{[(1/2a-2/3b)(1/2a+2/3b)^{3}-(frac{1}{4}a^{2}-frac{4}{9}b^{2})(frac{4}{9}b^2+frac{1}{4}a^2)]-frac{1}{3}ab(frac{1}{2}a^2+frac{1}{9}b^2)} eching

$$\{ [ ( \frac { 1 } { 2 } a - \frac { 2 } { 3 } b ) ( \frac { 1 } { 2 } a + \frac { 2 } { 3 } b ) ^ { 3 } - ( \frac { 1 } { 4 } a ^ { 2 } - \frac { 4 } { 9 } b ^ { 2 } ) ( \frac { 4 } { 9 } b ^ { 2 } + \frac { 1 } { 4 } a ^ { 2 } ) ] - \frac { 1 } { 3 } a b ( \frac { 1 } { 2 } a ^ { 2 } + \frac { 1 } { 9 } b ^ { 2 } ) \}$$

Baholash

$$-\frac{ab^{3}}{3}$$

Yechim qadamlari

\left(p+q\right)^{3}=p^{3}+3p^{2}q+3pq^{2}+q^{3} binom teoremasini \left(\frac{1}{2}a+\frac{2}{3}b\right)^{3} kengaytirilishi uchun ishlating.

$$\left(\frac{1}{2}a-\frac{2}{3}b\right)\left(\frac{1}{8}a^{3}+\frac{1}{2}a^{2}b+\frac{2}{3}ab^{2}+\frac{8}{27}b^{3}\right)-\left(\frac{1}{4}a^{2}-\frac{4}{9}b^{2}\right)\left(\frac{4}{9}b^{2}+\frac{1}{4}a^{2}\right)-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)$$

\frac{1}{2}a-\frac{2}{3}b ga \frac{1}{8}a^{3}+\frac{1}{2}a^{2}b+\frac{2}{3}ab^{2}+\frac{8}{27}b^{3} ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$\frac{1}{16}a^{4}+\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}-\left(\frac{1}{4}a^{2}-\frac{4}{9}b^{2}\right)\left(\frac{4}{9}b^{2}+\frac{1}{4}a^{2}\right)-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)$$

Hisoblang: \left(\frac{1}{4}a^{2}-\frac{4}{9}b^{2}\right)\left(\frac{4}{9}b^{2}+\frac{1}{4}a^{2}\right). Ko‘paytirish qoida yordamida turli kvadratlarga aylantirilishi mumkin: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{1}{16}a^{4}+\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}-\left(\left(\frac{1}{4}a^{2}\right)^{2}-\left(\frac{4}{9}b^{2}\right)^{2}\right)-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)$$

\left(\frac{1}{4}a^{2}\right)^{2} ni kengaytirish.

$$\frac{1}{16}a^{4}+\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}-\left(\left(\frac{1}{4}\right)^{2}\left(a^{2}\right)^{2}-\left(\frac{4}{9}b^{2}\right)^{2}\right)-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)$$

Daraja ko‘rsatkichini boshqa ko‘rsatkichga oshirish uchun, darajalarini ko‘paytiring. 2 va 2 ni ko‘paytirib, 4 ni oling.

$$\frac{1}{16}a^{4}+\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}-\left(\left(\frac{1}{4}\right)^{2}a^{4}-\left(\frac{4}{9}b^{2}\right)^{2}\right)-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)$$

2 daraja ko‘rsatkichini \frac{1}{4} ga hisoblang va \frac{1}{16} ni qiymatni oling.

$$\frac{1}{16}a^{4}+\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}-\left(\frac{1}{16}a^{4}-\left(\frac{4}{9}b^{2}\right)^{2}\right)-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)$$

\left(\frac{4}{9}b^{2}\right)^{2} ni kengaytirish.

$$\frac{1}{16}a^{4}+\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}-\left(\frac{1}{16}a^{4}-\left(\frac{4}{9}\right)^{2}\left(b^{2}\right)^{2}\right)-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)$$

Daraja ko‘rsatkichini boshqa ko‘rsatkichga oshirish uchun, darajalarini ko‘paytiring. 2 va 2 ni ko‘paytirib, 4 ni oling.

$$\frac{1}{16}a^{4}+\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}-\left(\frac{1}{16}a^{4}-\left(\frac{4}{9}\right)^{2}b^{4}\right)-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)$$

2 daraja ko‘rsatkichini \frac{4}{9} ga hisoblang va \frac{16}{81} ni qiymatni oling.

$$\frac{1}{16}a^{4}+\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}-\left(\frac{1}{16}a^{4}-\frac{16}{81}b^{4}\right)-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)$$

\frac{1}{16}a^{4}-\frac{16}{81}b^{4} teskarisini topish uchun har birining teskarisini toping.

$$\frac{1}{16}a^{4}+\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}-\frac{1}{16}a^{4}+\frac{16}{81}b^{4}-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)$$

0 ni olish uchun \frac{1}{16}a^{4} va -\frac{1}{16}a^{4} ni birlashtirish.

$$\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}+\frac{16}{81}b^{4}-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)$$

0 ni olish uchun -\frac{16}{81}b^{4} va \frac{16}{81}b^{4} ni birlashtirish.

$$\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)$$

-\frac{1}{3}ab ga \frac{1}{2}a^{2}+\frac{1}{9}b^{2} ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{1}{6}a^{3}b-\frac{1}{27}ab^{3}$$

0 ni olish uchun \frac{1}{6}a^{3}b va -\frac{1}{6}a^{3}b ni birlashtirish.

$$-\frac{8}{27}ab^{3}-\frac{1}{27}ab^{3}$$

-\frac{1}{3}ab^{3} ni olish uchun -\frac{8}{27}ab^{3} va -\frac{1}{27}ab^{3} ni birlashtirish.

$$-\frac{1}{3}ab^{3}$$

Kengaytirish

$$-\frac{ab^{3}}{3}$$

Yechim qadamlari

\left(p+q\right)^{3}=p^{3}+3p^{2}q+3pq^{2}+q^{3} binom teoremasini \left(\frac{1}{2}a+\frac{2}{3}b\right)^{3} kengaytirilishi uchun ishlating.

$$\left(\frac{1}{2}a-\frac{2}{3}b\right)\left(\frac{1}{8}a^{3}+\frac{1}{2}a^{2}b+\frac{2}{3}ab^{2}+\frac{8}{27}b^{3}\right)-\left(\frac{1}{4}a^{2}-\frac{4}{9}b^{2}\right)\left(\frac{4}{9}b^{2}+\frac{1}{4}a^{2}\right)-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)$$

\frac{1}{2}a-\frac{2}{3}b ga \frac{1}{8}a^{3}+\frac{1}{2}a^{2}b+\frac{2}{3}ab^{2}+\frac{8}{27}b^{3} ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$\frac{1}{16}a^{4}+\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}-\left(\frac{1}{4}a^{2}-\frac{4}{9}b^{2}\right)\left(\frac{4}{9}b^{2}+\frac{1}{4}a^{2}\right)-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)$$

Hisoblang: \left(\frac{1}{4}a^{2}-\frac{4}{9}b^{2}\right)\left(\frac{4}{9}b^{2}+\frac{1}{4}a^{2}\right). Ko‘paytirish qoida yordamida turli kvadratlarga aylantirilishi mumkin: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{1}{16}a^{4}+\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}-\left(\left(\frac{1}{4}a^{2}\right)^{2}-\left(\frac{4}{9}b^{2}\right)^{2}\right)-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)$$

\left(\frac{1}{4}a^{2}\right)^{2} ni kengaytirish.

$$\frac{1}{16}a^{4}+\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}-\left(\left(\frac{1}{4}\right)^{2}\left(a^{2}\right)^{2}-\left(\frac{4}{9}b^{2}\right)^{2}\right)-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)$$

Daraja ko‘rsatkichini boshqa ko‘rsatkichga oshirish uchun, darajalarini ko‘paytiring. 2 va 2 ni ko‘paytirib, 4 ni oling.

$$\frac{1}{16}a^{4}+\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}-\left(\left(\frac{1}{4}\right)^{2}a^{4}-\left(\frac{4}{9}b^{2}\right)^{2}\right)-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)$$

2 daraja ko‘rsatkichini \frac{1}{4} ga hisoblang va \frac{1}{16} ni qiymatni oling.

$$\frac{1}{16}a^{4}+\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}-\left(\frac{1}{16}a^{4}-\left(\frac{4}{9}b^{2}\right)^{2}\right)-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)$$

\left(\frac{4}{9}b^{2}\right)^{2} ni kengaytirish.

$$\frac{1}{16}a^{4}+\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}-\left(\frac{1}{16}a^{4}-\left(\frac{4}{9}\right)^{2}\left(b^{2}\right)^{2}\right)-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)$$

Daraja ko‘rsatkichini boshqa ko‘rsatkichga oshirish uchun, darajalarini ko‘paytiring. 2 va 2 ni ko‘paytirib, 4 ni oling.

$$\frac{1}{16}a^{4}+\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}-\left(\frac{1}{16}a^{4}-\left(\frac{4}{9}\right)^{2}b^{4}\right)-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)$$

2 daraja ko‘rsatkichini \frac{4}{9} ga hisoblang va \frac{16}{81} ni qiymatni oling.

$$\frac{1}{16}a^{4}+\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}-\left(\frac{1}{16}a^{4}-\frac{16}{81}b^{4}\right)-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)$$

\frac{1}{16}a^{4}-\frac{16}{81}b^{4} teskarisini topish uchun har birining teskarisini toping.

$$\frac{1}{16}a^{4}+\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}-\frac{1}{16}a^{4}+\frac{16}{81}b^{4}-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)$$

0 ni olish uchun \frac{1}{16}a^{4} va -\frac{1}{16}a^{4} ni birlashtirish.

$$\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{16}{81}b^{4}+\frac{16}{81}b^{4}-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)$$

0 ni olish uchun -\frac{16}{81}b^{4} va \frac{16}{81}b^{4} ni birlashtirish.

$$\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{1}{3}ab\left(\frac{1}{2}a^{2}+\frac{1}{9}b^{2}\right)$$

-\frac{1}{3}ab ga \frac{1}{2}a^{2}+\frac{1}{9}b^{2} ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$\frac{1}{6}a^{3}b-\frac{8}{27}ab^{3}-\frac{1}{6}a^{3}b-\frac{1}{27}ab^{3}$$

0 ni olish uchun \frac{1}{6}a^{3}b va -\frac{1}{6}a^{3}b ni birlashtirish.

$$-\frac{8}{27}ab^{3}-\frac{1}{27}ab^{3}$$

-\frac{1}{3}ab^{3} ni olish uchun -\frac{8}{27}ab^{3} va -\frac{1}{27}ab^{3} ni birlashtirish.

$$-\frac{1}{3}ab^{3}$$