y=2-1/2f(x+1) eching

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

f uchun yechish (complex solution)

$$\left\{\begin{matrix}f=\frac{2\left(2-y\right)}{x+1}\text{, }&x\neq -1\\f\in \mathrm{C}\text{, }&y=2\text{ and }x=-1\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

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

-\frac{1}{2}f ga x+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$2-\frac{1}{2}fx-\frac{1}{2}f=y$$

Ikkala tarafdan 2 ni ayirish.

$$-\frac{1}{2}fx-\frac{1}{2}f=y-2$$

f'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(-\frac{1}{2}x-\frac{1}{2}\right)f=y-2$$

Tenglama standart shaklda.

$$\frac{-x-1}{2}f=y-2$$

Ikki tarafini -\frac{1}{2}x-\frac{1}{2} ga bo‘ling.

$$\frac{2\times \frac{-x-1}{2}f}{-x-1}=\frac{2\left(y-2\right)}{-x-1}$$

-\frac{1}{2}x-\frac{1}{2} ga bo'lish -\frac{1}{2}x-\frac{1}{2} ga ko'paytirishni bekor qiladi.

$$f=\frac{2\left(y-2\right)}{-x-1}$$

y-2 ni -\frac{1}{2}x-\frac{1}{2} ga bo'lish.

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

x uchun yechish (complex solution)

$$\left\{\begin{matrix}x=\frac{4-f-2y}{f}\text{, }&f\neq 0\\x\in \mathrm{C}\text{, }&y=2\text{ and }f=0\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

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

-\frac{1}{2}f ga x+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$2-\frac{1}{2}fx-\frac{1}{2}f=y$$

Ikkala tarafdan 2 ni ayirish.

$$-\frac{1}{2}fx-\frac{1}{2}f=y-2$$

\frac{1}{2}f ni ikki tarafga qo’shing.

$$-\frac{1}{2}fx=y-2+\frac{1}{2}f$$

Tenglama standart shaklda.

$$\left(-\frac{f}{2}\right)x=\frac{f}{2}+y-2$$

Ikki tarafini -\frac{1}{2}f ga bo‘ling.

$$\frac{\left(-\frac{f}{2}\right)x}{-\frac{f}{2}}=\frac{\frac{f}{2}+y-2}{-\frac{f}{2}}$$

-\frac{1}{2}f ga bo'lish -\frac{1}{2}f ga ko'paytirishni bekor qiladi.

$$x=\frac{\frac{f}{2}+y-2}{-\frac{f}{2}}$$

y-2+\frac{f}{2} ni -\frac{1}{2}f ga bo'lish.

$$x=-\frac{2y+f-4}{f}$$

f uchun yechish

$$\left\{\begin{matrix}f=\frac{2\left(2-y\right)}{x+1}\text{, }&x\neq -1\\f\in \mathrm{R}\text{, }&y=2\text{ and }x=-1\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

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

-\frac{1}{2}f ga x+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$2-\frac{1}{2}fx-\frac{1}{2}f=y$$

Ikkala tarafdan 2 ni ayirish.

$$-\frac{1}{2}fx-\frac{1}{2}f=y-2$$

f'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(-\frac{1}{2}x-\frac{1}{2}\right)f=y-2$$

Tenglama standart shaklda.

$$\frac{-x-1}{2}f=y-2$$

Ikki tarafini -\frac{1}{2}x-\frac{1}{2} ga bo‘ling.

$$\frac{2\times \frac{-x-1}{2}f}{-x-1}=\frac{2\left(y-2\right)}{-x-1}$$

-\frac{1}{2}x-\frac{1}{2} ga bo'lish -\frac{1}{2}x-\frac{1}{2} ga ko'paytirishni bekor qiladi.

$$f=\frac{2\left(y-2\right)}{-x-1}$$

y-2 ni -\frac{1}{2}x-\frac{1}{2} ga bo'lish.

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

x uchun yechish

$$\left\{\begin{matrix}x=\frac{4-f-2y}{f}\text{, }&f\neq 0\\x\in \mathrm{R}\text{, }&y=2\text{ and }f=0\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

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

-\frac{1}{2}f ga x+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$2-\frac{1}{2}fx-\frac{1}{2}f=y$$

Ikkala tarafdan 2 ni ayirish.

$$-\frac{1}{2}fx-\frac{1}{2}f=y-2$$

\frac{1}{2}f ni ikki tarafga qo’shing.

$$-\frac{1}{2}fx=y-2+\frac{1}{2}f$$

Tenglama standart shaklda.

$$\left(-\frac{f}{2}\right)x=\frac{f}{2}+y-2$$

Ikki tarafini -\frac{1}{2}f ga bo‘ling.

$$\frac{\left(-\frac{f}{2}\right)x}{-\frac{f}{2}}=\frac{\frac{f}{2}+y-2}{-\frac{f}{2}}$$

-\frac{1}{2}f ga bo'lish -\frac{1}{2}f ga ko'paytirishni bekor qiladi.

$$x=\frac{\frac{f}{2}+y-2}{-\frac{f}{2}}$$

y-2+\frac{f}{2} ni -\frac{1}{2}f ga bo'lish.

$$x=-\frac{2y+f-4}{f}$$