ГДЗ учебник по алгебре 8 класс Макарычев. К параграфу 9 Номер 696

Решите уравнение:
а)

\frac{2 x + 1}{2 x - 1} - \frac{3 (2 x - 1)}{7 (2 x + 1)} + \frac{8}{1 - 4 x^{2}} = 0

;
б)

\frac{y}{y^{2} - 9} - \frac{1}{y^{2} + 3 y} + \frac{3}{6 y + 2 y^{2}} = 0

;
в)

\frac{2 y - 1}{14 y^{2} + 7 y} + \frac{8}{12 y^{2} - 3} = \frac{2 y + 1}{6 y^{2} - 3 y}

;
г)

\frac{3}{x^{2} - 9} - \frac{1}{9 - 6 x + x^{2}} = \frac{3}{2 x^{2} + 6 x}

;
д)

\frac{9 x + 12}{x^{3} - 64} - \frac{1}{x^{2} + 4 x + 16} = \frac{1}{x - 4}

;
е)

\frac{3}{8 y^{3} + 1} - \frac{1}{2 y + 1} = \frac{y + 3}{4 y^{2} - 2 y + 1}

;
ж)

\frac{32}{x^{3} - 2 x^{2} - x + 2} + \frac{1}{(x - 1) (x - 2)} = \frac{1}{x + 1}

з)

\frac{1}{3 (x - 4)} + \frac{1}{2 (x^{2} + 3)} + \frac{1}{x^{3} - 4 x^{2} + 3 x - 12} = 0

.

\frac{2 x + 1}{2 x - 1} - \frac{3 (2 x - 1)}{7 (2 x + 1)} + \frac{8}{1 - 4 x^{2}} = 0

\frac{2 x + 1}{2 x - 1} - \frac{3 (2 x - 1)}{7 (2 x + 1)} - \frac{8}{4 x^{2} - 1} = 0

|*7(2x + 1)(2x − 1)

{\begin{cases} 7 (2 x + 1)^{2} - 3 (2 x - 1)^{2} - 7 * 8 = 0 \\ x \neq \pm \frac{1}{2} \end{cases}

7 (4 x^{2} + 4 x + 1) - 3 (4 x^{2} - 4 x + 1) - 56 = 0

16 x^{2} + 40 x - 52 = 0

|:4

4 x^{2} + 10 x - 13 = 0

D = 5^{2} + 4 * 13 = 77

{\begin{cases} x = \frac{- 5 \pm \sqrt{77}}{4} \\ x \neq \pm \frac{1}{2} \end{cases}

Ответ:

x_{1, 2} = \frac{- 5 \pm \sqrt{77}}{4}

\frac{y}{y^{2} - 9} - \frac{1}{y^{2} + 3 y} + \frac{3}{6 y + 2 y^{2}} = 0

\frac{y}{(y - 3) (y + 3)} - \frac{1}{y (y + 3)} + \frac{3}{2 y (y + 3)} = 0

|*2y(y + 3)(y − 3)

{\begin{cases} 2 y^{2} - 2 (y - 3) + 3 (y - 3) = 0 \\ y \neq \pm 3 \end{cases}

2 y^{2} + y - 3 = 0

(2y + 3)(y − 1) = 0

{\begin{cases} y_{1} = - 1, 5, y_{2} = 1 \\ y \neq \pm 3 \end{cases}

Ответ:

y_{1} = - 1, 5

;

y_{2} = 1

.

\frac{2 y - 1}{14 y^{2} + 7 y} + \frac{8}{12 y^{2} - 3} = \frac{2 y + 1}{6 y^{2} - 3 y}

\frac{2 y - 1}{7 y (2 y + 1)} + \frac{8}{3 (4 y^{2} - 1)} = \frac{2 y + 1}{3 y (2 y - 1)}

|*21y(2y − 1)(2y + 1)

{\begin{cases} 3 (2 y - 1)^{2} + 8 * 7 y = 7 (2 y + 1)^{2} \\ y \neq \pm \frac{1}{2} \end{cases}

3 (4 y^{2} - 4 y + 1) + 56 y = 7 (4 y^{2} + 4 y + 1)

16 y^{2} - 16 y + 4 = 0

4 y^{2} - 4 y + 1 = 0

(2 y - 1)^{2} = 0

{\begin{cases} y = \frac{1}{2} \\ y \neq \pm \frac{1}{2} \end{cases}

Ответ: y = ∅ − пустое множество, нет решений.

\frac{3}{x^{2} - 9} - \frac{1}{9 - 6 x + x^{2}} = \frac{3}{2 x^{2} + 6 x}

\frac{3}{(x - 3) (x + 3)} - \frac{1}{(x - 3)^{2}} = \frac{3}{2 x (x + 3)} | * 2 x (x + 3) (x - 3)^{2}

{\begin{cases} 3 * 2 x (x - 3) - 2 x (x + 3) = 3 (x - 3)^{2} \\ x \neq \pm 3 \end{cases}

6 x^{2} - 18 x - 2 x^{2} - 6 x = 3 (x^{2} - 6 x + 9)

x^{2} - 6 x - 27 = 0

(x + 3)(x − 9) = 0

{\begin{cases} x_{1} = - 3, x_{2} = 9 \\ x \neq \pm 3 \end{cases}

Ответ: x = 9

\frac{9 x + 12}{x^{3} - 64} - \frac{1}{x^{2} + 4 x + 16} = \frac{1}{x - 4}

\frac{9 x + 12}{(x - 4) (x^{2} + 4 x + 16)} - \frac{1}{x^{2} + 4 x + 16} = \frac{1}{x - 4} | * (x - 4) (x^{2} + 4 x + 16)

{\begin{cases} (9 x + 12) - (x - 4) = x^{2} + 4 x + 16 \\ x \neq 4 \end{cases}

9 x + 12 - x + 4 - x^{2} - 4 x - 16 = 0

x^{2} - 4 x = 0

x(x − 4) = 0

{\begin{cases} x_{1} = 0, x_{2} = 4 \\ x \neq 4 \end{cases}

Ответ: x = 0

\frac{3}{8 y^{3} + 1} - \frac{1}{2 y + 1} = \frac{y + 3}{4 y^{2} - 2 y + 1}

\frac{3}{(2 y + 1) (4 y^{2} - 2 y + 1)} - \frac{1}{2 y + 1} = \frac{y + 3}{4 y^{2} - 2 y + 1} | * (2 y + 1) (4 y^{2} - 2 y + 1)

{\begin{cases} 3 - (4 y^{2} - 2 y + 1) = (y + 3) (2 y + 1) \\ y \neq - \frac{1}{2} \end{cases}

3 - 4 y^{2} + 2 y - 1 = 2 y^{2} + 7 y + 3

- 6 y^{2} - 5 y - 1 = 0

6 y^{2} + 5 y + 1 = 0

(3y + 1)(2y + 1) = 0

{\begin{cases} y_{1} = - \frac{1}{3}, y_{2} = - \frac{1}{2} \\ y \neq - \frac{1}{2} \end{cases}

Ответ:

y = - \frac{1}{3}

\frac{32}{x^{3} - 2 x^{2} - x + 2} + \frac{1}{(x - 1) (x - 2)} = \frac{1}{x + 1}

x^{3} - 2 x^{2} - x + 2 = x^{2} (x - 2) - (x - 2) = (x^{2} - 1) (x - 2)

\frac{32}{(x - 1) (x + 1) (x - 2)} + \frac{1}{(x - 1) (x - 2)} = \frac{1}{x + 1}

|*(x − 1)(x + 1)(x − 2)

{\begin{cases} 32 + x + 1 = (x - 1) (x - 2) \\ x \neq \pm 1; 2 \end{cases}

32 + x + 1 = x^{2} - x - 2 x + 2

x^{2} - 4 x - 31 = 0

D = 2^{2} + 31 = 4 + 31 = 35

{\begin{cases} x = 2 \pm \sqrt{35} \\ x \neq \pm 1; 2 \end{cases}

Ответ:

x_{1, 2} = 2 \pm \sqrt{35}

\frac{1}{3 (x - 4)} + \frac{1}{2 (x^{2} + 3)} + \frac{1}{x^{3} - 4 x^{2} + 3 x - 12} = 0

x^{3} - 4 x^{2} + 3 x - 12 = x^{2} (x - 4) + 3 (x - 4) = (x^{2} + 3) (x - 4)

\frac{1}{3 (x - 4)} + \frac{1}{2 (x^{2} + 3)} + \frac{1}{(x^{2} + 3) (x - 4)} = 0 | * 6 (x^{2} + 3) (x - 4)

{\begin{cases} 2 (x^{2} + 3) + 3 (x - 4) + 6 = 0 \\ x \neq 4 \end{cases}

2 x^{2} + 6 + 3 x - 12 + 6 = 0

2 x^{2} + 3 x = 0

x(2x + 3) = 0

{\begin{cases} x_{1} = 0, x_{2} = - 1, 5 \\ x \neq 4 \end{cases}

Ответ:

x_{1} = 0

;

x_{2} = - 1, 5

.

ГДЗ Алгебра 8 класс Макарычев, Миндюк, Нешков, Суворова