Упр.384 ГДЗ Колягин Ткачёва 8 класс (Алгебра)

1)1/4; 2)1/125; 3)2/3; 4)3/11; 5)-3/5; 6)-3 1/7. Сократить дробь:
1) (5a^2-35)/(a-v7);
2) (x^3-3x)/(x+v3);
3) (5x-5v3)/(3-x^2 );
4) (4va+vb)/(b-16a);
5) (9-2v3)/(3v6-2v2).

1. $$\frac14=0{,}25$$

2. $$\frac1{125}=0{,}008$$

3. $$\frac23=0{,}(6)$$

4. $$\frac3{11}=0{,}(27)$$

5. $$-\frac35=-0{,}6$$

6. $$-3\frac17=-3{,}(142857)$$

1. $$\frac{5a^2-35}{a-\sqrt7}=\frac{5(a^2-7)}{a-\sqrt7}=\frac{5(a-\sqrt7)(a+\sqrt7)}{a-\sqrt7}=5(a+\sqrt7)$$

2. $$\frac{x^3-3x}{x+\sqrt3}=\frac{x(x^2-3)}{x+\sqrt3}=\frac{x(x-\sqrt3)(x+\sqrt3)}{x+\sqrt3}=x(x-\sqrt3)$$

3. $$\frac{5x-5\sqrt3}{3-x^2}=\frac{5(x-\sqrt3)}{(\sqrt3-x)(\sqrt3+x)}=-\frac5{\sqrt3+x}$$

4. $$\frac{4\sqrt a+\sqrt b}{b-16a}=\frac{4\sqrt a+\sqrt b}{(\sqrt b-4\sqrt a)(\sqrt b+4\sqrt a)}=\frac1{\sqrt b-4\sqrt a}$$

5. $$\frac{9-2\sqrt3}{3\sqrt6-2\sqrt2}=\frac{\sqrt3(3\sqrt3-2)}{\sqrt2(3\sqrt3-2)}=\frac{\sqrt3}{\sqrt2}=\sqrt{\frac32}$$

Ответ

$$0{,}25;\ 0{,}008;\ 0{,}(6);\ 0{,}(27);\ -0{,}6;\ -3{,}(142857)$$

$$5(a+\sqrt7);\ x(x-\sqrt3);\ -\frac5{\sqrt3+x};\ \frac1{\sqrt b-4\sqrt a};\ \sqrt{\frac32}$$