(5t-1)^2+(2t-2)(-t+4) eching

$$( 5 t - 1 ) ^ { 2 } + ( 2 t - 2 ) ( - t + 4 )$$

Baholash

$$23t^{2}-7$$

Yechim qadamlari

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(5t-1\right)^{2} kengaytirilishi uchun ishlating.

$$25t^{2}-10t+1+\left(2t-2\right)\left(-t+4\right)$$

2t-2 ga -t+4 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$25t^{2}-10t+1+2t\left(-t\right)+8t-2\left(-t\right)-8$$

2 hosil qilish uchun -2 va -1 ni ko'paytirish.

$$25t^{2}-10t+1+2t\left(-t\right)+8t+2t-8$$

10t ni olish uchun 8t va 2t ni birlashtirish.

$$25t^{2}-10t+1+2t\left(-t\right)+10t-8$$

0 ni olish uchun -10t va 10t ni birlashtirish.

$$25t^{2}+1+2t\left(-t\right)-8$$

-7 olish uchun 1 dan 8 ni ayirish.

$$25t^{2}-7+2t\left(-t\right)$$

t^{2} hosil qilish uchun t va t ni ko'paytirish.

$$25t^{2}-7+2t^{2}\left(-1\right)$$

-2 hosil qilish uchun 2 va -1 ni ko'paytirish.

$$25t^{2}-7-2t^{2}$$

23t^{2} ni olish uchun 25t^{2} va -2t^{2} ni birlashtirish.

$$23t^{2}-7$$

Kengaytirish

$$23t^{2}-7$$

Yechim qadamlari

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(5t-1\right)^{2} kengaytirilishi uchun ishlating.

$$25t^{2}-10t+1+\left(2t-2\right)\left(-t+4\right)$$

2t-2 ga -t+4 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$25t^{2}-10t+1+2t\left(-t\right)+8t-2\left(-t\right)-8$$

2 hosil qilish uchun -2 va -1 ni ko'paytirish.

$$25t^{2}-10t+1+2t\left(-t\right)+8t+2t-8$$

10t ni olish uchun 8t va 2t ni birlashtirish.

$$25t^{2}-10t+1+2t\left(-t\right)+10t-8$$

0 ni olish uchun -10t va 10t ni birlashtirish.

$$25t^{2}+1+2t\left(-t\right)-8$$

-7 olish uchun 1 dan 8 ni ayirish.

$$25t^{2}-7+2t\left(-t\right)$$

t^{2} hosil qilish uchun t va t ni ko'paytirish.

$$25t^{2}-7+2t^{2}\left(-1\right)$$

-2 hosil qilish uchun 2 va -1 ni ko'paytirish.

$$25t^{2}-7-2t^{2}$$

23t^{2} ni olish uchun 25t^{2} va -2t^{2} ni birlashtirish.

$$23t^{2}-7$$