(11x+3)^2-(6x+1)(6x-1) eching

$$( 11 x + 3 ) ^ { 2 } - ( 6 x + 1 ) ( 6 x - 1 )$$

Baholash

$$85x^{2}+66x+10$$

Yechim qadamlari

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(11x+3\right)^{2} kengaytirilishi uchun ishlating.

$$121x^{2}+66x+9-\left(6x+1\right)\left(6x-1\right)$$

Hisoblang: \left(6x+1\right)\left(6x-1\right). Ko‘paytirish qoida yordamida turli kvadratlarga aylantirilishi mumkin: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. 1 kvadratini chiqarish.

$$121x^{2}+66x+9-\left(\left(6x\right)^{2}-1\right)$$

\left(6x\right)^{2} ni kengaytirish.

$$121x^{2}+66x+9-\left(6^{2}x^{2}-1\right)$$

2 daraja ko‘rsatkichini 6 ga hisoblang va 36 ni qiymatni oling.

$$121x^{2}+66x+9-\left(36x^{2}-1\right)$$

36x^{2}-1 teskarisini topish uchun har birining teskarisini toping.

$$121x^{2}+66x+9-36x^{2}+1$$

85x^{2} ni olish uchun 121x^{2} va -36x^{2} ni birlashtirish.

$$85x^{2}+66x+9+1$$

10 olish uchun 9 va 1'ni qo'shing.

$$85x^{2}+66x+10$$

Kengaytirish

$$85x^{2}+66x+10$$

Yechim qadamlari

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(11x+3\right)^{2} kengaytirilishi uchun ishlating.

$$121x^{2}+66x+9-\left(6x+1\right)\left(6x-1\right)$$

Hisoblang: \left(6x+1\right)\left(6x-1\right). Ko‘paytirish qoida yordamida turli kvadratlarga aylantirilishi mumkin: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. 1 kvadratini chiqarish.

$$121x^{2}+66x+9-\left(\left(6x\right)^{2}-1\right)$$

\left(6x\right)^{2} ni kengaytirish.

$$121x^{2}+66x+9-\left(6^{2}x^{2}-1\right)$$

2 daraja ko‘rsatkichini 6 ga hisoblang va 36 ni qiymatni oling.

$$121x^{2}+66x+9-\left(36x^{2}-1\right)$$

36x^{2}-1 teskarisini topish uchun har birining teskarisini toping.

$$121x^{2}+66x+9-36x^{2}+1$$

85x^{2} ni olish uchun 121x^{2} va -36x^{2} ni birlashtirish.

$$85x^{2}+66x+9+1$$

10 olish uchun 9 va 1'ni qo'shing.

$$85x^{2}+66x+10$$