Question

ПЖ СРОЧНО 50 БАЛЛОВ !!!!!!!!!!!!!!

Answer 1

$\displaystyle\bf\\1)\\\\6x^{2} -7x-24 < 0\\\\6\Big(x-2\frac{2}{3} \Big)\Big(x+1,5\Big) < 0\\\\\Big(x-2\frac{2}{3} \Big)\Big(x+1,5\Big) < 0\\\\\\+ + + + + \Big(-1,5\Big) - - - - - \Big(2\frac{2}{3} \Big) + + + + + \\\\\\Otvet \ : \ x\in\Big(-1,5 \ ; \ 2\frac{2}{3} \Big)\\\\\\2)\\\\\sqrt{28}\Big(\sqrt{14 } -\sqrt{7} \Big)-2\sqrt{98} =\sqrt{28\cdot 14} -\sqrt{28\cdot 7}-2\sqrt{49\cdot 2} =\\\\\\=\sqrt{2\cdot 14^{2} } -\sqrt{7^{2} \cdot 2^{2} } -2\sqrt{7^{2} \cdot 2} =$

$\displaystyle\bf\\=14\sqrt{2} -7\cdot 2-14\sqrt{2} =\boxed{-14}\\\\\\3)\\\\\left \{ {{x-4y=3} \atop {x^{2} -21y=28}} \right.\\\\\\\left \{ {{x=4y+3} \atop {(4y+3)^{2} }-21y=28} \right. \\\\\\\left \{ {{x=4y+3} \atop {16y^{2} +24y+9-21y=28}} \right. \\\\\\\left \{ {{x=4y+3} \atop {16y^{2} +3y-19=0}} \right. \\\\\\16y^{2}+3y-19=0\\\\D=3^{2} -4\cdot 16\cdot(-19)=9+1216=1225=35^{2}\\\\\\y_{1} =\frac{-3-35}{32} =-1,1875\\\\\\y_{2} =\frac{-3+35}{32} =1$

$\displaystyle\bf\\x_{1} =4\cdot (-1,1875)+3=-4,75+3=-1,75\\\\x_{2} =4\cdot 1+3=7\\\\\\Otvet \ : \ (-1,75 \ ; \ -1,1875) \ , \ (7 \ ; \ 1)$