(25p-15q)^2-(15p-25q)^2 eching

$$( 25 p - 15 q ) ^ { 2 } - ( 15 p - 25 q ) ^ { 2 }$$

Baholash

$$400\left(p^{2}-q^{2}\right)$$

Yechim qadamlari

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(25p-15q\right)^{2} kengaytirilishi uchun ishlating.

$$625p^{2}-750pq+225q^{2}-\left(15p-25q\right)^{2}$$

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(15p-25q\right)^{2} kengaytirilishi uchun ishlating.

$$625p^{2}-750pq+225q^{2}-\left(225p^{2}-750pq+625q^{2}\right)$$

225p^{2}-750pq+625q^{2} teskarisini topish uchun har birining teskarisini toping.

$$625p^{2}-750pq+225q^{2}-225p^{2}+750pq-625q^{2}$$

400p^{2} ni olish uchun 625p^{2} va -225p^{2} ni birlashtirish.

$$400p^{2}-750pq+225q^{2}+750pq-625q^{2}$$

0 ni olish uchun -750pq va 750pq ni birlashtirish.

$$400p^{2}+225q^{2}-625q^{2}$$

-400q^{2} ni olish uchun 225q^{2} va -625q^{2} ni birlashtirish.

$$400p^{2}-400q^{2}$$

Kengaytirish

$$400p^{2}-400q^{2}$$

Yechim qadamlari

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(25p-15q\right)^{2} kengaytirilishi uchun ishlating.

$$625p^{2}-750pq+225q^{2}-\left(15p-25q\right)^{2}$$

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(15p-25q\right)^{2} kengaytirilishi uchun ishlating.

$$625p^{2}-750pq+225q^{2}-\left(225p^{2}-750pq+625q^{2}\right)$$

225p^{2}-750pq+625q^{2} teskarisini topish uchun har birining teskarisini toping.

$$625p^{2}-750pq+225q^{2}-225p^{2}+750pq-625q^{2}$$

400p^{2} ni olish uchun 625p^{2} va -225p^{2} ni birlashtirish.

$$400p^{2}-750pq+225q^{2}+750pq-625q^{2}$$

0 ni olish uchun -750pq va 750pq ni birlashtirish.

$$400p^{2}+225q^{2}-625q^{2}$$

-400q^{2} ni olish uchun 225q^{2} va -625q^{2} ni birlashtirish.

$$400p^{2}-400q^{2}$$