1. Muammo: $x=\frac{\sqrt{a+2b}+\sqrt{a-2b}}{\sqrt{a+2b}-\sqrt{a-2b}}$ bo'lsa, $bx^2-ax+b$ ifodasini soddalashtiring. 2. Avvalo, $x$ ni soddalashtirish uchun kasrni quyidagi tarzda ko'paytiramiz: kasrning yuqori va pastki qismini $\sqrt{a+2b}+\sqrt{a-2b}$ ga ko'paytiramiz. 3. Shunday qilib, $$x=\frac{(\sqrt{a+2b}+\sqrt{a-2b})^2}{(\sqrt{a+2b})^2-(\sqrt{a-2b})^2}$$ 4. Yuqoridagi ifodani ochamiz: $$\text{Yuoqri qism} = (\sqrt{a+2b})^2 + 2\sqrt{a+2b}\sqrt{a-2b} + (\sqrt{a-2b})^2 = (a+2b) + 2\sqrt{(a+2b)(a-2b)} + (a-2b)$$ 5. Pastki qism: $$ (\sqrt{a+2b})^2 - (\sqrt{a-2b})^2 = (a+2b) - (a-2b) = 4b $$ 6. Yuqori qismni soddalashtiramiz: $$ (a+2b) + (a-2b) + 2\sqrt{a^2 - (2b)^2} = 2a + 2\sqrt{a^2 - 4b^2} $$ 7. Shunday qilib, $$ x = \frac{2a + 2\sqrt{a^2 - 4b^2}}{4b} = \frac{a + \sqrt{a^2 - 4b^2}}{2b} $$ 8. Endi $bx^2 - ax + b$ ni hisoblaymiz: $$ bx^2 - ax + b = b\left(\frac{a + \sqrt{a^2 - 4b^2}}{2b}\right)^2 - a\left(\frac{a + \sqrt{a^2 - 4b^2}}{2b}\right) + b $$ 9. Kvadratni ochamiz: $$ \left(\frac{a + \sqrt{a^2 - 4b^2}}{2b}\right)^2 = \frac{(a + \sqrt{a^2 - 4b^2})^2}{4b^2} = \frac{a^2 + 2a\sqrt{a^2 - 4b^2} + (a^2 - 4b^2)}{4b^2} = \frac{2a^2 + 2a\sqrt{a^2 - 4b^2} - 4b^2}{4b^2} $$ 10. Endi ifodani to'liq yozamiz: $$ b \cdot \frac{2a^2 + 2a\sqrt{a^2 - 4b^2} - 4b^2}{4b^2} - a \cdot \frac{a + \sqrt{a^2 - 4b^2}}{2b} + b $$ 11. Soddalashtiramiz: $$ \frac{b(2a^2 + 2a\sqrt{a^2 - 4b^2} - 4b^2)}{4b^2} - \frac{a(a + \sqrt{a^2 - 4b^2})}{2b} + b = \frac{2a^2 + 2a\sqrt{a^2 - 4b^2} - 4b^2}{4b} - \frac{a^2 + a\sqrt{a^2 - 4b^2}}{2b} + b $$ 12. Har bir qismni umumiy maxrajga keltiramiz ($4b$): $$ \frac{2a^2 + 2a\sqrt{a^2 - 4b^2} - 4b^2}{4b} - \frac{2a^2 + 2a\sqrt{a^2 - 4b^2}}{4b} + \frac{4b^2}{4b} $$ 13. Qo'shamiz va ayiramiz: $$ \frac{2a^2 + 2a\sqrt{a^2 - 4b^2} - 4b^2 - 2a^2 - 2a\sqrt{a^2 - 4b^2} + 4b^2}{4b} = \frac{0}{4b} = 0 $$ 14. Natija: $bx^2 - ax + b = 0$. Demak, ifoda soddalashtirilganda natija $0$ ga teng.