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extr90
@extr90
July 2022
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√(2х-1)+ √(х-2)=√(х+1)
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selmash21
Это иррациональное уравнение. Решаем методом возведения в квадрат обеих частей уравнения
1 votes
Thanks 1
natali15medved
Verified answer
Возводим в квадрат
2х-1+2!/(2х-1)(х-2) +х-2=х+1
3х-3+2!/(2х^2-4х-х+2)=х+1
2!/(2х^2-5х+2)=х+1-3х+3
2!/(2х^2-5х+2)=-2х+4
2!/(2х^2-5х+2)=2(2-х)
!/(2х^2-5х+2)=2-х
2х^2-5х+2=(2-х)^2
2х^2-5х+2=4-4х+х^2
2х^2-5х+2-4+4х-х^2=0
х^2-х-2=0
д=1-4×(-2)=9
х1=(1-3)/2=-1
х2=(1+3)/2=2
1 votes
Thanks 1
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Answers & Comments
Verified answer
Возводим в квадрат2х-1+2!/(2х-1)(х-2) +х-2=х+1
3х-3+2!/(2х^2-4х-х+2)=х+1
2!/(2х^2-5х+2)=х+1-3х+3
2!/(2х^2-5х+2)=-2х+4
2!/(2х^2-5х+2)=2(2-х)
!/(2х^2-5х+2)=2-х
2х^2-5х+2=(2-х)^2
2х^2-5х+2=4-4х+х^2
2х^2-5х+2-4+4х-х^2=0
х^2-х-2=0
д=1-4×(-2)=9
х1=(1-3)/2=-1
х2=(1+3)/2=2