Question

Решите хотя бы 2 плиз

Answer 1

$1)\;\frac{2x^2-3x}3-\frac{3x^2-12}5=0\\\\\frac{5\cdot(2x^2-3x)-3\cdot(3x^2-12)}{15}=0\\\\10x^2-15x-9x^2+36=0\\x^2-15x+36=0\\D=225-4\cdot1\cdot36=225-144=81\\x_{1,2}=\frac{15\pm9}2\\x_1=3,\;x_2=12$

$2)\;\frac{4x^2-2x}5-\frac{3x^2-3}4=0\\\\\frac{4\cdot(4x^2-2x)-5\cdot(3x^2-3)}{20}=0\\\\16x^2-8x-15x^2+15=0\\x^2-8x+15=0\\D=64-4\cdot15=64-60=4\\x_{1,2}=\frac{8\pm2}2\\x_1=3,\;x_2=5$

$3)\;(x-2)^2+(x-5)^2=5\\x^2-4x+4+x^2-10x+25=5\\2x^2-14x+24=0\;\;\div2\\x^2-7x+12=0\\D=49-4\cdot1\cdot12=49-48=1\\x_{1,2}=\frac{7\pm1}2\\x_1=3,\;x_2=4$

$4)\;(x-3)^2+(x-4)^2=5\\x^2-6x+9+x^2-8x+16=5\\2x^2-14x+20=0\;\;\;\div2\\x^2-7x+10=0\\D=49-4\cdot1\cdot10=49-40=9\\x_{1,2}=\frac{7\pm3}2\\x_1=2,\;x_2=5$