Помогите решить под номером 3 и 5.

Замена [latex] \sqrt{ \frac{3x+2}{2x-3} } =y[/latex] y + 1/y = 2,5 = 2 + 1/2 [latex]y1=\sqrt{ \frac{3x+2}{2x-3} } =2[/latex] [latex]\frac{3x+2}{2x-3}=4 [/latex] [latex] \frac{3x+2-4(2x-3)}{2x-3} = \frac{3x+2-8x+12}{2x-3} = \frac{-5x+14}{2x-3} =0[/latex] x1=14/5 = 2,8 [latex]y2=\sqrt{ \frac{3x+2}{2x-3} } =\ \frac{1}{2} [/latex] [latex]\frac{3x+2}{2x-3}= \frac{1}{4} [/latex] [latex] \frac{4(3x+2)-(2x-3)}{4(2x-3)} = \frac{12x+8-2x+3}{4(2x-3)} = \frac{10x+11}{4(2x-3)} =0[/latex] x2 = -11/10 = -1,1 Ответ: x1 = 2,8; x2 = -1,1 5) [latex] \left \{ {{(x-y)^2-x+y=0} \atop {x^2y^2-xy-2=0}} \right. [/latex] [latex] \left \{ {{(x-y)^2-(x-y)=0} \atop {(xy)^2 - xy - 2=0}} \right. [/latex] Замена x - y = a; xy = b [latex] \left \{ {{a^2-a=a(a - 1)=0} \atop {b^2-b-2=0}} \right. [/latex] [latex] \left \{ {{a1=x-y=0;a2=x-y=1} \atop {b1=xy=-1;b2=xy=2}} \right. [/latex] Получаем 4 системы: а) [latex] \left \{ {{x-y=0} \atop {xy=-1}} \right. [/latex] Решений нет, потому что из 1 уравнения x = y, а из 2 х и у имеют разные знаки б) [latex] \left \{ {{x-y=0} \atop {xy=2}} \right. [/latex] x = y = √2 в) [latex] \left \{ {{x-y=1} \atop {xy=-1}} \right. [/latex] [latex] \left \{ {{x=y+1} \atop {(y+1)*y=-1}} \right. [/latex] y^2 + y + 1 = 0 Решений нет г) [latex] \left \{ {{x-y=1} \atop {xy=2}} \right. [/latex] x = 2; y = 1 x = -1; y = -2 Ответ: x1 = y1 = √2; x2 = 2; y2 = 1; x3 = -1; y3 = -2