{left(1/2x(x+1)right)}^2 eching

$${ \left( \frac{ 1 }{ 2 } x(x+1) \right) }^{ 2 }$$

Baholash

$$\frac{\left(x\left(x+1\right)\right)^{2}}{4}$$

Yechim qadamlari

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

$$\left(\frac{1}{2}x^{2}+\frac{1}{2}x\right)^{2}$$

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(\frac{1}{2}x^{2}+\frac{1}{2}x\right)^{2} kengaytirilishi uchun ishlating.

$$\frac{1}{4}\left(x^{2}\right)^{2}+\frac{1}{2}x^{2}x+\frac{1}{4}x^{2}$$

Daraja ko‘rsatkichini boshqa ko‘rsatkichga oshirish uchun, darajalarini ko‘paytiring. 2 va 2 ni ko‘paytirib, 4 ni oling.

$$\frac{1}{4}x^{4}+\frac{1}{2}x^{2}x+\frac{1}{4}x^{2}$$

Ayni asosning daraja ko‘rsatkichlarini ko‘paytirish uchun ularning darajalarini qo‘shing. 2 va 1 ni qo‘shib, 3 ni oling.

$$\frac{1}{4}x^{4}+\frac{1}{2}x^{3}+\frac{1}{4}x^{2}$$

Kengaytirish

$$\frac{x^{4}}{4}+\frac{x^{3}}{2}+\frac{x^{2}}{4}$$

Yechim qadamlari

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

$$\left(\frac{1}{2}x^{2}+\frac{1}{2}x\right)^{2}$$

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(\frac{1}{2}x^{2}+\frac{1}{2}x\right)^{2} kengaytirilishi uchun ishlating.

$$\frac{1}{4}\left(x^{2}\right)^{2}+\frac{1}{2}x^{2}x+\frac{1}{4}x^{2}$$

Daraja ko‘rsatkichini boshqa ko‘rsatkichga oshirish uchun, darajalarini ko‘paytiring. 2 va 2 ni ko‘paytirib, 4 ni oling.

$$\frac{1}{4}x^{4}+\frac{1}{2}x^{2}x+\frac{1}{4}x^{2}$$

Ayni asosning daraja ko‘rsatkichlarini ko‘paytirish uchun ularning darajalarini qo‘shing. 2 va 1 ni qo‘shib, 3 ni oling.

$$\frac{1}{4}x^{4}+\frac{1}{2}x^{3}+\frac{1}{4}x^{2}$$