Решебник по алгебре 8 класс Мерзляк Задание 663

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\[1)\ (2x - 5)(x + 2) = 18\]

\[2x^{2} + 4x - 5x - 10 - 18 = 0\]

\[2x^{2} - x - 28 = 0\]

\[D = 1 + 4 \cdot 2 \cdot 28 = 225\]

\[x = \frac{1 \pm \sqrt{225}}{4} = \frac{1 \pm 15}{4}\]

\[x_{1} = 4,\ \ x_{2} = - \frac{7}{2} = - 3,5\]

\[Ответ:\ x = - 3,5;x = 4.\]

\[16x^{2} - 24x + 9 + 9x^{2} - 1 - 9 =\]

\[= 0\]

\[25x^{2} - 24x - 1 = 0\]

\[D = 576 + 100 = 676\]

\[x = \frac{24 \pm \sqrt{676}}{50} = \frac{24 \pm 26}{50}\]

\[x_{1} = 1,\ \ x_{2} = - 0,04\]

\[Ответ:\ \ x = - 0,04;x = 1.\]

\[3)\ (x + 3)^{2} - (2x - 1)^{2} = 16\]

\[- 3x^{2} + 10x - 8 = 0\]

\[D = 100 - 4 \cdot 3 \cdot 8 = 100 - 96 =\]

\[= 4\]

\[x = \frac{- 10 \pm \sqrt{4}}{- 6} = \frac{- 10 \pm 2}{- 6}\]

\[x_{1} = 2,\ \ x_{2} = \frac{4}{3} = 1\frac{1}{3}\]

\[Ответ:x = 2;\ \ x = 1\frac{1}{3}.\]

\[4)\ (x - 6)^{2} - 2x(x + 3) =\]

\[= 30 - 12x\]

\[- x^{2} - 6x + 6 = 0\]

\[x^{2} + 6x - 6 = 0\]

\[D = 36 + 24 = 60\]

\[x = \frac{- 6 \pm \sqrt{60}}{2} = \frac{- 6 \pm 2\sqrt{15}}{2} =\]

\[= - 3 \pm \sqrt{15}\]

\[Ответ:\ x = - 3 \pm \sqrt{15}.\]

\[- 3x^{2} + 27x - 54 = 0\ \ \ \ |\ :( - 3)\]

\[x^{2} - 9x + 18 = 0\]

\[D = 81 - 4 \cdot 18 = 9\]

\[x = \frac{9 \pm \sqrt{9}}{2} = \frac{9 \pm 3}{2}\]

\[x_{1} = 6,\ \ x_{2} = 3\]

\[Ответ:x = 3;x = 6.\]

\[6)\ (2x - 1)(2x + 1) - x(1 - x) =\]

\[= 2x(x + 1)\]

\[4x^{2} - 1 - x + x^{2} - 2x^{2} - 2x =\]

\[= 0\]

\[3x^{2} - 3x - 1 = 0\]

\[D = 9 + 12 = 21\]

\[x = \frac{3 \pm \sqrt{21}}{6}\]

\[Ответ:\ \ x = \frac{3 \pm \sqrt{21}}{6}.\]