Question

media aritmetica a trei numere rationale pozitive este egala cu 17 . media aritmetica a ultimelor doua este 16.stiind ca al doileanumar este egal cu o treime din al treilea aflati numerele

Answer 1

$M_{aritmetica} = \frac{\frac{a}{b} + \frac{c}{d} + \frac{e}{f} }{3}=17$

$\frac{a}{b} + \frac{c}{d} + \frac{e}{f} =17*3$

$\frac{a}{b} + \frac{c}{d} + \frac{e}{f} =51$

$M_{aritmetica} = \frac{\frac{c}{d} + \frac{e}{f} }{2}=16$

$\frac{c}{d} + \frac{e}{f} =16*2$

$\frac{c}{d} + \frac{e}{f} =32$

$\frac{b}{c} = \frac{1}{3} * \frac{e}{f}$

$\frac{1}{3} * \frac{e}{f} + \frac{e}{f} =32$

$\frac{e}{f} + 3*\frac{e}{f} =32*3$

$4*\frac{e}{f} =32*3$

$\frac{e}{f} =\frac{96}{4}$

$\frac{c}{d} + \frac{96}{4} =32$

$\frac{c}{d} =32-\frac{96}{4}$

$\frac{c}{d} =\frac{32*4-96}{4}$

$\frac{c}{d} =\frac{128-96}{4}$

$\frac{c}{d} =\frac{32}{4}$

$\frac{a}{b} + \frac{c}{d} + \frac{e}{f} =51$

$\frac{a}{b} + \frac{96}{4} + \frac{32}{4} =51$

$\frac{a}{b} + \frac{96+32}{4}=51$

$\frac{a}{b} + \frac{128}{4}=51$

$\frac{a}{b} =51-\frac{128}{4}$

$\frac{a}{b} =\frac{51*4-128}{4}$

$\frac{a}{b} =\frac{204-128}{4}$

$\frac{a}{b} =\frac{76}{4}$

Numerele sunt: $\frac{a}{b} =\frac{76}{4};\frac{c}{d} =\frac{32}{4};\frac{e}{f} =\frac{96}{4}$

Verificare:

$M_{aritmetica} = \frac{\frac{a}{b} + \frac{c}{d} + \frac{e}{f} }{3}=\frac{\frac{76}{4} + \frac{32}{4} + \frac{96}{4} }{3}=\frac{19+8+24}{3}=\frac{51}{3}=17$