Kvadrat ildizlar

🟢 — masalani tezlashtiradi · 🟡 — ilg'or daraja · 🔵 — chuqur o'rganish (olimpiada darajasi)

Irratsional ifodalar (surds)

$\sqrt{ab} = \sqrt{a}\,\sqrt{b} \qquad \sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}} \qquad \sqrt{a}\cdot\sqrt{a} = a \qquad m\sqrt{a} + n\sqrt{a} = (m + n)\sqrt{a}$

⚠️ $\sqrt{a + b} \neq \sqrt{a} + \sqrt{b}$. Masalan, $\sqrt{9 + 16} = 5$, lekin $\sqrt{9} + \sqrt{16} = 7$.

Maxrajni irratsionallikdan qutqarish

Qo'shma ifoda: $(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b$

$\frac{k}{\sqrt{b}} = \frac{k\sqrt{b}}{b} \qquad \frac{1}{a + \sqrt{b}} = \frac{a - \sqrt{b}}{a^2 - b} \qquad \frac{1}{\sqrt{a} + \sqrt{b}} = \frac{\sqrt{a} - \sqrt{b}}{a - b}$

Misol: $\dfrac{6}{3 - \sqrt{3}} = \dfrac{6(3 + \sqrt{3})}{9 - 3} = 3 + \sqrt{3}$

🟢 Ichma-ich ildizni soddalashtirish

$x + y = a$ va $xy = b$ bo'lsa:

$\sqrt{a + 2\sqrt{b}} = \sqrt{x} + \sqrt{y} \qquad \sqrt{a - 2\sqrt{b}} = \sqrt{x} - \sqrt{y} \quad (x > y)$

Misol: $\sqrt{5 + 2\sqrt{6}} = \sqrt{3} + \sqrt{2}$, chunki $3 + 2 = 5$, $3 \cdot 2 = 6$.

🟢 Maxrajni qutqarishning qo'shimcha shakli

$\frac{1}{\sqrt{a} - \sqrt{b}} = \frac{\sqrt{a} + \sqrt{b}}{a - b}$