Question

1.Разложите на множители квадратный трехчлен:

а) х^2-10х+21;

б) 5у^2+9у-2.

В) 2x^2-x-1

г) -2x^2+4x+6

д) 6x^2-18x+12

2. Сократите дробь
а) 7x-2+4^2 / 1 -16x^2
б) 2x^2-5x-3 / x^2+x-6

Answer 1

Разложить квадратный трехчлен можно по формуле

$a {x}^{2} + bx + c = a(x - x_1)(x -x_2)$, где $x_1$ и $x_2$ — корни квадратного уравнения $a {x}^{2} + bx + c = 0$.

#1.

а) ${x}^{2} - 10x + 21 = (x - 7)(x - 3)$

Решение:

${x}^{2} - 10x + 21 = 0$

$D = {b}^{2} - 4ac = {( - 10)}^{2} - 4 \times 21 = 100 - 84 = 16 = {4}^{2}$

$x_1 = \frac{ - b + \sqrt{D} }{2a} = \frac{10 + 4}{2} = \frac{14}{2} = 7$

$x_2 = \frac{ - b - \sqrt{D} }{2a} = \frac{10 - 4}{2} = \frac{6}{2} = 3$

б) $5 {y}^{2} + 9y - 2 = 5(x - \frac{1}{5} )(x - ( - 2)) = (5x - 1)(x + 2)$

Решение:

$5 {y}^{2} + 9y - 2 = 0$

$D = {b}^{2} - 4ac = {9}^{2} - 4 \times 5 \times ( - 2) = 81 + 40 = 121 = {11}^{2}$

$y_1 = \frac{ - b + \sqrt{D} }{2a} = \frac{ - 9 + 11}{2 \times 5} = \frac{2}{10} = \frac{1}{5}$

$y_2 = \frac{ - b - \sqrt{D} }{2a} = \frac{ - 9 - 11}{2 \times 5} = \frac{ - 20}{10} = - 2$

в) $2 {x}^{2} - x - 1 = 2(x - \frac{1}{2} )(x - ( - 1)) = (2x - 1)(x + 1)$

Решение:

$2 {x}^{2} - x - 1 = 0$

$D = {b}^{2} - 4ac = {( - 1)}^{2} - 4 \times 2 \times ( - 1) = 1 + 8 = 9 = {3}^{2}$

$x_1 = \frac{ - b + \sqrt{D} }{2a} = \frac{ - 1 + 3}{2 \times 2} = \frac{2}{4} = \frac{1}{2}$

$x_2 = \frac{ - b - \sqrt{D} }{2a} = \frac{ - 1 - 3}{2 \times 2} = \frac{ - 4}{4} = - 1$

г) $- 2 {x}^{2} + 4x + 6 = - 2(x - ( - 1))(x - 3) = - 2(x + 1)(x - 3) = (x + 1)(6 - 2x)$

Решение:

$- 2 {x}^{2} + 4x + 6 = 0$

${x}^{2} - 2x - 3 = 0$

$D = {b}^{2} - 4ac = { (- 2)}^{2} - 4 \times ( - 3) = 4 + 12 = 16 = {4}^{2}$

$x_1 = \frac{ - b + \sqrt{D} }{2a} = \frac{2 + 4}{2} = \frac{6}{2} = 3$

$x_2 = \frac{ - b - \sqrt{D} }{2a} = \frac{2 - 4}{2} = \frac{ - 2}{2} = - 1$

д) $6 {x}^{2} - 18x + 12 = 6(x - 2)(x - 1) = (x - 2)(6x - 6)$

Решение:

$6 {x}^{2} - 18x + 12 = 0$

${x}^{2} - 3x + 2 = 0$

$D = {b}^{2} - 4ac = {( - 3)}^{2} - 4 \times 2 = 9 - 8 = 1$

$x_1 = \frac{ - b - \sqrt{D} }{2a} = \frac{3 + 1}{2} = \frac{4}{2} = 2$

$x_2 = \frac{ - b - \sqrt{D} }{2a} = \frac{3 - 1}{2} = \frac{2}{2} = 1$

#2.

а) $\frac{7x - 2 + 4 {x}^{2} }{1 - 16 {x}^{2} } = \frac{(4x - 1)(x + 2)}{ {1}^{2} - {(4x)}^{2} } = \frac{(4x - 1)(x + 2)}{(1 - 4x)(1 + 4x)} = \frac{(4x - 1)(x + 2)}{ - (4x - 1)(1 + 4x)} = - \frac{x + 2}{1 + 4x}$

Разложение на множители числителя:

$7x - 2 + 4 {x}^{2} = 4 {x}^{2} + 7x - 2 = 4(x - \frac{1}{4} )(x - ( - 2)) = (4x - 1)(x + 2)$

$4 {x}^{2} + 7x - 2 = 0$

$D = {b}^{2} - 4ac = {7}^{2} - 4 \times 4 \times ( - 2) = 49 + 32 = 81 = {9}^{2}$

$x_1 = \frac{ - b + \sqrt{D} }{2a} = \frac{ - 7 + 9}{2 \times 4} = \frac{2}{8} = \frac{1}{4}$

$x_2 = \frac{ - b - \sqrt{D} }{2a} = \frac{ - 7 - 9}{2 \times 4} = \frac{ - 16}{8} = - 2$

б) $\frac{2 {x}^{2} - 5x - 3}{ {x}^{2} + x - 6 }$

тут у меня не получается. Выражение точно выглядит так?

Получается так:

$2 {x}^{2} - 5x - 3 = 0$

$D = {b}^{2} - 4ac = {(-5)}^{2} - 4 \times 2 \times ( - 3) = 25 + 24 = 49 = {7}^{2}$

$x_1 = \frac{ - b + \sqrt{D} }{2a} = \frac{ 5 + 7}{2 \times 2} = \frac{12}{4} = 3$

$x_2 = \frac{ - b - \sqrt{D} }{2a} = \frac{ 5 - 7}{2 \times 2} = \frac{-2}{4} = - \frac{1}{2}$

$2 {x}^{2} - 5x - 3 = 2(x-3)(x+\frac{1}{2}) = (x-3)(2x+1)$.

${x}^{2} + x - 6 = 0$

$D = {b}^{2} - 4ac = {1}^{2} - 4 \times ( - 6) = 1 + 24 = 25 = {5}^{2}$

$x_1 = \frac{ - b + \sqrt{D} }{2a} = \frac{ - 1+ 5}{2} = \frac{4}{2} = 2$

$x_1 = \frac{ - b - \sqrt{D} }{2a} = \frac{ - 1- 5}{2} = \frac{-6}{2} = - 3$

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

Обобщаем:

$\frac{2 {x}^{2} - 5x - 3}{ {x}^{2} + x - 6 } = \frac{(x-3)(2x+1)}{ (x-2)(x+3) }$