Question

решите неравенства.
2 и 4.
ПОЖАЛУЙСТА!!!!

Answer 1

$(5x-3)^2+(12x+5)^2 \leq (7-13x)^2+34x^2+17x+410\\25x^2-30x+9+144x^2+120x+25 \leq 49-182x+169x^2+34x^2+\\+17x+410\\169x^2+90x+34 \leq 459-165x+203x^2\\-2x^2+15x-25 \leq 0\\2x^2-15x+25 \geq 0\\2x^2-15x+25=0\\D=225-200=25\\x_1=\frac{15+5}{4} =5\\x_2=2,5\\x \leq 2,5;x \geq 5$
$3x(x+2)-(4-x)(4+x) \leq 5(x^2+1)-4(1,5x-1)\\3x^2+6x-16+x^2 \leq 5x^2+9-6x\\-x^2+12x-25 \leq 0\\x^2-12x+25 \geq 0\\x^2-12x+25=0\\D=144-100=44\\x_1=\frac{12+ \sqrt{44} }{2} =6+ \sqrt{11} \\x_2=6- \sqrt{11} \\x \leq 6- \sqrt{11} ;x \geq 6+ \sqrt{11}$