Question

comparați fractiile ordinare 2^27+2^27 supra 2^27+2^28 si 2^29-2^26 supra 2^26+2^29 ​

Answer 1

$\it \dfrac{2^{27}+2^{27}}{2^{27}+2^{28}} =\dfrac{2\cdot2^{27}}{2^{27}+2\cdot2^{27}}=\dfrac{2\cdot2^{27}}{3\cdot2^{27}}=\dfrac{2}{3}\\ \\ \\ \dfrac{2^{29}-2^{26}}{2^{26}+2^{29}} =\dfrac{2^{26}(2^3-1)}{2^{26}(1+2^3)}=\dfrac{7}{9}$

$\it \dfrac{^{3)}2}{\ \ 3}=\dfrac{6}{9}<\dfrac{7}{9}$

Deci, prima fracție este mai mică decât a doua.