ПОМОГИТЕ ПОЖАЛУЙСТА))). Created by shusam. algebra-ru

ПОМОГИТЕ ПОЖАЛУЙСТА)))...

shusam @shusam

June 2022 1 3 Report

ПОМОГИТЕ ПОЖАЛУЙСТА)))

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Luluput

Verified answer

$\frac{10-2y}{4y-5} + \frac{3y+16}{4y+3} = \frac{10-2y}{4y-5} * \frac{3y+16}{4y+3}$
$\frac{(10-2y)(4y+3)+(3y+16)(4y-5)}{(4y-5)(4y+3)} = \frac{(10-2y)(3y+16)}{(4y-5)(4y+3)}$
ОДЗ:
$4y-5 \neq 0$
$y \neq 1.25$
$4y+3 \neq 0$
$y \neq -0.75$

$(10-2y)(4y+3)+(3y+16)(4y-5)} = (10-2y)(3y+16)}$
$40y+30-8y^2-6y+12y^2-15y+64y-80 =30y+160 -6y^2-32y$
$4y^2+83y-50 =-6y^2-2y+160$
$4y^2+83y-50 +6y^2+2y-160=0$
$10y^2+85y-210=0$
$2y^2+17y-42=0$
$D=17^2-4*2*(-42)=625$
$y_1= \frac{-17+25}{4}=2$
$y_2= \frac{-17-25}{4}=-10.5$

$y_1^2+y_2^2=2^2+(-10.5)^2=4+110.25=114.25$

Ответ: 114.25

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