f(x)=(x+2x^3)^4a=-1 eching

$$f ( x ) = ( x + 2 x ^ { 3 } ) ^ { 4 } a = - 1$$

a uchun yechish

$$a=-\frac{1}{\left(2x^{3}+x\right)^{4}}$$

$$x\neq 0$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tenglama standart shaklda.

$$\left(2x^{3}+x\right)^{4}a=-1$$

Ikki tarafini \left(x+2x^{3}\right)^{4} ga bo‘ling.

$$\frac{\left(2x^{3}+x\right)^{4}a}{\left(2x^{3}+x\right)^{4}}=-\frac{1}{\left(2x^{3}+x\right)^{4}}$$

\left(x+2x^{3}\right)^{4} ga bo'lish \left(x+2x^{3}\right)^{4} ga ko'paytirishni bekor qiladi.

$$a=-\frac{1}{\left(2x^{3}+x\right)^{4}}$$

-1 ni \left(x+2x^{3}\right)^{4} ga bo'lish.

$$a=-\frac{1}{\left(x\left(2x^{2}+1\right)\right)^{4}}$$

x uchun yechish

$$x=\frac{\sqrt[3]{6\sqrt{6+81\sqrt{-\frac{1}{a}}}+54\sqrt[4]{-\frac{1}{a}}}+\sqrt[3]{-6\sqrt{6+81\sqrt{-\frac{1}{a}}}+54\sqrt[4]{-\frac{1}{a}}}}{6}$$

$$x=\frac{\sqrt[3]{6\sqrt{6+81\sqrt{-\frac{1}{a}}}-54\sqrt[4]{-\frac{1}{a}}}-\sqrt[3]{6\sqrt{6+81\sqrt{-\frac{1}{a}}}+54\sqrt[4]{-\frac{1}{a}}}}{6}\text{, }a<0$$