cos60^circ/1+sin60^circ+1/tan30^circ eching

$$\frac { \cos 60 ^ { \circ } } { 1 + \sin 60 ^ { \circ } } + \frac { 1 } { \tan 30 ^ { \circ } }$$

Baholash

$$2$$

Qisqa yechim qadamlari

Trigonometrik qiymatlar jadvaldan \cos(60) qiymatini oling.

$$\frac{\frac{1}{2}}{1+\sin(60)}+\frac{1}{\tan(30)}$$

Trigonometrik qiymatlar jadvaldan \sin(60) qiymatini oling.

$$\frac{\frac{1}{2}}{1+\frac{\sqrt{3}}{2}}+\frac{1}{\tan(30)}$$

Ifodalarni qo‘shish yoki ayirish uchun ularni yoyib, maxrajlarini bir xil qiling. 1 ni \frac{2}{2} marotabaga ko'paytirish.

$$\frac{\frac{1}{2}}{\frac{2}{2}+\frac{\sqrt{3}}{2}}+\frac{1}{\tan(30)}$$

\frac{2}{2} va \frac{\sqrt{3}}{2} da bir xil maxraji bor, ularning suratini qo‘shish orqali qo‘shing.

$$\frac{\frac{1}{2}}{\frac{2+\sqrt{3}}{2}}+\frac{1}{\tan(30)}$$

\frac{1}{2} ni \frac{2+\sqrt{3}}{2} ga bo'lish \frac{1}{2} ga k'paytirish \frac{2+\sqrt{3}}{2} ga qaytarish.

$$\frac{2}{2\left(2+\sqrt{3}\right)}+\frac{1}{\tan(30)}$$

Trigonometrik qiymatlar jadvaldan \tan(30) qiymatini oling.

$$\frac{2}{2\left(2+\sqrt{3}\right)}+\frac{1}{\frac{\sqrt{3}}{3}}$$

1 ni \frac{\sqrt{3}}{3} ga bo'lish 1 ga k'paytirish \frac{\sqrt{3}}{3} ga qaytarish.

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

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

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

\sqrt{3} kvadrati – 3.

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

3 va 3 ni qisqartiring.

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

Ifodalarni qo‘shish yoki ayirish uchun ularni yoyib, maxrajlarini bir xil qiling. \sqrt{3} ni \frac{2\left(2+\sqrt{3}\right)}{2\left(2+\sqrt{3}\right)} marotabaga ko'paytirish.

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

\frac{2}{2\left(2+\sqrt{3}\right)} va \frac{\sqrt{3}\times 2\left(2+\sqrt{3}\right)}{2\left(2+\sqrt{3}\right)} da bir xil maxraji bor, ularning suratini qo‘shish orqali qo‘shing.

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

2+\sqrt{3}\times 2\left(2+\sqrt{3}\right) ichidagi ko‘paytirishlarni bajaring.

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

2+4\sqrt{3}+6 hisob-kitobini qiling.

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

2\left(2+\sqrt{3}\right) ni kengaytirish.

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

\frac{8+4\sqrt{3}}{2\sqrt{3}+4} maxrajini 2\sqrt{3}-4 orqali surat va maxrajini koʻpaytirish orqali ratsionallashtiring.

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

Hisoblang: \left(2\sqrt{3}+4\right)\left(2\sqrt{3}-4\right). Ko‘paytirish qoida yordamida turli kvadratlarga aylantirilishi mumkin: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

\left(2\sqrt{3}\right)^{2} ni kengaytirish.

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

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

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

\sqrt{3} kvadrati – 3.

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

12 hosil qilish uchun 4 va 3 ni ko'paytirish.

$$\frac{\left(8+4\sqrt{3}\right)\left(2\sqrt{3}-4\right)}{12-4^{2}}$$

2 daraja ko‘rsatkichini 4 ga hisoblang va 16 ni qiymatni oling.

$$\frac{\left(8+4\sqrt{3}\right)\left(2\sqrt{3}-4\right)}{12-16}$$

-4 olish uchun 12 dan 16 ni ayirish.

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

8+4\sqrt{3} ga 2\sqrt{3}-4 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$\frac{-32+8\left(\sqrt{3}\right)^{2}}{-4}$$

\sqrt{3} kvadrati – 3.

$$\frac{-32+8\times 3}{-4}$$

24 hosil qilish uchun 8 va 3 ni ko'paytirish.

$$\frac{-32+24}{-4}$$

-8 olish uchun -32 va 24'ni qo'shing.

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

2 ni olish uchun -8 ni -4 ga bo‘ling.

$$2$$