Question

aflati x apartine Q+ din proportiile:
a)x/6=15A/27,iar 15A se divide cu 9
b)3/46=x/2A,iar 2A este numar prim,2A mai mic si egal decat 25
c)2A5/35=36/x,2A5 se divide cu 9
d)2²⁰¹⁹+2²⁰²⁰+2²⁰²¹=32⁴⁰⁴/0,5,x nu este egal cu 0
e)x/1+2+3+...+102=0,(3)/103​

Answer 1

Explicație pas cu pas:

a)

$\overline {15a} : 9 \iff (1 + 5 + a) : 9 \\ \iff (6 + a) : 9 \implies a = 3$

$\frac{x}{6} = \frac{153}{27} \iff x = \frac{6 \cdot 153}{27} \\ \implies x = \bf 34$

b)

$\overline {2a}\leqslant 25 \implies \overline {2a} = 23$

$\frac{3}{46} = \frac{x}{23} \iff x = \frac{3 \cdot 23}{46} \\ \implies x = \bf \frac{3}{2}$

c)

$\overline {2a5} : 9 \iff (2 + a + 5) : 9 \\ \iff (7 + a) : 9 \implies a = 2$

$\frac{225}{35} = \frac{36}{x} \iff x = \frac{35 \cdot 36}{225} \\ \implies x = \bf \frac{28}{5}$

d)

${2}^{2019} + {2}^{2020} + {2}^{2021} = {2}^{2019}(1 + 2 + 4) = 7 \cdot {2}^{2019}$

$\frac{{32}^{404}}{0.5} = \frac{{( {2}^{5} )}^{404}}{ \frac{1}{2} } = 2 \cdot {2}^{2020} = {2}^{2021}$

(unde este x??)

e)

$\frac{x}{1 + 2 + 3 + ... + 102} = \frac{0.3}{103} \\ \frac{x}{ \frac{102 \cdot 103}{2} } = \frac{ \frac{3}{9} }{103} \iff \frac{2x}{102 \cdot 103} = \frac{1}{3 \cdot 103} \\ x = \frac{102}{2 \cdot 3} \implies x = \bf 17$