Упр.389 ГДЗ Колягин Ткачёва 8 класс (Алгебра)

1) Вычислим значения с точностью до $$0{,}001$$:

$$\sqrt{8}=2{,}828427\ldots \approx 2{,}828$$

$$\sqrt{13}=3{,}605551\ldots \approx 3{,}606$$

$$\sqrt{6{,}6}=2{,}569046\ldots \approx 2{,}569$$

$$\sqrt{4{,}3}=2{,}073644\ldots \approx 2{,}074$$

$$\sqrt{0{,}5}=0{,}707107\ldots \approx 0{,}707$$

$$\sqrt{0{,}05}=0{,}223607\ldots \approx 0{,}224$$

2) Упростим выражения.

$$\left(\sqrt{ab}-\frac{ab}{a+\sqrt{ab}}\right):\frac{a^2b}{a-b}$$
$$=\frac{\sqrt{ab}(a+\sqrt{ab})-ab}{a+\sqrt{ab}}\cdot\frac{a-b}{a^2b}$$
$$=\frac{a\sqrt{ab}}{a+\sqrt{ab}}\cdot\frac{a-b}{a^2b}$$
$$=\frac{\sqrt{ab}(a-b)}{(a+\sqrt{ab})a^2b}$$
$$=\frac{\sqrt{a}-\sqrt{b}}{a\sqrt{b}}.$$

$$\left(\frac{a+\sqrt{b}}{a-\sqrt{b}}+\frac{a-\sqrt{b}}{a+\sqrt{b}}\right)\cdot\frac{a-\sqrt{b}}{a^2+b}$$
$$=\frac{(a+\sqrt{b})^2+(a-\sqrt{b})^2}{(a-\sqrt{b})(a+\sqrt{b})}\cdot\frac{a-\sqrt{b}}{a^2+b}$$
$$=\frac{2(a^2+b)}{a^2-b}\cdot\frac{a-\sqrt{b}}{a^2+b}$$
$$=\frac{2}{a+\sqrt{b}}.$$

$$\left(\frac{c-\sqrt{d}}{c+\sqrt{d}}-\frac{c+\sqrt{d}}{c-\sqrt{d}}\right):\frac{2c\sqrt{d}}{c+\sqrt{d}}$$
$$=\frac{(c-\sqrt{d})^2-(c+\sqrt{d})^2}{(c+\sqrt{d})(c-\sqrt{d})}\cdot\frac{c+\sqrt{d}}{2c\sqrt{d}}$$
$$=\frac{-4c\sqrt{d}}{(c-\sqrt{d})(c+\sqrt{d})}\cdot\frac{c+\sqrt{d}}{2c\sqrt{d}}$$
$$=-\frac{2}{c-\sqrt{d}}=\frac{2}{\sqrt{d}-c}.$$

$$\left(2+\sqrt{b}\right)\left(\frac{2}{\sqrt{b}+2}-\frac{2}{2-\sqrt{b}}+\frac{2b}{4-b}\right)$$
$$=\left(2+\sqrt{b}\right)\cdot\frac{2(2-\sqrt{b})-2(2+\sqrt{b})+2b}{(2-\sqrt{b})(2+\sqrt{b})}$$
$$=\left(2+\sqrt{b}\right)\cdot\frac{4-2\sqrt{b}-4-2\sqrt{b}+2b}{4-b}$$
$$=\left(2+\sqrt{b}\right)\cdot\frac{-4\sqrt{b}+2b}{4-b}$$
$$=\left(2+\sqrt{b}\right)\cdot\frac{-2\sqrt{b}(2-\sqrt{b})}{(2-\sqrt{b})(2+\sqrt{b})}$$
$$=-2\sqrt{b}.$$

Ответ

1) $$2{,}828$$; $$3{,}606$$; $$2{,}569$$; $$2{,}074$$; $$0{,}707$$; $$0{,}224$$.
2) $$\frac{\sqrt{a}-\sqrt{b}}{a\sqrt{b}}$$; $$\frac{2}{a+\sqrt{b}}$$; $$\frac{2}{\sqrt{d}-c}$$; $$-2\sqrt{b}$$.