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sendo x=4,333. e y=3,444. calcule o valor de:

1)9. x-18. y

2)x. y

3)x/y

4)y/x

5)x. y/x+y

6)x: y/x+y

7)x. y/x: y

valeu galera. é desespero mesmo. SerginhoHunter



RESOLVENDO

Malu Black.

\( x=4,333.=4+\frac39=4+\frac13=\frac{12+1}{3}=\frac{13}3\\\\y=3,444.=3+\frac49=\frac{27+4}9=\frac{31}9\\\\ 1)9x-18y=9\cdot \frac{13}3-18\cdot \frac{31}9=3\cdot 13-2\cdot 31=39-62=-23\\\\ 2)x. y=\frac{13}3\cdot \frac{31}9=\frac{403}{27}\\\\ 3)x/y=\frac{13}3\cdot \frac9{31}=\frac{39}{31}\\\\ 4)y/x=\frac{31}9 \cdot \frac{13}3=\frac{403}{27} \)

\( 5)x. y/x+y=y+y=2y=2 \cdot \frac{31}9=\frac{62}9 \)

\( 6)x: y/x+y=\frac{x}{y} \cdot \frac1{x}+y=\frac1{y}+y=\frac9{31}+\frac{31}9=\frac{81+961}{279}=\frac{1042}{279}\\\\ 7)x. y/x: y=xy/x: y=y: y=1 \)



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