sqrt{4-x}quadf(x)=1/sqrt{x+5} eching

$$\sqrt { 4 - x } \quad f ( x ) = \frac { 1 } { \sqrt { x + 5 } }$$

f uchun yechish (complex solution)

$$f=\frac{\left(4-x\right)^{-\frac{1}{2}}\left(x+5\right)^{-\frac{1}{2}}}{x}$$

$$x\neq -5\text{ and }x\neq 0\text{ and }x\neq 4$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tenglama standart shaklda.

$$\sqrt{4-x}xf=\frac{1}{\sqrt{x+5}}$$

Ikki tarafini \sqrt{4-x}x ga bo‘ling.

$$\frac{\sqrt{4-x}xf}{\sqrt{4-x}x}=\frac{\left(x+5\right)^{-\frac{1}{2}}}{\sqrt{4-x}x}$$

\sqrt{4-x}x ga bo'lish \sqrt{4-x}x ga ko'paytirishni bekor qiladi.

$$f=\frac{\left(x+5\right)^{-\frac{1}{2}}}{\sqrt{4-x}x}$$

\left(x+5\right)^{-\frac{1}{2}} ni \sqrt{4-x}x ga bo'lish.

$$f=\frac{\left(4-x\right)^{-\frac{1}{2}}\left(x+5\right)^{-\frac{1}{2}}}{x}$$

f uchun yechish

$$f=\frac{1}{x\sqrt{\left(4-x\right)\left(x+5\right)}}$$

$$x\neq 0\text{ and }x>-5\text{ and }x<4$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tenglama standart shaklda.

$$\sqrt{4-x}xf=\frac{1}{\sqrt{x+5}}$$

Ikki tarafini \sqrt{4-x}x ga bo‘ling.

$$\frac{\sqrt{4-x}xf}{\sqrt{4-x}x}=\frac{1}{\sqrt{x+5}\sqrt{4-x}x}$$

\sqrt{4-x}x ga bo'lish \sqrt{4-x}x ga ko'paytirishni bekor qiladi.

$$f=\frac{1}{\sqrt{x+5}\sqrt{4-x}x}$$

\frac{1}{\sqrt{x+5}} ni \sqrt{4-x}x ga bo'lish.

$$f=\frac{1}{x\sqrt{\left(4-x\right)\left(x+5\right)}}$$