помогите,[tex] \sqrt{x+3} \sqrt{x+6} \ \textless \ x+4[/tex] пожалуйста, решить. Created by AleksandraSavina. matematika-ru

(х+3)(х+6)<(х+4)²х²+6х+3х+18<x²+8x+16x²-x²+6x+3x-8x<16-18x< - 22<2-2<2

tex] пожалуйста, решить...

AleksandraSavina @AleksandraSavina

July 2022 1 18 Report

помогите,[tex] \sqrt{x+3} \sqrt{x+6} \ \textless \ x+4[/tex] пожалуйста, решить

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гюйс

Verified answer

$\sqrt{x+3} \sqrt{x+6} \ \textless \ x+4
$
(х+3)(х+6)<(х+4)²
х²+6х+3х+18<x²+8x+16
x²-x²+6x+3x-8x<16-18
x< - 2
$\sqrt{ -2+3} \sqrt{-2+6}\ \textless \ -2+4$
$\sqrt{1} \sqrt{4}\ \textless \ 2$
2<2
-2<2

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