-2left(x^+1right)^2left(x^2+1right)left(x-1right) eching

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

Baholash

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

Yechim qadamlari

1 daraja ko‘rsatkichini x ga hisoblang va x ni qiymatni oling.

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

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

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

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

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

-2x^{2}-4x-2 ga x^{2}+1 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$\left(-2x^{4}-4x^{2}-4x^{3}-4x-2\right)\left(x-1\right)$$

-2x^{4}-4x^{2}-4x^{3}-4x-2 ga x-1 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

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

Kengaytirish

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

Yechim qadamlari

1 daraja ko‘rsatkichini x ga hisoblang va x ni qiymatni oling.

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

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

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

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

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

-2x^{2}-4x-2 ga x^{2}+1 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$\left(-2x^{4}-4x^{2}-4x^{3}-4x-2\right)\left(x-1\right)$$

-2x^{4}-4x^{2}-4x^{3}-4x-2 ga x-1 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

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