f^-1(x)=2x^2-5x+1 eching

$$f ^ { - 1 } ( x ) = 2 x ^ { 2 } - 5 x + 1$$

f uchun yechish

$$f=-\frac{x}{-2x^{2}+5x-1}$$

$$x\neq 0\text{ and }x\neq \frac{\sqrt{17}+5}{4}\text{ and }x\neq \frac{5-\sqrt{17}}{4}$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Shartlarni qayta saralash.

$$\frac{1}{f}x=2x^{2}-5x+1$$

f qiymati 0 teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini f ga ko'paytirish.

$$1x=2x^{2}f-5xf+f$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$2x^{2}f-5xf+f=1x$$

Shartlarni qayta saralash.

$$2fx^{2}-5fx+f=x$$

f'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(2x^{2}-5x+1\right)f=x$$

Ikki tarafini 2x^{2}-5x+1 ga bo‘ling.

$$\frac{\left(2x^{2}-5x+1\right)f}{2x^{2}-5x+1}=\frac{x}{2x^{2}-5x+1}$$

2x^{2}-5x+1 ga bo'lish 2x^{2}-5x+1 ga ko'paytirishni bekor qiladi.

$$f=\frac{x}{2x^{2}-5x+1}$$

f qiymati 0 teng bo‘lmaydi.

$$f=\frac{x}{2x^{2}-5x+1}\text{, }f\neq 0$$

x uchun yechish (complex solution)

$$x=-\frac{\sqrt{17f^{2}+10f+1}-5f-1}{4f}$$

$$x=\frac{\sqrt{17f^{2}+10f+1}+5f+1}{4f}\text{, }f\neq 0$$

x uchun yechish

$$x=-\frac{\sqrt{17f^{2}+10f+1}-5f-1}{4f}$$

$$x=\frac{\sqrt{17f^{2}+10f+1}+5f+1}{4f}\text{, }f\leq \frac{-2\sqrt{2}-5}{17}\text{ or }\left(f\neq 0\text{ and }f\geq \frac{2\sqrt{2}-5}{17}\right)$$