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AleksandraSavina
@AleksandraSavina
July 2022
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помогите,[tex] \sqrt{x+3} \sqrt{x+6} \ \textless \ x+4[/tex] пожалуйста, решить
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гюйс
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(х+3)(х+6)<(х+4)²
х²+6х+3х+18<x²+8x+16
x²-x²+6x+3x-8x<16-18
x< - 2
2<2
-2<2
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Verified answer
(х+3)(х+6)<(х+4)²
х²+6х+3х+18<x²+8x+16
x²-x²+6x+3x-8x<16-18
x< - 2
2<2
-2<2