}\ \ (A)$