f(x+2)-f(x-1)=26/3f(x) eching

$$f ( x + 2 ) - f ( x - 1 ) = \frac { 26 } { 3 } f ( x )$$

f uchun yechish (complex solution)

$$\left\{\begin{matrix}\\f=0\text{, }&\text{unconditionally}\\f\in \mathrm{C}\text{, }&x=\frac{9}{26}\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

f ga x+2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$fx+2f-f\left(x-1\right)=\frac{26}{3}fx$$

f ga x-1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$fx+2f-\left(fx-f\right)=\frac{26}{3}fx$$

fx-f teskarisini topish uchun har birining teskarisini toping.

$$fx+2f-fx+f=\frac{26}{3}fx$$

0 ni olish uchun fx va -fx ni birlashtirish.

$$2f+f=\frac{26}{3}fx$$

3f ni olish uchun 2f va f ni birlashtirish.

$$3f=\frac{26}{3}fx$$

Ikkala tarafdan \frac{26}{3}fx ni ayirish.

$$3f-\frac{26}{3}fx=0$$

f'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(3-\frac{26}{3}x\right)f=0$$

Tenglama standart shaklda.

$$\left(-\frac{26x}{3}+3\right)f=0$$

0 ni 3-\frac{26}{3}x ga bo'lish.

$$f=0$$

x uchun yechish (complex solution)

$$\left\{\begin{matrix}\\x=\frac{9}{26}\text{, }&\text{unconditionally}\\x\in \mathrm{C}\text{, }&f=0\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

f ga x+2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$fx+2f-f\left(x-1\right)=\frac{26}{3}fx$$

f ga x-1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$fx+2f-\left(fx-f\right)=\frac{26}{3}fx$$

fx-f teskarisini topish uchun har birining teskarisini toping.

$$fx+2f-fx+f=\frac{26}{3}fx$$

0 ni olish uchun fx va -fx ni birlashtirish.

$$2f+f=\frac{26}{3}fx$$

3f ni olish uchun 2f va f ni birlashtirish.

$$3f=\frac{26}{3}fx$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$\frac{26}{3}fx=3f$$

Tenglama standart shaklda.

$$\frac{26f}{3}x=3f$$

Ikki tarafini \frac{26}{3}f ga bo‘ling.

$$\frac{3\times \frac{26f}{3}x}{26f}=\frac{3\times 3f}{26f}$$

\frac{26}{3}f ga bo'lish \frac{26}{3}f ga ko'paytirishni bekor qiladi.

$$x=\frac{3\times 3f}{26f}$$

3f ni \frac{26}{3}f ga bo'lish.

$$x=\frac{9}{26}$$

f uchun yechish

$$\left\{\begin{matrix}\\f=0\text{, }&\text{unconditionally}\\f\in \mathrm{R}\text{, }&x=\frac{9}{26}\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

f ga x+2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$fx+2f-f\left(x-1\right)=\frac{26}{3}fx$$

f ga x-1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$fx+2f-\left(fx-f\right)=\frac{26}{3}fx$$

fx-f teskarisini topish uchun har birining teskarisini toping.

$$fx+2f-fx+f=\frac{26}{3}fx$$

0 ni olish uchun fx va -fx ni birlashtirish.

$$2f+f=\frac{26}{3}fx$$

3f ni olish uchun 2f va f ni birlashtirish.

$$3f=\frac{26}{3}fx$$

Ikkala tarafdan \frac{26}{3}fx ni ayirish.

$$3f-\frac{26}{3}fx=0$$

f'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(3-\frac{26}{3}x\right)f=0$$

Tenglama standart shaklda.

$$\left(-\frac{26x}{3}+3\right)f=0$$

0 ni 3-\frac{26}{3}x ga bo'lish.

$$f=0$$

x uchun yechish

$$\left\{\begin{matrix}\\x=\frac{9}{26}\text{, }&\text{unconditionally}\\x\in \mathrm{R}\text{, }&f=0\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

f ga x+2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$fx+2f-f\left(x-1\right)=\frac{26}{3}fx$$

f ga x-1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$fx+2f-\left(fx-f\right)=\frac{26}{3}fx$$

fx-f teskarisini topish uchun har birining teskarisini toping.

$$fx+2f-fx+f=\frac{26}{3}fx$$

0 ni olish uchun fx va -fx ni birlashtirish.

$$2f+f=\frac{26}{3}fx$$

3f ni olish uchun 2f va f ni birlashtirish.

$$3f=\frac{26}{3}fx$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$\frac{26}{3}fx=3f$$

Tenglama standart shaklda.

$$\frac{26f}{3}x=3f$$

Ikki tarafini \frac{26}{3}f ga bo‘ling.

$$\frac{3\times \frac{26f}{3}x}{26f}=\frac{3\times 3f}{26f}$$

\frac{26}{3}f ga bo'lish \frac{26}{3}f ga ko'paytirishni bekor qiladi.

$$x=\frac{3\times 3f}{26f}$$

3f ni \frac{26}{3}f ga bo'lish.

$$x=\frac{9}{26}$$