Числа и выражения: тождества, свойства корней и степеней

$$\frac{1}{a}=\frac{1}{b}-\frac{1}{c}$$

$$\frac{1}{c}=\frac{1}{b}-\frac{1}{a}$$

$$\frac{1}{c}=\frac{a-b}{ab}$$

$$c=\frac{ab}{a-b}$$

$$x=\frac{2}{yz}$$

$$x^3-2x^2-4x+8=x^2(x-2)-4(x-2)=(x-2)(x^2-4)=(x-2)(x-2)(x+2)=(x-2)^2(x+2)$$

$$x^4+x^3-x-1=x^3(x+1)-(x+1)=(x+1)(x^3-1)=(x+1)(x-1)(x^2+1+x)=(x^2-1)(x^2+x+1)$$

$$x^2+4x+4=(x+2)^2$$

$$(\sqrt{7}-2+2)^2=(\sqrt{7})^2=7$$

$$x^2-6x+9=(x-3)^2$$

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

$$\frac{x^2-25}{x+5}-\frac{x^2-5x}{x-5}=\frac{(x-5)(x+5)}{x+5}-\frac{x(x-5)}{x-5}=x-5-x=-5$$

$$\frac{5}{4}=\sqrt{\frac{25}{16}}=\sqrt{1\frac{9}{16}}<\sqrt{1.69}=1.3<\sqrt{3}$$

$$\frac{5}{4}; 1.3; \sqrt{3}$$ — в порядке возрастания

$$3.2=\sqrt{10.24} >\sqrt{5} >\sqrt{\frac{36}{49}}=\frac{6}{7}$$

$$3.2; \sqrt{5}; \frac{6}{7}$$ — в порядке убывания.

$$n^2=4km$$

$$n=2\sqrt{km}$$

$$\sqrt[3]{125\cdot0.027}=\sqrt[3]{5^3\cdot(0.3)^3}=\sqrt[3]{5^3}\cdot\sqrt[3]{(0.3)^3}=5\cdot0.3=1.5$$

$$\sqrt[4]{625\cdot0.0016}=\sqrt[4]{5^4}\cdot\sqrt[4]{(0.2)^4}=5\cdot0.2=1$$

$$\sqrt[3]{0.008\cdot27}=\sqrt[3]{(0.2)^3}\cdot\sqrt[3]{3^3}=0.2\cdot3=0.6$$

$$\sqrt[4]{0.0625\cdot81}=\sqrt[4]{(0.5)^4}\cdot\sqrt[4]{3^4}=0.5\cdot3=1.5$$

$$\sqrt[5]{2^{20}b^{15}}=2^{\frac{20}{5}}b^{\frac{15}{5}}=2^4b^3$$

$$\sqrt[4]{27a}\cdot\sqrt[4]{3a^3}=\sqrt[4]{27\cdot3}\cdot\sqrt[4]{a^4}=3a$$

$$\sqrt[4]{(-3)^2\cdot2}\cdot\sqrt[4]{8\cdot9}=\sqrt[4]{81\cdot16}=3\cdot2=6$$

$$\frac{\sqrt[5]{192t}}{\sqrt[5]{6t^{11}}}=\sqrt[5]{\frac{192t}{6t^{11}}}=\sqrt[5]{\frac{32}{t^{10}}}=\frac{\sqrt[5]{2^5}}{\sqrt[5]{t^{10}}}=\frac{2}{t^2}$$

$$\sqrt[3]{16ab^{12}}:\sqrt[3]{2a^4b^9}=\sqrt[3]{\frac{16ab^{16}}{2a^4b^9}}=\sqrt[3]{\frac{2^3b^3}{a^3}}=\frac{2b}{a}$$

$$\frac{\sqrt[8]{16a^5b^7}}{\sqrt{2ab}}+2\sqrt[8]{ab^3}=\sqrt[8]{\frac{16a^5b^7}{2^4a^4b^4}}+2\sqrt[8]{ab^3}=\sqrt[8]{ab^3}+2\sqrt[8]{ab^3}=3\sqrt[8]{ab^3}$$