4(p+q)^2-(2p+q)^2 eching

$$4 ( p + q ) ^ { 2 } - ( 2 p + q ) ^ { 2 }$$

Baholash

$$q\left(4p+3q\right)$$

Yechim qadamlari

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

$$4\left(p^{2}+2pq+q^{2}\right)-\left(2p+q\right)^{2}$$

4 ga p^{2}+2pq+q^{2} ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$4p^{2}+8pq+4q^{2}-\left(2p+q\right)^{2}$$

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

$$4p^{2}+8pq+4q^{2}-\left(4p^{2}+4pq+q^{2}\right)$$

4p^{2}+4pq+q^{2} teskarisini topish uchun har birining teskarisini toping.

$$4p^{2}+8pq+4q^{2}-4p^{2}-4pq-q^{2}$$

0 ni olish uchun 4p^{2} va -4p^{2} ni birlashtirish.

$$8pq+4q^{2}-4pq-q^{2}$$

4pq ni olish uchun 8pq va -4pq ni birlashtirish.

$$4pq+4q^{2}-q^{2}$$

3q^{2} ni olish uchun 4q^{2} va -q^{2} ni birlashtirish.

$$4pq+3q^{2}$$

Kengaytirish

$$4pq+3q^{2}$$

Yechim qadamlari

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

$$4\left(p^{2}+2pq+q^{2}\right)-\left(2p+q\right)^{2}$$

4 ga p^{2}+2pq+q^{2} ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$4p^{2}+8pq+4q^{2}-\left(2p+q\right)^{2}$$

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

$$4p^{2}+8pq+4q^{2}-\left(4p^{2}+4pq+q^{2}\right)$$

4p^{2}+4pq+q^{2} teskarisini topish uchun har birining teskarisini toping.

$$4p^{2}+8pq+4q^{2}-4p^{2}-4pq-q^{2}$$

0 ni olish uchun 4p^{2} va -4p^{2} ni birlashtirish.

$$8pq+4q^{2}-4pq-q^{2}$$

4pq ni olish uchun 8pq va -4pq ni birlashtirish.

$$4pq+4q^{2}-q^{2}$$

3q^{2} ni olish uchun 4q^{2} va -q^{2} ni birlashtirish.

$$4pq+3q^{2}$$