Question

Simplificat fracțiune xy00xy pe 610061 ; ab+ba pe xy+yx ; abc+bca+cab pe xyz+yzx+zxy

Answer 1

  
[tex]\displaystyle \\ a)~~ \frac{\overline{xy00xy}}{610061} = \frac{10001 \cdot \overline{xy}}{10001 \cdot 61} = \frac{\overline{xy}}{61} \\ \\ \\ b)~~ \frac{\overline{ab}+\overline{ba}}{\overline{xy}+\overline{yx}}= \frac{10a+b+10b+a}{10x+y+10y+x}= \frac{11a+11b}{11x+11y} = \frac{11(a+b)}{11(x+y)} =\frac{a+b}{x+y} \\ \\ \\ c)~~ \frac{\overline{abc}+\overline{bca}+\overline{cab}}{\overline{xyz}+\overline{yzx}+\overline{zxy}}= [/tex]


[tex]\displaystyle \\ =\frac{ 100a+10b+c +100b+10c+a+ 100c+10a+b}{ 100x+10y+z +100y+10z+x+100z+10x+y}= \\ \\ \\ = \frac{111a+111b+111c}{111x+111y+111z}= \frac{111(a+b+c)}{111(x+y+z)}= \frac{a+b+c}{x+y+z}[/tex]