tex] равна их произведению?...

July 2022 1 1 Report

При каком значении игрик(у) разность дробей [tex] \frac{1}{y} [/tex] и [tex] \frac{y}{y+1} [/tex] равна их произведению?

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nafanya2014

Verified answer

ОДЗ:
{y≠0
{y+1≠0

$\frac{1}{y} - \frac{y}{y+1} =\frac{1}{y} \cdot \frac{y}{y+1} \\ \\ \frac{y+1-y}{y(y+1)}= \frac{1}{y+1} \\ \\ \frac{1}{y(y+1)}= \frac{1}{y+1}$

y+1=y(y+1)
(y+1)-y(y+1)=0
(y+1)(1-y)=0
y+1≠0
1-y=0
y=1

О т в е т. у=1

2 votes Thanks 1

yana01111 Спасибо

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