(x^2+1)^2-6(x^2+1)+8= eching

$$( x ^ { 2 } + 1 ) ^ { 2 } - 6 ( x ^ { 2 } + 1 ) + 8 =$$

Baholash

$$x^{4}-4x^{2}+3$$

Yechim qadamlari

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

$$\left(x^{2}\right)^{2}+2x^{2}+1-6\left(x^{2}+1\right)+8$$

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

$$x^{4}+2x^{2}+1-6\left(x^{2}+1\right)+8$$

-6 ga x^{2}+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$x^{4}+2x^{2}+1-6x^{2}-6+8$$

-4x^{2} ni olish uchun 2x^{2} va -6x^{2} ni birlashtirish.

$$x^{4}-4x^{2}+1-6+8$$

-5 olish uchun 1 dan 6 ni ayirish.

$$x^{4}-4x^{2}-5+8$$

3 olish uchun -5 va 8'ni qo'shing.

$$x^{4}-4x^{2}+3$$

Kengaytirish

$$x^{4}-4x^{2}+3$$

Yechim qadamlari

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

$$\left(x^{2}\right)^{2}+2x^{2}+1-6\left(x^{2}+1\right)+8$$

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

$$x^{4}+2x^{2}+1-6\left(x^{2}+1\right)+8$$

-6 ga x^{2}+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$x^{4}+2x^{2}+1-6x^{2}-6+8$$

-4x^{2} ni olish uchun 2x^{2} va -6x^{2} ni birlashtirish.

$$x^{4}-4x^{2}+1-6+8$$

-5 olish uchun 1 dan 6 ni ayirish.

$$x^{4}-4x^{2}-5+8$$

3 olish uchun -5 va 8'ni qo'shing.

$$x^{4}-4x^{2}+3$$