Question

срочно 7,8 и 9!!!!!!!!
плачу 100.

Answer 1

7.

$\displaystyle\bf\\\left \{ {{ {x}^{2} + 2 {y}^{2} = 5 } \atop {y - x = 1 }} \right. \\ \displaystyle\bf\\\left \{ {{ {x}^{2} + 2(x + 1) {}^{2} = 5} \atop {y= x + 1 }} \right. \\ \\ {x}^{2} + 2( {x}^{2} + 2x + 1) = 5 \\ {x}^{2} + 2 {x}^{2} + 4x + 2 - 5 = 0 \\ 3 {x}^{2} + 4x - 3 = 0 \\ a = 3\\ b = 4 \\ c = - 3 \\ D = {b}^{2} - 4ac = {4}^{2} - 4 \times 3 \times ( - 3) = \\ = 16 + 36 = 52 \: \: \: ( \sqrt{D} = 2 \sqrt{13}) \\ x_{1} = \frac{ - 4 - 2 \sqrt{13} }{2 \times 3} = \frac{2( - 2 - \sqrt{13}) }{2 \times 3} = \frac{ - 2 - \sqrt{13} }{3} \\ x_{2} = \frac{ - 4 + 2 \sqrt{13} }{2 \times 3} = \frac{2( - 2 + \sqrt{13}) }{2 \times 3} = \frac{ - 2 + \sqrt{13} }{3} \\ \\ y_{1} = \frac{ - 2 - \sqrt{13} }{3} + 1 = \frac{ - 2 - \sqrt{13} + 3 }{3} = \frac{1 - \sqrt{13} }{3} \\ y_{2} = \frac{ - 2 + \sqrt{13} }{3} + 1 = \frac{ - 2 + \sqrt{13} + 3}{3} = \frac{1 + \sqrt{13} }{3} \\ \\ otvet \: \: \: ( \frac{ - 2 - \sqrt{13} }{3} ; \: \frac{1 - \sqrt{13} }{3} ) \: \: \: and \: \: \: \\ ( \frac{ - 2 + \sqrt{13} }{3} ; \: \frac{1 + \sqrt{13} }{3} )$

8.

20 см - гипотенуза

х см - один катет

х + 4 см - другой катет

По теореме Пифагора:

${x}^{2} + (x + 4) {}^{2} = 20 {}^{2} \\ {x}^{2} + {x}^{2} + 8x + 16 -400= 0 \\ 2 {x}^{2} + 8x - 384 = 0 \\ {x}^{2} + 4x -192= 0 \\ a = 1 \\ b = 4 \\ c = - 192 \\ D = {b}^{2} - 4ac = {4}^{2} - 4 \times 1 \times ( - 192) = \\ = 16 +7 68 = 784 \: \: \: ( \sqrt{D} = 28) \\ x_{1} = \frac{ - 4 - 28}{2 \times 1} = - \frac{32}{2} = - 16\\ x_{2} = \frac{ - 4 + 28}{2 \times 1} = \frac{24}{2} = 12$

Первый корень не подходит, т.к. сторона не может быть отрицательной

$x + 4 = 12 + 4 = 16$

Ответ: 12 см и 16 см

9.

$\displaystyle\bf y = \sqrt{6 {x}^{2} + 5x - 4 } - \frac{1}{ \sqrt{3 - 2x} } \\ \\ \left \{ {{6 {x}^{2} + 5x - 4 \geqslant 0 } \atop {3 - 2x > 0 }} \right. \\ \\ 1) \: 6 {x}^{2} + 5x - 4 \geqslant 0 \\ 6 {x}^{2} + 5x - 4 = 0 \\ a = 6\\ b =5 \\ c = - 4 \\ D = {b}^{2} - 4ac = {5}^{2} - 4 \times 6 \times ( - 4) = \\ = 25 + 96 = 121 \\ x_{1} = \frac{ - 5 - 11}{2 \times 6} = - \frac{16}{12} = - \frac{4}{3} = - 1 \frac{1}{3} \\ x_{2} = \frac{ - 5 + 11}{2 \times 6} = \frac{6}{12} = 0.5 \\ \\{ax}^{2} + bx + c = a(x - x_{1})(x - x_{2}) \\ 6 {x}^{2} + 5x - 4 = 6(x + 1 \frac{1}{3} )(x - 0.5) \\ \\ (x + 1 \frac{1}{3} )(x - 0.5) \geqslant 0 \\ + + + [ - 1 \frac{1}{3} ] - - - [0.5] + + + \\ x \leqslant - 1 \frac{1}{3} \: \: \: and \: \: \: x \geqslant 0.5 \\ \\ 2) \: 3 - 2x > 0 \\ - 2x > - 3 \: \: | \div ( - 2) \\ x < 1.5 \\ \displaystyle\bf\\3) \: \left \{ {{x \leqslant - 1 \frac{1}{3} \: \: \: and \: \: \: x \geqslant 0.5} \atop {x < 1.5 }} \right. \\ \\ otvet \: \: \: x \: \epsilon\: ( - \propto; \: - 1 \frac{1}{3} ]U[0.5; \: 1.5)$