Question

ДАЮ 80 балов
Розв’яжіть ірраціональне рівняння за методом заміни.
Решите иррациональное уравнение за методом замены.

Answer 1

$\displaystyle\bf\\\sqrt{2x^{2} -3x+7} +\sqrt{2x^{2} -3x+2} =5\\\\2x^{2} -3x+2=m \ , \ m > 0 \ \ \ \Rightarrow \ \ \ 2x^{2} -3x+7=m+5\\\\\Big(\sqrt{m+5} +\sqrt{m}\Big)^{2} =5^{2} \\\\m+5+2\sqrt{m(m+5)} +m=25\\\\2m+2\sqrt{m(m+5)} =20\\\\\Big(\sqrt{m(m+5)} \Big)^{2}=\Big(10-m\Big) ^{2} \\\\\\m(m+5)=100-20m+m^{2} \\\\m^{2} +5m=100-20m+m^{2} \\\\m^{2}+5m+20m-m^{2} =100\\\\25m=100\\\\m=4$

$\displaystyle\bf\\2x^{2} -3x+2=4\\\\2x^{2} -3x-2=0\\\\D=(-3)^{2} -4\cdot 2\cdot(-2)=9+16=25=5^{2} \\\\\\x_{1} =\frac{3+5}{4} =\frac{8}{4} =2\\\\\\x_{2} =\frac{3-5}{4} =-0,5 < 0p-neyd\\\\\\Proverka:\\\\\sqrt{2\cdot 2^{2} -3\cdot 2+7} +\sqrt{2\cdot 2^{2} -3\cdot 2+2} =5\\\\\sqrt{2\cdot 4-6+7} +\sqrt{2\cdot 4-6+2} =5\\\\\sqrt{8+1} +\sqrt{8-4}=5\\\\\sqrt{9} +\sqrt{4}=5\\\\3+2=5\\\\5=5-verno\\\\Otvet \ : \ 2$