$\sqrt[3]{-1} = -1, \qquad \sqrt[3]{-8} = -2, \qquad \sqrt{4} = 2$

Za koje $x$ je $\sqrt{5 - x}$ definisan?

$5 - x \geq 0 \implies x \leq 5, \qquad x \in (-\infty, 5]$

Za koje $x$ je $\sqrt{-9x}$ definisan?

$-9x \geq 0 \implies x \leq 0, \qquad x \in (-\infty, 0]$

Za koje $x$ je $\sqrt{-36a^2}$ definisan?

Pošto je $a^2 \geq 0$ uvek, sledi $-36a^2 \leq 0$. Jedina mogućnost je $a = 0$.

Oblast definisanosti $f(x) = \sqrt{\dfrac{2-x}{x+1}}$:

Tražimo $\dfrac{2-x}{x+1} \geq 0$ uz $x \neq -1$. Rešenjem nejednačine dobijamo $x \in (-1, 2]$.

$\sqrt{48a^5b^{12}c^3} = \sqrt{16 \cdot 3 \cdot a^4 \cdot a \cdot b^{12} \cdot c^2 \cdot c} = 4a^2b^6c\sqrt{3ac}$

$\sqrt[3]{27c^4} = \sqrt[3]{3^3 \cdot c^3 \cdot c} = 3c\sqrt[3]{c}$

$\sqrt{8a^7(a+b)^{4n}} = \sqrt{4 \cdot 2 \cdot a^6 \cdot a \cdot ((a+b)^{2n})^2} = 2a^3(a+b)^{2n}\sqrt{2a}$

$5\sqrt{7} = \sqrt{25 \cdot 7} = \sqrt{175}$

$\frac{1}{3}\sqrt{27} = \sqrt{\frac{1}{9} \cdot 27} = \sqrt{3}$

$x\sqrt{\frac{1}{x}} = \sqrt{x^2 \cdot \frac{1}{x}} = \sqrt{x}, \quad x > 0$

$(x-1)\sqrt{\frac{1}{x^2-1}} = \sqrt{\frac{(x-1)^2}{(x-1)(x+1)}} = \sqrt{\frac{x-1}{x+1}}, \quad x > 1$

Koji je veći: $\sqrt{2}$ ili $\sqrt[3]{3}$?

NZS za 2 i 3 je 6: $\sqrt{2} = \sqrt[6]{2^3} = \sqrt[6]{8}, \qquad \sqrt[3]{3} = \sqrt[6]{3^2} = \sqrt[6]{9}$

Pošto je $9 > 8$, sledi $\sqrt[3]{3} > \sqrt{2}$.

Koji je veći: $3\sqrt{5}$ ili $5\sqrt{3}$?

$3\sqrt{5} = \sqrt{9 \cdot 5} = \sqrt{45}, \qquad 5\sqrt{3} = \sqrt{25 \cdot 3} = \sqrt{75}$

Pošto je $75 > 45$, sledi $5\sqrt{3} > 3\sqrt{5}$.

$2\sqrt{18} + 3\sqrt{8} - \sqrt{50} + 3\sqrt{32} = 6\sqrt{2} + 6\sqrt{2} - 5\sqrt{2} + 12\sqrt{2} = 19\sqrt{2}$

$5\sqrt[3]{4} + 2\sqrt[3]{32} - \sqrt[3]{108} = 5\sqrt[3]{4} + 4\sqrt[3]{4} - 3\sqrt[3]{4} = 6\sqrt[3]{4}$

$\frac{a}{\sqrt[m]{b}} = \frac{a \cdot \sqrt[m]{b^{m-1}}}{\sqrt[m]{b^m}} = \frac{a\sqrt[m]{b^{m-1}}}{b}$

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

$\frac{5}{\sqrt{3}+2} \cdot \frac{\sqrt{3}-2}{\sqrt{3}-2} = \frac{5(\sqrt{3}-2)}{3-4} = \frac{5(\sqrt{3}-2)}{-1} = 10 - 5\sqrt{3}$

$\frac{6}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$

Prvo množimo sa $(\sqrt{2}+\sqrt{3})-\sqrt{5}$, pa imenilac postaje $2\sqrt{6}$. Zatim ponovo racionalizujemo.

Konačno: $\dfrac{2\sqrt{3} + 3\sqrt{2} - \sqrt{30}}{2}$.

$\frac{1}{\sqrt[3]{a}+\sqrt[3]{b}} = \frac{\sqrt[3]{a^2}-\sqrt[3]{ab}+\sqrt[3]{b^2}}{a+b}$

$\sqrt{(-7)^2} = |-7| = 7 \qquad \text{(ne } -7\text{!)}$

Uprostiti $\sqrt{x^2}$ za $x \leq 0$:

$\sqrt{x^2} = |x| = -x \quad \text{(jer je } x \leq 0\text{)}$

Uprostiti $\sqrt{x^6}$ za $x < 0$:

$\sqrt{x^6} = \sqrt{(x^3)^2} = |x^3| = -x^3 \quad \text{(jer je } x^3 < 0 \text{ kada } x < 0\text{)}$

Uprostiti $\sqrt{x^2-6x+9} + \sqrt{x^2+6x+9}$ za $-3 \leq x \leq 3$:

$= \sqrt{(x-3)^2} + \sqrt{(x+3)^2} = |x-3| + |x+3|$

Za $-3 \leq x \leq 3$: $x-3 \leq 0$, pa $|x-3| = 3-x$; i $x+3 \geq 0$, pa $|x+3| = x+3$.

$= (3-x) + (x+3) = 6$

$\sqrt{7 \pm 4\sqrt{3}} = \sqrt{4 \pm 4\sqrt{3} + 3} = \sqrt{(2 \pm \sqrt{3})^2} = 2 \pm \sqrt{3}$

$\sqrt{5 - 2\sqrt{6}}: \quad \text{tražimo } p^2 + q^2 = 5, \; pq = \sqrt{6} \implies p = \sqrt{3}, q = \sqrt{2}$

$\sqrt{5 - 2\sqrt{6}} = \sqrt{(\sqrt{3}-\sqrt{2})^2} = \sqrt{3} - \sqrt{2}$

$\sqrt{4 \pm 2\sqrt{3}} = \sqrt{4 \pm \sqrt{12}}: \quad C = \sqrt{16-12} = 2$

$\sqrt{\frac{4+2}{2}} \pm \sqrt{\frac{4-2}{2}} = \sqrt{3} \pm 1$

Dokazati: $(x^{\frac{1}{3}} - y^{\frac{1}{3}})(x^{\frac{2}{3}} + x^{\frac{1}{3}}y^{\frac{1}{3}} + y^{\frac{2}{3}}) = x - y$

Smenom $a = x^{1/3}$, $b = y^{1/3}$ prepoznajemo razliku kubova:

$(a-b)(a^2+ab+b^2) = a^3 - b^3 = x - y \checkmark$

Uprostiti $\dfrac{x-1}{x + x^{1/4}}$ za $x = 16$:

$x^{1/4} = 2, \quad x^{1/2} = 4, \quad x^{3/4} = 8$

$\frac{16-1}{16+2} = \frac{15}{18} = \frac{5}{6}$

$A = \sqrt[3]{10+6\sqrt{3}} - \sqrt[3]{10-6\sqrt{3}}$

Proveravamo: $(1+\sqrt{3})^3 = 1 + 3\sqrt{3} + 9 + 3\sqrt{3} = 10 + 6\sqrt{3}$ ✓

$A = (1+\sqrt{3}) - (1-\sqrt{3}) = 2\sqrt{3}$