f(x)^2/3+(mn^2)^2/3=1 eching

$$f ( x ) ^ { 2 / 3 } + ( m n ^ { 2 } ) ^ { 2 / 3 } = 1$$

f uchun yechish (complex solution)

$$\left\{\begin{matrix}f=x^{-\frac{2}{3}}\left(1-\left(mn^{2}\right)^{\frac{2}{3}}\right)\text{, }&x\neq 0\\f\in \mathrm{C}\text{, }&m=\frac{1}{n^{2}}\text{ and }n\neq 0\text{ and }x=0\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

\left(mn^{2}\right)^{\frac{2}{3}} ni kengaytirish.

$$fx^{\frac{2}{3}}+m^{\frac{2}{3}}\left(n^{2}\right)^{\frac{2}{3}}=1$$

Daraja ko‘rsatkichini boshqa ko‘rsatkichga oshirish uchun, darajalarini ko‘paytiring. 2 va \frac{2}{3} ni ko‘paytirib, \frac{4}{3} ni oling.

$$fx^{\frac{2}{3}}+m^{\frac{2}{3}}n^{\frac{4}{3}}=1$$

Ikkala tarafdan m^{\frac{2}{3}}n^{\frac{4}{3}} ni ayirish.

$$fx^{\frac{2}{3}}=1-m^{\frac{2}{3}}n^{\frac{4}{3}}$$

Shartlarni qayta saralash.

$$x^{\frac{2}{3}}f=-m^{\frac{2}{3}}n^{\frac{4}{3}}+1$$

Tenglama standart shaklda.

$$x^{\frac{2}{3}}f=1-m^{\frac{2}{3}}n^{\frac{4}{3}}$$

Ikki tarafini x^{\frac{2}{3}} ga bo‘ling.

$$\frac{x^{\frac{2}{3}}f}{x^{\frac{2}{3}}}=\frac{1-m^{\frac{2}{3}}n^{\frac{4}{3}}}{x^{\frac{2}{3}}}$$

x^{\frac{2}{3}} ga bo'lish x^{\frac{2}{3}} ga ko'paytirishni bekor qiladi.

$$f=\frac{1-m^{\frac{2}{3}}n^{\frac{4}{3}}}{x^{\frac{2}{3}}}$$

-m^{\frac{2}{3}}n^{\frac{4}{3}}+1 ni x^{\frac{2}{3}} ga bo'lish.

$$f=x^{-\frac{2}{3}}\left(1-m^{\frac{2}{3}}n^{\frac{4}{3}}\right)$$

f uchun yechish

$$\left\{\begin{matrix}f=-\frac{m^{\frac{2}{3}}n^{\frac{4}{3}}-1}{x^{\frac{2}{3}}}\text{, }&x\neq 0\\f\in \mathrm{R}\text{, }&n\neq 0\text{ and }x=0\text{ and }|m|=\frac{1}{n^{2}}\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

\left(mn^{2}\right)^{\frac{2}{3}} ni kengaytirish.

$$fx^{\frac{2}{3}}+m^{\frac{2}{3}}\left(n^{2}\right)^{\frac{2}{3}}=1$$

Daraja ko‘rsatkichini boshqa ko‘rsatkichga oshirish uchun, darajalarini ko‘paytiring. 2 va \frac{2}{3} ni ko‘paytirib, \frac{4}{3} ni oling.

$$fx^{\frac{2}{3}}+m^{\frac{2}{3}}n^{\frac{4}{3}}=1$$

Ikkala tarafdan m^{\frac{2}{3}}n^{\frac{4}{3}} ni ayirish.

$$fx^{\frac{2}{3}}=1-m^{\frac{2}{3}}n^{\frac{4}{3}}$$

Shartlarni qayta saralash.

$$x^{\frac{2}{3}}f=-m^{\frac{2}{3}}n^{\frac{4}{3}}+1$$

Tenglama standart shaklda.

$$x^{\frac{2}{3}}f=1-m^{\frac{2}{3}}n^{\frac{4}{3}}$$

Ikki tarafini x^{\frac{2}{3}} ga bo‘ling.

$$\frac{x^{\frac{2}{3}}f}{x^{\frac{2}{3}}}=\frac{1-m^{\frac{2}{3}}n^{\frac{4}{3}}}{x^{\frac{2}{3}}}$$

x^{\frac{2}{3}} ga bo'lish x^{\frac{2}{3}} ga ko'paytirishni bekor qiladi.

$$f=\frac{1-m^{\frac{2}{3}}n^{\frac{4}{3}}}{x^{\frac{2}{3}}}$$

m uchun yechish (complex solution)

$$\left\{\begin{matrix}m=\frac{\left(1-x^{\frac{2}{3}}f\right)^{3}}{n^{2}}\text{, }&\left(n\neq 0\text{ and }arg(1-x^{\frac{2}{3}}f)<\frac{2\pi }{3}\right)\text{ or }\left(n\neq 0\text{ and }f=x^{-\frac{2}{3}}\text{ and }x\neq 0\right)\\m\in \mathrm{C}\text{, }&f=x^{-\frac{2}{3}}\text{ and }x\neq 0\text{ and }n=0\end{matrix}\right,$$

m uchun yechish

$$\left\{\begin{matrix}m=-\frac{1}{n^{2}}\text{; }m=\frac{1}{n^{2}}\text{, }&x=0\text{ and }n\neq 0\\m\in \mathrm{R}\text{, }&f=\frac{1}{x^{\frac{2}{3}}}\text{ and }n=0\text{ and }x\neq 0\\m=-\frac{\left(1-x^{\frac{2}{3}}f\right)^{\frac{3}{2}}}{n^{2}}\text{; }m=\frac{\left(1-x^{\frac{2}{3}}f\right)^{\frac{3}{2}}}{n^{2}}\text{, }&n\neq 0\text{ and }f\leq \frac{1}{x^{\frac{2}{3}}}\text{ and }x\neq 0\end{matrix}\right,$$