Question

REPEDE VA ROG!!!!!!!​

Answer 1

 

$\displaystyle\bf\\a=2^{1990}-2^{1989}-2^{1988}\\\\Se~cere~x~din~proportia\!:~~~\frac{a}{x}=\frac{4^{993}}{0,25}.\\\\Rezolvare\!:\\\\a=2^{1990}-2^{1989}-2^{1988}\\\\a=2^{1988+2}-2^{1988+1}-2^{1988}\\\\a=2^{1988}\big(2^2-2^1-1\big)\\\\a=2^{1988}\big(\underbrace{4-2-1}_{=1}\big)\\\\\boxed{\bf a=2^{1988}}$

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$\displaystyle\bf\\\frac{a}{x}=\frac{4^{993}}{0,25}\\\\\\\frac{a}{x}=\frac{\bigg(2^2\bigg)^{993}}{0,25}\\\\\\\frac{a}{x}=\frac{\bigg(2^2\bigg)^{993}}{0,25}\\\\\\\frac{a}{x}=\frac{2^{2\times993}}{0,25}\\\\\\\frac{a}{x}=\frac{2^{1986}}{0,25}\\\\\\a=2^{1988}\\\\\frac{2^{1988}}{x}=\frac{2^{1986}}{0,25}\\\\\\x=\frac{2^{1988}\times0,\!25}{2^{1986}}\\\\\\x=2^{1988-1986}\times0,\!25\\\\x=2^2\times0,\!25\\\\x=2^2\times\frac{1}{4} \\\\\\x=\frac{2^2}{4}\\\\x=\frac{4}{4}\\\\\boxed{\boxed{\bf x=1}}$