(x-y)^3=A(x+y).text{provethat}(2x+y)dy/dx=x+2y eching

$$( x - y ) ^ { 3 } = A ( x + y ) . \text { prove that } ( 2 x + y ) \frac { d y } { d x } = x + 2 y$$

A uchun yechish

$$\left\{\begin{matrix}A=\frac{\left(x-y\right)^{3}}{x+y}\text{, }&x\neq -y\\A\in \mathrm{R}\text{, }&x=0\text{ and }y=0\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

\left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} binom teoremasini \left(x-y\right)^{3} kengaytirilishi uchun ishlating.

$$x^{3}-3x^{2}y+3xy^{2}-y^{3}=A\left(x+y\right)$$

A ga x+y ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$x^{3}-3x^{2}y+3xy^{2}-y^{3}=Ax+Ay$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$Ax+Ay=x^{3}-3x^{2}y+3xy^{2}-y^{3}$$

A'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(x+y\right)A=x^{3}-3x^{2}y+3xy^{2}-y^{3}$$

Tenglama standart shaklda.

$$\left(x+y\right)A=x^{3}+3xy^{2}-y^{3}-3yx^{2}$$

Ikki tarafini x+y ga bo‘ling.

$$\frac{\left(x+y\right)A}{x+y}=\frac{\left(x-y\right)^{3}}{x+y}$$

x+y ga bo'lish x+y ga ko'paytirishni bekor qiladi.

$$A=\frac{\left(x-y\right)^{3}}{x+y}$$