(sqrt{6}-sqrt{8})*(2sqrt{2}+sqrt{6}) eching

$$( \sqrt { 6 } - \sqrt { 8 } ) \times ( 2 \sqrt { 2 } + \sqrt { 6 } )$$

Baholash

$$-2$$

Yechim qadamlari

Faktor: 8=2^{2}\times 2. \sqrt{2^{2}\times 2} koʻpaytmasining kvadrat ildizini \sqrt{2^{2}}\sqrt{2} kvadrat ildizlarining koʻpaytmasi sifatida qayta yozing. 2^{2} ning kvadrat ildizini chiqarish.

$$\left(\sqrt{6}-2\sqrt{2}\right)\left(2\sqrt{2}+\sqrt{6}\right)$$

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

$$\left(\sqrt{6}\right)^{2}-\left(2\sqrt{2}\right)^{2}$$

\sqrt{6} kvadrati – 6.

$$6-\left(2\sqrt{2}\right)^{2}$$

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

$$6-2^{2}\left(\sqrt{2}\right)^{2}$$

2 daraja ko‘rsatkichini 2 ga hisoblang va 4 ni qiymatni oling.

$$6-4\left(\sqrt{2}\right)^{2}$$

\sqrt{2} kvadrati – 2.

$$6-4\times 2$$

8 hosil qilish uchun 4 va 2 ni ko'paytirish.

$$6-8$$

-2 olish uchun 6 dan 8 ni ayirish.

$$-2$$

Omil

$$-2$$