6x(x-1)^2-3x*(x+1)^2 eching

$$6 x ( x - 1 ) ^ { 2 } - 3 x \cdot ( x + 1 ) ^ { 2 }$$

Baholash

$$3x\left(x^{2}-6x+1\right)$$

Yechim qadamlari

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

$$6x\left(x^{2}-2x+1\right)-3x\left(x+1\right)^{2}$$

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

$$6x^{3}-12x^{2}+6x-3x\left(x+1\right)^{2}$$

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

$$6x^{3}-12x^{2}+6x-3x\left(x^{2}+2x+1\right)$$

-3x ga x^{2}+2x+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$6x^{3}-12x^{2}+6x-3x^{3}-6x^{2}-3x$$

3x^{3} ni olish uchun 6x^{3} va -3x^{3} ni birlashtirish.

$$3x^{3}-12x^{2}+6x-6x^{2}-3x$$

-18x^{2} ni olish uchun -12x^{2} va -6x^{2} ni birlashtirish.

$$3x^{3}-18x^{2}+6x-3x$$

3x ni olish uchun 6x va -3x ni birlashtirish.

$$3x^{3}-18x^{2}+3x$$

Kengaytirish

$$3x^{3}-18x^{2}+3x$$

Yechim qadamlari

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

$$6x\left(x^{2}-2x+1\right)-3x\left(x+1\right)^{2}$$

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

$$6x^{3}-12x^{2}+6x-3x\left(x+1\right)^{2}$$

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

$$6x^{3}-12x^{2}+6x-3x\left(x^{2}+2x+1\right)$$

-3x ga x^{2}+2x+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$6x^{3}-12x^{2}+6x-3x^{3}-6x^{2}-3x$$

3x^{3} ni olish uchun 6x^{3} va -3x^{3} ni birlashtirish.

$$3x^{3}-12x^{2}+6x-6x^{2}-3x$$

-18x^{2} ni olish uchun -12x^{2} va -6x^{2} ni birlashtirish.

$$3x^{3}-18x^{2}+6x-3x$$

3x ni olish uchun 6x va -3x ni birlashtirish.

$$3x^{3}-18x^{2}+3x$$