y=axsqrt{b} eching

$$y = a x \sqrt { b }$$

a uchun yechish (complex solution)

$$\left\{\begin{matrix}a=\frac{b^{-\frac{1}{2}}y}{x}\text{, }&b\neq 0\text{ and }x\neq 0\\a\in \mathrm{C}\text{, }&\left(b=0\text{ or }x=0\right)\text{ and }y=0\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$ax\sqrt{b}=y$$

Tenglama standart shaklda.

$$\sqrt{b}xa=y$$

Ikki tarafini x\sqrt{b} ga bo‘ling.

$$\frac{\sqrt{b}xa}{\sqrt{b}x}=\frac{y}{\sqrt{b}x}$$

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

$$a=\frac{y}{\sqrt{b}x}$$

y ni x\sqrt{b} ga bo'lish.

$$a=\frac{b^{-\frac{1}{2}}y}{x}$$

a uchun yechish

$$\left\{\begin{matrix}a=\frac{y}{\sqrt{b}x}\text{, }&x\neq 0\text{ and }b>0\\a\in \mathrm{R}\text{, }&\left(y=0\text{ and }x=0\text{ and }b>0\right)\text{ or }\left(b=0\text{ and }y=0\right)\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$ax\sqrt{b}=y$$

Tenglama standart shaklda.

$$\sqrt{b}xa=y$$

Ikki tarafini x\sqrt{b} ga bo‘ling.

$$\frac{\sqrt{b}xa}{\sqrt{b}x}=\frac{y}{\sqrt{b}x}$$

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

$$a=\frac{y}{\sqrt{b}x}$$

b uchun yechish

$$\left\{\begin{matrix}b=\left(\frac{y}{ax}\right)^{2}\text{, }&x\neq 0\text{ and }a\neq 0\text{ and }\left(a<0\text{ or }x<0\text{ or }y\geq 0\right)\text{ and }\left(x<0\text{ or }a>0\text{ or }y\leq 0\right)\text{ and }\left(a<0\text{ or }x>0\text{ or }y\leq 0\right)\text{ and }\left(x>0\text{ or }a>0\text{ or }y\geq 0\right)\\b\geq 0\text{, }&\left(y=0\text{ and }x=0\right)\text{ or }\left(x\neq 0\text{ and }y=0\text{ and }a=0\right)\end{matrix}\right,$$

Ildizli tenglamalarni yechish qadamlari

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$ax\sqrt{b}=y$$

Ikki tarafini ax ga bo‘ling.

$$\frac{ax\sqrt{b}}{ax}=\frac{y}{ax}$$

ax ga bo'lish ax ga ko'paytirishni bekor qiladi.

$$\sqrt{b}=\frac{y}{ax}$$

Tenglamaning ikkala taraf kvadratini chiqarish.

$$b=\frac{y^{2}}{\left(ax\right)^{2}}$$

b uchun yechish (complex solution)

$$\left\{\begin{matrix}b=\left(\frac{y}{ax}\right)^{2}\text{, }&x\neq 0\text{ and }a\neq 0\text{ and }\left(|-arg(y)+arg(\sqrt{\left(\frac{y}{ax}\right)^{2}}ax)|<\pi \text{ or }y=0\right)\\b\in \mathrm{C}\text{, }&\left(x=0\text{ or }a=0\right)\text{ and }y=0\end{matrix}\right,$$