Решебник по алгебре 9 класс Макарычев Задание 935

 

\[\boxed{\text{935\ (935).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ 0,3x(x + 13) - 2x(0,9 - 0,2x) =\]

\[= 0\]

\[0,3x^{2} + 3,9x - 1,8x + 0,4x^{2} = 0\]

\[0,7x^{2} + 2,1x = 0\]

\[0,7x(x + 3) = 0\]

\[x_{1} = 0,\ \ x_{2} = - 3.\]

\[Ответ:x = 0;\ x = - 3.\]

\[\textbf{б)}\ 1,5x(x + 4) - x(7 - 0,5x) =\]

\[= 0,5 \cdot (10 - 2x)\]

\[1,5x^{2} + 6x - 7x + 0,5x^{2} = 5 - x\]

\[2x^{2} = 5\]

\[x^{2} = 2,5\]

\[x = \pm \sqrt{2,5}\]

\[Ответ:x = \pm \sqrt{2,5}.\]

\[\textbf{в)}\ \frac{(2x + 1)^{2}}{25} - \frac{x - 1}{3} = x\]

\[3 \cdot \left( 4x^{2} + 4x + 1 \right) - 25x + 25 =\]

\[= 75x\]

\[12x^{2} - 88x + 28 = 0\]

\[3x^{2} - 22x + 7 = 0\]

\[D = 484 - 84 = 400,\]

\[x_{1} = \frac{22 + 20}{6} = 7,\]

\[x_{2} = \frac{22 - 20}{6} = \frac{1}{3}.\]

\[Ответ:x = 7;x = \frac{1}{3}.\]

\[\textbf{г)}\ \frac{(3x + 2)^{2}}{11} - \frac{x + 5}{4} = x²\]

\[4 \cdot \left( 9x^{2} + 12x + 4 \right) - 11x - 55 =\]

\[= 44x^{2}\]

\[- 8x^{2} + 37x - 39 = 0\]

\[8x^{2} - 37x + 39 = 0\]

\[D = 1369 - 1248\]

\[x_{1} = \frac{37 + 11}{16} = 3,\ \ \]

\[x_{2} = \frac{37 - 11}{6} = \frac{13}{8} = 1\frac{5}{8}.\]

\[Ответ:x = 3;\ \ x = 1\frac{5}{8}.\]

\[\textbf{д)}\ \frac{(2 - x)^{2}}{3} - 2x = \frac{(7 + 2x)^{2}}{5}\]

\[5 \cdot \left( 4 - 4x + x^{2} \right) - 30x =\]

\[= 3 \cdot \left( 49 + 28x + 4x^{2} \right)\]

\[- 7x^{2} - 134x - 127 = 0\]

\[7x² + 134x + 127 = 0\]

\[D = 17956 - 3556 = 14\ 400 =\]

\[= 120^{2}\]

\[x_{1} = \frac{- 134 + 120}{14} = - 1,\ \ \]

\[x_{2} = \frac{- 134 - 120}{14} = - 18\frac{1}{7}.\]

\[Ответ:\ x = - 1;\ x = - 18\frac{1}{7}.\]

\[\textbf{е)}\frac{(6 - x)^{2}}{8} + x = 7 - \frac{(2x - 1)^{2}}{3}\]

\[3 \cdot \left( 36 - 12x + x^{2} \right) + 24x =\]

\[= 168 - 8 \cdot \left( 4x^{2} - 4x + 1 \right)\]

\[35x^{2} - 44x - 52 = 0\]

\[D = 1936 + 7280 = 9216\]

\[x_{1} = \frac{44 + 96}{70} = 2,\ \ \]

\[x_{2} = \frac{44 - 96}{70} = - \frac{26}{35}.\]

\[Ответ:\ \ x = 2;\ x = - \frac{26}{35}.\]