frac{sqrt{5}+sqrt{3}}{sqrt{5}-sqrt{3}}-frac{sqrt{5}-sqrt{3}}{sqrt{5}+sqrt{3}}=2sqrt{15} eching

$$\frac{ \sqrt{ 5 } + \sqrt{ 3 } }{ \sqrt{ 5 } - \sqrt{ 3 } } - \frac{ \sqrt{ 5 } - \sqrt{ 3 } }{ \sqrt{ 5 } + \sqrt{ 3 } } = 2 \sqrt{ 15 }$$

Tasdiqlamoq

Yechim qadamlari

\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} maxrajini \sqrt{5}+\sqrt{3} orqali surat va maxrajini koʻpaytirish orqali ratsionallashtiring.

$$\frac{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=2\sqrt{15}$$

Hisoblang: \left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right). Ko‘paytirish qoida yordamida turli kvadratlarga aylantirilishi mumkin: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}\right)^{2}-\left(\sqrt{3}\right)^{2}}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=2\sqrt{15}$$

\sqrt{5} kvadratini chiqarish. \sqrt{3} kvadratini chiqarish.

$$\frac{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{5-3}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=2\sqrt{15}$$

2 olish uchun 5 dan 3 ni ayirish.

$$\frac{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{2}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=2\sqrt{15}$$

\left(\sqrt{5}+\sqrt{3}\right)^{2} hosil qilish uchun \sqrt{5}+\sqrt{3} va \sqrt{5}+\sqrt{3} ni ko'paytirish.

$$\frac{\left(\sqrt{5}+\sqrt{3}\right)^{2}}{2}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=2\sqrt{15}$$

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(\sqrt{5}+\sqrt{3}\right)^{2} kengaytirilishi uchun ishlating.

$$\frac{\left(\sqrt{5}\right)^{2}+2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{2}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=2\sqrt{15}$$

\sqrt{5} kvadrati – 5.

$$\frac{5+2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{2}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=2\sqrt{15}$$

\sqrt{5} va \sqrt{3} ni koʻpaytirish uchun kvadrat ildiz ichidagi sonlarni koʻpaytiring.

$$\frac{5+2\sqrt{15}+\left(\sqrt{3}\right)^{2}}{2}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=2\sqrt{15}$$

\sqrt{3} kvadrati – 3.

$$\frac{5+2\sqrt{15}+3}{2}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=2\sqrt{15}$$

8 olish uchun 5 va 3'ni qo'shing.

$$\frac{8+2\sqrt{15}}{2}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=2\sqrt{15}$$

4+\sqrt{15} natijani olish uchun 8+2\sqrt{15} ning har bir ifodasini 2 ga bo‘ling.

$$4+\sqrt{15}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=2\sqrt{15}$$

\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}} maxrajini \sqrt{5}-\sqrt{3} orqali surat va maxrajini koʻpaytirish orqali ratsionallashtiring.

$$4+\sqrt{15}-\frac{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}=2\sqrt{15}$$

Hisoblang: \left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right). Ko‘paytirish qoida yordamida turli kvadratlarga aylantirilishi mumkin: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$4+\sqrt{15}-\frac{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}\right)^{2}-\left(\sqrt{3}\right)^{2}}=2\sqrt{15}$$

\sqrt{5} kvadratini chiqarish. \sqrt{3} kvadratini chiqarish.

$$4+\sqrt{15}-\frac{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{5-3}=2\sqrt{15}$$

2 olish uchun 5 dan 3 ni ayirish.

$$4+\sqrt{15}-\frac{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{2}=2\sqrt{15}$$

\left(\sqrt{5}-\sqrt{3}\right)^{2} hosil qilish uchun \sqrt{5}-\sqrt{3} va \sqrt{5}-\sqrt{3} ni ko'paytirish.

$$4+\sqrt{15}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{2}=2\sqrt{15}$$

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(\sqrt{5}-\sqrt{3}\right)^{2} kengaytirilishi uchun ishlating.

$$4+\sqrt{15}-\frac{\left(\sqrt{5}\right)^{2}-2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{2}=2\sqrt{15}$$

\sqrt{5} kvadrati – 5.

$$4+\sqrt{15}-\frac{5-2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{2}=2\sqrt{15}$$

\sqrt{5} va \sqrt{3} ni koʻpaytirish uchun kvadrat ildiz ichidagi sonlarni koʻpaytiring.

$$4+\sqrt{15}-\frac{5-2\sqrt{15}+\left(\sqrt{3}\right)^{2}}{2}=2\sqrt{15}$$

\sqrt{3} kvadrati – 3.

$$4+\sqrt{15}-\frac{5-2\sqrt{15}+3}{2}=2\sqrt{15}$$

8 olish uchun 5 va 3'ni qo'shing.

$$4+\sqrt{15}-\frac{8-2\sqrt{15}}{2}=2\sqrt{15}$$

4-\sqrt{15} natijani olish uchun 8-2\sqrt{15} ning har bir ifodasini 2 ga bo‘ling.

$$4+\sqrt{15}-\left(4-\sqrt{15}\right)=2\sqrt{15}$$

4-\sqrt{15} teskarisini topish uchun har birining teskarisini toping.

$$4+\sqrt{15}-4-\left(-\sqrt{15}\right)=2\sqrt{15}$$

-\sqrt{15} ning teskarisi \sqrt{15} ga teng.

$$4+\sqrt{15}-4+\sqrt{15}=2\sqrt{15}$$

0 olish uchun 4 dan 4 ni ayirish.

$$\sqrt{15}+\sqrt{15}=2\sqrt{15}$$

2\sqrt{15} ni olish uchun \sqrt{15} va \sqrt{15} ni birlashtirish.

$$2\sqrt{15}=2\sqrt{15}$$

Ikkala tarafdan 2\sqrt{15} ni ayirish.

$$2\sqrt{15}-2\sqrt{15}=0$$

0 ni olish uchun 2\sqrt{15} va -2\sqrt{15} ni birlashtirish.

$$0=0$$

0 va 0 ni taqqoslang.

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