(9x2+6x))=0?...

July 2022 1 4 Report

между какими целыми числами находится каждый корень уравнения (2/(9x2-4)) -(1/(9x2-6x))+((3x-4)/(9x2+6x))=0?

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Verified answer

2/[(3x-2)93x+2)]-1/[3x(3x-2)]+(3x-4)/[3x(3x+2)]=0
ОЗ 3x(3x-2)(3x+2)≠0⇒x≠0;x≠2/3;x≠-2/3
2*3x-(3x+2)+(3x-4)(3x-2)=0
6x-3x-2+9x²-6x-12x+8=0
9x²-15x+6=0
3x²-5x+2=0
D=25-24=1
x1=(5-1)/6=1/6
x2=(5+1)/6=1
0<1/6<1
0<1<2

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Answers & Comments

Verified answer

perevedite tekst srochno5db29bfd2c6ffb39c939f77a7d06390f 33995

chto takoe ploskost napishite opredeleniya

kakaya massa sernoj kisloty potrebuetsya dlya polucheniya 297 g sulfata ammoniya iz

chto znachit reshit zadachu v dannoj proporcii

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