$$ \displaystyle \begin{align} a^{\frac{1}{n}} = \left\{\begin{array}{l11} n = 2 & \Rightarrow & \toggle{\sqrt[2]{a}}{\sqrt{a}}\endtoggle \\ n = 3 & \Rightarrow & \sqrt[3]{a} \\ n = 4 & \Rightarrow & \sqrt[4]{a} \\ n = 5 & \Rightarrow & \sqrt[5]{a} \\ ... & ... & ... \\ n = k & \Rightarrow & \sqrt[k]{a} \\ \end{array} \right. \end{align}$$

$$ \sqrt[n]{0} = 0$$

$$ \begin{array}{1c1l} \text{(I)} & \sqrt[2]{0} = 0 \\ \text{(II)} & \sqrt[15]{0} = 0 \end{array}$$

$$ \sqrt[n]{a^m} = a^{\frac{m}{n}}$$

$$ \begin{array}{1c1l} \text{(I)} & \sqrt{3^3} = 3^{\frac32} \\ \text{(II)} & \sqrt[15]{7^{15}} = 7^{\frac{15}{15}} = 7 \end{array}$$

$$ a^m = \sqrt[n]{a^{n \times m}}$$

$$ \begin{array}{1c1l} \text{(I)} & 2\sqrt{3^3} = \sqrt{2^{1\times2} \times 3^3} \\ \text{(II)} & 2^2\sqrt[3]{2^{15}} = \sqrt[3]{2^{2\times3} \times 2^{15}} \end{array}$$

$$ \sqrt{a^2} = |a|$$

$$ \begin{array}{1c1l} \text{(I)} & \sqrt{16} = \sqrt{4^2} = |4| = 4 \\ \text{(II)} & \sqrt{121} = \sqrt{11^2} = |11| = 11 \\ \text{(III)} & \sqrt{64} = \sqrt{8^2} = 8 \:\:\: \color{green} \unicode{x2714} \\ & \sqrt{64} = \sqrt{8^2} = \pm 8 \:\:\: \color{crimson} \unicode{x2718} \end{array}$$

$$ \sqrt[n]{a} \times \sqrt[n]{b} = \sqrt[n]{a \times b}$$

$$ \begin{array}{1c1l} \text{(I)} & \sqrt[3]{2} \times \sqrt[3]{5} = \sqrt[3]{2 \times 5} \\ \text{(II)} & \sqrt[5]{11} \times \sqrt[5]{3} = \sqrt[5]{11 \times 3} \end{array}$$

$$ \frac{\sqrt[n]{a}}{\sqrt[n]{b}} = \sqrt[n]{\frac{a}{b}}$$

$$ \begin{array}{1c1l} \text{(I)} & \frac{\sqrt[3]{2}}{\sqrt[3]{5}} = \sqrt[3]{\frac25} \\ \text{(II)} & \frac{\sqrt[5]{11}}{\sqrt[5]{3}} = \sqrt[5]{\frac{11}{3}} \end{array}$$

$$ (p \times \sqrt[n]{a}) \pm (r \times \sqrt[n]{a}) = (p \pm r)\times \sqrt[n]{a}$$

$$ \begin{array}{1c1l} \text{(I)} & 2 \sqrt[3]{5} - 5 \sqrt[3]{5} = (2 - 5) \sqrt[3]{5} \\ \text{(II)} & 5 \sqrt[5]{11} + 7 \sqrt[5]{11} = (5 + 7) \sqrt[5]{11} \end{array}$$

$$ \begin{array}{1l1l} \sqrt[n]{a^n} = a & n = 2m + 1 \:,\: m \in \mathbb {Z}^{+} \\ \sqrt[n]{a^n} = |a| & n = 2m \:,\: m \in \mathbb {Z}^{+} \end{array}$$

$$ \begin{array}{1c1l} \text{(I)} & \sqrt[6]{(x + 1)^6} = |x + 1| \\ \text{(II)} & \sqrt[5]{(x + 1)^5} = x + 1 \end{array}$$

$$ \sqrt[m]{\sqrt[n]{\sqrt[r]{a}}} = \sqrt[m \cdot n \cdot r]{a}$$

$$ \begin{array}{1c1l} \text{(I)} & \sqrt[3]{\sqrt{\sqrt[5]{2}}} = \sqrt[3\times2\times5]{2} \\ \text{(II)} & \sqrt{\sqrt{\sqrt{5}}} = \sqrt[8]{5} \end{array}$$

$$ \sqrt[n]{a \times \sqrt[n]{a \times \sqrt[n]{a \times {...}}}} = \sqrt[n-1]{a}$$

$$ \begin{align} \sqrt[n]{a \times \sqrt[n]{a \times \sqrt[n]{a \times {...}}}} &= x \\ \sqrt[n]{a \times x} &= x \\ a \times x &= x^n \\ a &= x^{n - 1} \end{align}$$ $$ \begin{align} \Rightarrow x = \sqrt[n-1]{a} \end{align}$$

$$ \sqrt[n]{a \div \sqrt[n]{a \div \sqrt[n]{a \div {...}}}} = \sqrt[n+1]{a}$$

$$ \begin{align} \sqrt[n]{a \div \sqrt[n]{a \div \sqrt[n]{a \div {...}}}} &= x \\ \sqrt[n]{a \div x} &= x \\ a \div x &= x^n \\ a &= x^{n + 1} \end{align}$$ $$ \begin{align} \Rightarrow x = \sqrt[n+1]{a} \end{align}$$

$$ \sqrt{a + \sqrt{a + \sqrt{a + ...}}} = \frac{1 + \sqrt{1+4a}}{2}$$

$$ \begin{align} \sqrt{a + \sqrt{a + \sqrt{a + ...}}} &= x \\ \sqrt{a + x} &= x \end{align}$$ $$ \begin{align} x^2 - x - a = 0 \end{align}$$ $$ \begin{align} \Delta &= (-1)^2 - (4)(1)(-a) \\ \Delta &= 1 + 4a \end{align}$$ $$ \begin{align} x_{1,2} = \frac{1 \pm \sqrt{1 + 4a}}{2} \end{align}$$ $$ \begin{align} x > 0 \Rightarrow x = \frac{1 + \sqrt{1 + 4a}}{2} \end{align}$$

Çarpma ve bölme işleminde olduğu gibi aynı işlemi n. mertebeden kök için düşündüğümüzde konu oldukça karmaşıklaşıyor ve karşımıza aşağıdaki gibi bir denklem çıkıyor. Bu denklemin çözümü oldukça zor olup, kapalı ifadelerin yüksek mertebeden kısmi türevlerinin alınması gerekir. O yüzden şimdilik o konuya girmiyorum.

$$ \begin{align} x^n - x - a =0 \end{align}$$

$$ \sqrt{a - \sqrt{a - \sqrt{a - ...}}} = \frac{-1 + \sqrt{1+4a}}{2}$$

$$ \begin{align} \sqrt{a - \sqrt{a - \sqrt{a - ...}}} &= x \\ \sqrt{a - x} &= x \end{align}$$ $$ \begin{align} x^2 + x - a = 0 \end{align}$$ $$ \begin{align} \Delta &= (1)^2 - (4)(1)(-a) \\ \Delta &= 1 + 4a \end{align}$$ $$ \begin{align} x_{1,2} = \frac{-1 \pm \sqrt{1 + 4a}}{2} \end{align}$$ $$ \begin{align} x > 0 \Rightarrow x = \frac{-1 + \sqrt{1 + 4a}}{2} \end{align}$$

Çarpma ve bölme işleminde olduğu gibi aynı işlemi n. mertebeden kök için düşündüğümüzde konu oldukça karmaşıklaşıyor ve karşımıza aşağıdaki gibi bir denklem çıkıyor. Bu denklemin çözümü oldukça zor olup, kapalı ifadelerin yüksek mertebeden kısmi türevlerinin alınması gerekir. O yüzden şimdilik o konuya girmiyorum.

$$ \begin{align} x^n + x - a =0 \end{align}$$

$$ \underbrace {\sqrt{\lower 8pt {x \pm 2} \sqrt{\lower 6pt {y}}} \lower 6pt {\:= \sqrt{a} \pm \sqrt{b}}}_{\color{red} y = a \times b,\:\:\: x = a + b}$$

$$ \begin{align} \sqrt{x \pm 2 \sqrt{y}} &= \sqrt{a} \pm \sqrt{b} \\ \left(\sqrt{x \pm 2 \sqrt{y}}\right)^2 &= \left( \sqrt{a} \pm \sqrt{b} \right)^2 \\ x \pm 2 \sqrt{y} &= a + b \pm 2\sqrt{ab} \end{align}$$ $$ \begin{align} \Rightarrow x &= a + b \\ \Rightarrow y &= a \times b \end{align}$$

$$ \begin{align} \frac{(\sqrt{3}-1)\cdot(\sqrt{4+2\sqrt{3}})\cdot(\sqrt{1-\frac{\sqrt{3}}{2}})}{2} \end{align}$$

$$ \begin{align} &= \frac{(\sqrt{3}-1)(\sqrt{3}+1) \cdot \sqrt{1-\frac{\sqrt{3}}{2}}}{2} \\ &= \frac{2 \cdot \sqrt{1-\frac{\sqrt{3}}{2}}}{2} \\ &= \frac{\toggle{\sqrt{4-2\sqrt{3}}}{\sqrt{3}-1}\endtoggle}{2} \\ &=\frac{(\sqrt{3}-1)\cdot(\sqrt{3}+1)}{2 \cdot (\sqrt{3}+1)} \\ &= \frac{2}{2 \cdot (\sqrt{3}+1)} = \frac{1}{\sqrt{3}+1} \end{align}$$

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

$$ \begin{align} \frac{a-b}{a\sqrt{b}+b\sqrt{a}} &= 1 \\ \frac{\left(\sqrt{a} - \sqrt{b}\right)\left(\sqrt{a} + \sqrt{b}\right)}{\sqrt{ab}\left(\sqrt{a} + \sqrt{b}\right)} &= 1 \\ \frac{\sqrt{a} - \sqrt{b}}{\sqrt{ab}} &= 1 \\ \sqrt{a} - \sqrt{b} &= \sqrt{ab} \\ \frac{\sqrt{a}}{\sqrt{b}} - \frac{\sqrt{b}}{\sqrt{b}} &= \frac{\sqrt{a}\sqrt{b}}{\sqrt{b}} \\ \frac{\sqrt{a}}{\sqrt{b}} - 1 &= \sqrt{a} \\ \frac{\sqrt{a}}{\sqrt{b}} &= \sqrt{a} + 1 \\ \frac{\sqrt{a}}{\sqrt{a} + 1} &= \sqrt{b} \end{align}$$ $$ \begin{align} \Rightarrow b = \frac{a}{\left(\sqrt{a} + 1\right)^2} \end{align}$$

$$ \begin{align} \sqrt{147 \times 102 - 98 \times 151} = ? \end{align}$$

$$ \begin{align} &= \sqrt{147 \times 102 - 98 \times 151} \\ &= \sqrt{147 \times 98 + 147 \times 4 - 98 \times 151} \\ &= \sqrt{98 \times (147-151) + 147 \times 4} \\ &= \sqrt{4 \times (147-98)} \\ &= \sqrt{4 \times 49} \\ &= 2 \times 7 = 14 \end{align}$$

$$ \begin{align} A &= \sqrt[7]{\frac{\sqrt{7}-1}{\sqrt{3}+1}} \\ B &= \sqrt[7]{\frac{3\sqrt{3}-3}{\sqrt{7}+1}} \end{align}$$

$$ \begin{align} B \cdot \frac{1}{A} &= \sqrt[7]{\frac{3\sqrt{3}-3}{\sqrt{7}+1}} \cdot \frac{1}{\sqrt[7]{\frac{\sqrt{7}-1}{\sqrt{3}+1}}} \\ &= \sqrt[7]{\frac{3 \cdot (\sqrt{3}-1)}{\sqrt{7}+1}} \cdot \sqrt[7]{\frac{\sqrt{3}+1}{\sqrt{7}-1}} \\ &= \sqrt[7]{\frac{3 \cdot 2}{7 - 1}} = \sqrt[7]{1} = 1 \end{align}$$ $$ \begin{align} \frac{B}{A} = 1 \Rightarrow B = A \end{align}$$

$$ \begin{align} \sqrt{\frac{a^2}{b^2}-2+\frac{b^2}{a^2}} = a-b \end{align}$$

$$ \begin{align} \sqrt{\frac{a^2}{b^2}-2+\frac{b^2}{a^2}} &= a - b \\ \sqrt{\left(\frac{a}{b}-\frac{b}{a}\right)^2} &= a - b \\ \left|\frac{a}{b}-\frac{b}{a}\right| &= a - b \end{align}$$ $$ \begin{align} \frac{a}{b}-\frac{b}{a} = a - b \:\:\: \vee \:\:\: \frac{a}{b}-\frac{b}{a} = -(a - b) \end{align}$$ $$ \begin{align} &\frac{a}{b}-\frac{b}{a} = a - b \\ &\Rightarrow \frac{a^2-b^2}{a \cdot b} = a - b \\ &\Rightarrow \frac{(a-b) \cdot (a+b)}{a \cdot b} = a - b \\ &\Rightarrow \underbrace{a+b = a \cdot b}_{a=2\:,\:b=2} \end{align}$$ $$ \begin{align} &\frac{a}{b}-\frac{b}{a} = -(a - b) \\ &\Rightarrow \frac{a^2-b^2}{a \cdot b} = -(a - b) \\ &\Rightarrow \frac{(a-b) \cdot (a+b)}{a \cdot b} = -(a - b) \\ &\Rightarrow \underbrace{a+b = -a \cdot b}_{a=-2\:,\:b=-2} \end{align}$$ $$ \begin{align} \Rightarrow min(a+b) &= -4 \end{align}$$

$$ \begin{align} \sum_{n=2}^{169} \frac{(-1)^n}{\sqrt{n}-\sqrt{n-1}} = ? \end{align}$$

$$ \toggle{\begin{align} n=2 &\Rightarrow \frac{1}{\sqrt{2}-\sqrt{1}} \\ &\Rightarrow (+\sqrt{2}+\sqrt{1}) \\ n=3 &\Rightarrow \frac{-1}{\sqrt{3}-\sqrt{2}} \\ &\Rightarrow (-\sqrt{3}-\sqrt{2}) \\ n=4 &\Rightarrow \frac{1}{\sqrt{4}-\sqrt{3}} \\ &\Rightarrow(+\sqrt{4}+\sqrt{3}) \\ n=5 &\Rightarrow \frac{-1}{\sqrt{5}-\sqrt{4}} \\ &\Rightarrow (-\sqrt{5}-\sqrt{4}) \\ n=6 &\Rightarrow \frac{1}{\sqrt{6}-\sqrt{5}} \\ &\Rightarrow(+\sqrt{6}+\sqrt{5}) \\ &......\\ &......\\ &......\\ n=169 &\Rightarrow \frac{-1}{\sqrt{169}-\sqrt{168}} \\ &\Rightarrow(-\sqrt{169}-\sqrt{168}) \end{align}}{\begin{align} n=2 &\Rightarrow \frac{1}{\sqrt{2}-\sqrt{1}} \\ &\Rightarrow(\cancel{+\sqrt{2}}+\sqrt{1}) \\ n=3 &\Rightarrow \frac{-1}{\sqrt{3}-\sqrt{2}} \\ &\Rightarrow (\cancel{-\sqrt{3}}-\cancel{\sqrt{2}}) \\ n=4 &\Rightarrow \frac{1}{\sqrt{4}-\sqrt{3}} \\ &\Rightarrow (\cancel{+\sqrt{4}}+\cancel{\sqrt{3}}) \\ n=5 &\Rightarrow \frac{-1}{\sqrt{5}-\sqrt{4}} \\ &\Rightarrow (\cancel{-\sqrt{5}}-\cancel{\sqrt{4}}) \\ n=6 &\Rightarrow \frac{1}{\sqrt{6}-\sqrt{5}} \\ &\Rightarrow (\cancel{+\sqrt{6}}+\cancel{\sqrt{5}}) \\ &......\\ &......\\ &......\\ n=169 &\Rightarrow \frac{-1}{\sqrt{169}-\sqrt{168}} \\ &\Rightarrow (-\sqrt{169}\cancel{-\sqrt{168}}) \end{align}}\endtoggle$$ $$ \begin{align} \sum_{n=2}^{169} \frac{(-1)^n}{\sqrt{n}-\sqrt{n-1}} &= 1 - \sqrt{169} \\ &= 1 - \sqrt{13^2} \\ &= 1 - 13 = -12 \end{align}$$