Ответ:
[tex]4[/tex]
Объяснение:
ОДЗ:
[tex]x>3[/tex]
[tex]log_6(x-3)=1-log_6(x+2)\\\\log_6(x-3)+log_6(x+2)=1\\\\log_6((x-3)(x+2))=log_66\\\\(x-3)(x+2)=6\\\\x^2+2x-3x-6-6=0\\\\x^2-x-12=0\\\\D=(-1)^2-4\cdot 1\cdot (-12)=1+48=49\\\\\sqrt{D}=\sqrt{49}=7\\\\x_1=\frac{1-7}{2\cdot 1}=\frac{-6}{2}=-3 < 0\\\\x_2=\frac{1+7}{2\cdot 1}=\frac{8}{2}=4[/tex]
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Answers & Comments
Ответ:
[tex]4[/tex]
Объяснение:
ОДЗ:
[tex]x>3[/tex]
[tex]log_6(x-3)=1-log_6(x+2)\\\\log_6(x-3)+log_6(x+2)=1\\\\log_6((x-3)(x+2))=log_66\\\\(x-3)(x+2)=6\\\\x^2+2x-3x-6-6=0\\\\x^2-x-12=0\\\\D=(-1)^2-4\cdot 1\cdot (-12)=1+48=49\\\\\sqrt{D}=\sqrt{49}=7\\\\x_1=\frac{1-7}{2\cdot 1}=\frac{-6}{2}=-3 < 0\\\\x_2=\frac{1+7}{2\cdot 1}=\frac{8}{2}=4[/tex]