Вопрос в картинках...

Question

Решите задачу:

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Answer 1

0\\\\ D=9-4*1*5<0\\\\ \frac{1}{log_{3}2}+log_{3}(x^2+3x+5)=2\\\\ log_{3}(x^2+3x+5)=2-\frac{1}{log_{3}2}\\\\ x^2+3x+5=3^{2-\frac{1}{log_{3}2}}\\\\ x^2+3x+5=\frac{9}{3^{\frac{1}{log_{3}2}}}\\\\ x^2+3x+5 - \frac{9}{3^{\frac{1}{log_{3}2}}}=0\\\\ D=9-4*1*(5-\frac{9}{3^{\frac{1}{log_{3}2}}}<0" alt="log_{2}3+log_{3}(x^2+3x+5)=2\\\\ x^2+3x+5>0\\\\ D=9-4*1*5<0\\\\ \frac{1}{log_{3}2}+log_{3}(x^2+3x+5)=2\\\\ log_{3}(x^2+3x+5)=2-\frac{1}{log_{3}2}\\\\ x^2+3x+5=3^{2-\frac{1}{log_{3}2}}\\\\ x^2+3x+5=\frac{9}{3^{\frac{1}{log_{3}2}}}\\\\ x^2+3x+5 - \frac{9}{3^{\frac{1}{log_{3}2}}}=0\\\\ D=9-4*1*(5-\frac{9}{3^{\frac{1}{log_{3}2}}}<0" align="absmiddle" class="latex-formula">
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