a=sqrt{6-sqrt{35}}-sqrt{6+sqrt{35}}S eching

$$a = \sqrt { 6 - \sqrt { 35 } } - \sqrt { 6 + \sqrt { 35 } } S$$

S uchun yechish

$$S=\frac{\sqrt{10}a-\sqrt{14}a+12-2\sqrt{35}}{2}$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$\sqrt{6-\sqrt{35}}-\sqrt{6+\sqrt{35}}S=a$$

Ikkala tarafdan \sqrt{6-\sqrt{35}} ni ayirish.

$$-\sqrt{6+\sqrt{35}}S=a-\sqrt{6-\sqrt{35}}$$

Tenglama standart shaklda.

$$\left(-\sqrt{\sqrt{35}+6}\right)S=a-\sqrt{6-\sqrt{35}}$$

Ikki tarafini -\sqrt{6+\sqrt{35}} ga bo‘ling.

$$\frac{\left(-\sqrt{\sqrt{35}+6}\right)S}{-\sqrt{\sqrt{35}+6}}=\frac{a+\frac{\sqrt{10}}{2}-\frac{\sqrt{14}}{2}}{-\sqrt{\sqrt{35}+6}}$$

-\sqrt{6+\sqrt{35}} ga bo'lish -\sqrt{6+\sqrt{35}} ga ko'paytirishni bekor qiladi.

$$S=\frac{a+\frac{\sqrt{10}}{2}-\frac{\sqrt{14}}{2}}{-\sqrt{\sqrt{35}+6}}$$

a-\frac{\sqrt{14}}{2}+\frac{\sqrt{10}}{2} ni -\sqrt{6+\sqrt{35}} ga bo'lish.

$$S=-\frac{\frac{\sqrt{14}-\sqrt{10}}{2}\left(2a+\sqrt{10}-\sqrt{14}\right)}{2}$$

a uchun yechish

$$a=\frac{-\left(\sqrt{10}+\sqrt{14}\right)S+\sqrt{14}-\sqrt{10}}{2}$$

Yechim qadamlari

Shartlarni qayta saralash.

$$a=-\sqrt{\sqrt{35}+6}S+\sqrt{-\sqrt{35}+6}$$