Question

Решите уравнение
$x^{2} \sqrt{x} - 26x^\frac{13}{10} - 27x^\frac{1}{10} = 0$

Answer 1

$x^2*x ^{\frac{1}{2} }-26x^{ \frac{13}{10} }-27x^{ \frac{1}{10} }=0 \\ \\ x^{ \frac{5}{2} }-26x^{ \frac{13}{10} }-27x^{ \frac{1}{10} }=0 \\ \\ x^{ \frac{1}{10} }(x^{ \frac{12}{5} }-26x^{ \frac{6}{5} }-27)=0 \\ \\ x^{ \frac{1}{10} }( x^{ \frac{12}{5} }+x^{ \frac{6}{5}}-27(x^{ \frac{6}{5}}-1))=0 \\ \\ x^{ \frac{1}{10}}(x^{ \frac{6}{5}}*(x^{ \frac{6}{5}}+1 } )-27(x^{ \frac{6}{5} }+1))=0 \\ \\ x^{ \frac{1}{10} }(x^{ \frac{6}{5}}+1)(x^{ \frac{6}{5}}-27)=0 \\ \\ x_1=0 \\ \\ x^{ \frac{6}{5}}=-1$
$x \in \emptyset \\ \\ x^{ \frac{6}{5}}=27 \\ \sqrt[5]{x^6} =27 \\ x^6=27^5 \\ x_2=9 \sqrt{3}$

x1=0; x2=9√3