(sqrt{6}-sqrt{2})^2-frac{sqrt{6}-sqrt{2}}{sqrt{6}+sqrt{2}} eching

$$( \sqrt { 6 } - \sqrt { 2 } ) ^ { 2 } - \frac { \sqrt { 6 } - \sqrt { 2 } } { \sqrt { 6 } + \sqrt { 2 } }$$

Baholash

$$6-3\sqrt{3}\approx 0,803847577$$

Yechim qadamlari

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(\sqrt{6}-\sqrt{2}\right)^{2} kengaytirilishi uchun ishlating.

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

\sqrt{6} kvadrati – 6.

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

Faktor: 6=2\times 3. \sqrt{2\times 3} koʻpaytmasining kvadrat ildizini \sqrt{2}\sqrt{3} kvadrat ildizlarining koʻpaytmasi sifatida qayta yozing.

$$6-2\sqrt{2}\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^{2}-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}$$

2 hosil qilish uchun \sqrt{2} va \sqrt{2} ni ko'paytirish.

$$6-2\times 2\sqrt{3}+\left(\sqrt{2}\right)^{2}-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}$$

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

$$6-4\sqrt{3}+\left(\sqrt{2}\right)^{2}-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}$$

\sqrt{2} kvadrati – 2.

$$6-4\sqrt{3}+2-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}$$

8 olish uchun 6 va 2'ni qo'shing.

$$8-4\sqrt{3}-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}$$

\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}} maxrajini \sqrt{6}-\sqrt{2} orqali surat va maxrajini koʻpaytirish orqali ratsionallashtiring.

$$8-4\sqrt{3}-\frac{\left(\sqrt{6}-\sqrt{2}\right)\left(\sqrt{6}-\sqrt{2}\right)}{\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{6}-\sqrt{2}\right)}$$

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

$$8-4\sqrt{3}-\frac{\left(\sqrt{6}-\sqrt{2}\right)\left(\sqrt{6}-\sqrt{2}\right)}{\left(\sqrt{6}\right)^{2}-\left(\sqrt{2}\right)^{2}}$$

\sqrt{6} kvadratini chiqarish. \sqrt{2} kvadratini chiqarish.

$$8-4\sqrt{3}-\frac{\left(\sqrt{6}-\sqrt{2}\right)\left(\sqrt{6}-\sqrt{2}\right)}{6-2}$$

4 olish uchun 6 dan 2 ni ayirish.

$$8-4\sqrt{3}-\frac{\left(\sqrt{6}-\sqrt{2}\right)\left(\sqrt{6}-\sqrt{2}\right)}{4}$$

\left(\sqrt{6}-\sqrt{2}\right)^{2} hosil qilish uchun \sqrt{6}-\sqrt{2} va \sqrt{6}-\sqrt{2} ni ko'paytirish.

$$8-4\sqrt{3}-\frac{\left(\sqrt{6}-\sqrt{2}\right)^{2}}{4}$$

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(\sqrt{6}-\sqrt{2}\right)^{2} kengaytirilishi uchun ishlating.

$$8-4\sqrt{3}-\frac{\left(\sqrt{6}\right)^{2}-2\sqrt{6}\sqrt{2}+\left(\sqrt{2}\right)^{2}}{4}$$

\sqrt{6} kvadrati – 6.

$$8-4\sqrt{3}-\frac{6-2\sqrt{6}\sqrt{2}+\left(\sqrt{2}\right)^{2}}{4}$$

Faktor: 6=2\times 3. \sqrt{2\times 3} koʻpaytmasining kvadrat ildizini \sqrt{2}\sqrt{3} kvadrat ildizlarining koʻpaytmasi sifatida qayta yozing.

$$8-4\sqrt{3}-\frac{6-2\sqrt{2}\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^{2}}{4}$$

2 hosil qilish uchun \sqrt{2} va \sqrt{2} ni ko'paytirish.

$$8-4\sqrt{3}-\frac{6-2\times 2\sqrt{3}+\left(\sqrt{2}\right)^{2}}{4}$$

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

$$8-4\sqrt{3}-\frac{6-4\sqrt{3}+\left(\sqrt{2}\right)^{2}}{4}$$

\sqrt{2} kvadrati – 2.

$$8-4\sqrt{3}-\frac{6-4\sqrt{3}+2}{4}$$

8 olish uchun 6 va 2'ni qo'shing.

$$8-4\sqrt{3}-\frac{8-4\sqrt{3}}{4}$$

2-\sqrt{3} natijani olish uchun 8-4\sqrt{3} ning har bir ifodasini 4 ga bo‘ling.

$$8-4\sqrt{3}-\left(2-\sqrt{3}\right)$$

2-\sqrt{3} teskarisini topish uchun har birining teskarisini toping.

$$8-4\sqrt{3}-2+\sqrt{3}$$

6 olish uchun 8 dan 2 ni ayirish.

$$6-4\sqrt{3}+\sqrt{3}$$

-3\sqrt{3} ni olish uchun -4\sqrt{3} va \sqrt{3} ni birlashtirish.

$$6-3\sqrt{3}$$

Omil

$$3 {(2 - \sqrt{3})} = 0,803847577$$