[latex]1. \sqrt{3x-2}* \sqrt{x+2}=5-x [/latex] [latex]2. \sqrt{4+2x \sqrt{4-x^2} } =6-2x[/latex] Распишите ,пожалуйста , подробней.

[latex]1)[/latex] [latex] \sqrt{3x-2} * \sqrt{x+2}=5-x [/latex] ОДЗ: [latex] \left \{ {{3x-2 \geq 0} \atop {x+2 \geq 0}}\atop {5-x \geq 0} \right. [/latex] [latex] \left \{ {{x \geq \frac{2}{3} } \atop {x \geq -2}}\atop {x \leq 5} \right. [/latex] [latex]x[/latex] ∈ [latex][ \frac{2}{3};5] [/latex] [latex](\sqrt{3x-2} * \sqrt{x+2})^2=(5-x)^2 [/latex] [latex]({3x-2} )* ({x+2})=25+ x^{2} -10x[/latex] [latex]3 x^{2} +6x-2x-4=25+ x^{2} -10x[/latex] [latex]3 x^{2} +4x-4-25-x^{2}+10x=0[/latex] [latex]2 x^{2} +14x-29=0[/latex] [latex]D_1=( \frac{b}{2})^2-ac= 7^2-2*(-29)=49+58=107[/latex] [latex]x_1= \frac{- \frac{b}{2}+ \sqrt{D_1} }{a}= \frac{-7+ \sqrt{107} }{2} [/latex] [latex]x_2= \frac{- \frac{b}{2}- \sqrt{D_1} }{a}= \frac{-7- \sqrt{107} }{2}\ \textless \ 0 [/latex]  - не подходит  Ответ:   [latex]\frac{-7+ \sqrt{107} }{2} [/latex] [latex]2)[/latex] [latex] \sqrt{4+2x \sqrt{4- x^{2} } } =6-2x[/latex] ОДЗ:  [latex] \left \{ {{4- x^{2} \geq 0} \atop {6-2x \geq 0}} \right. [/latex] [latex] \left \{ {{(2-x)(2+x) \geq 0} \atop {-2x \geq -6}} \right. [/latex] [latex] \left \{ {{(2-x)(2+x) \geq 0} \atop {x \leq 3}} \right. [/latex]           -                     +                   -  --------------[-2]---------------[2]------------------                      ////////////////// --------------------------------------------[3]-------- ///////////////////////////////////////////// [latex]x[/latex] ∈ [latex][-2;2][/latex] [latex] \sqrt{(x+ \sqrt{4- x^{2} } )^2} =6-2x[/latex] [latex]|x+ \sqrt{4- x^{2} } |=6-2x[/latex] [latex]x+ \sqrt{4- x^{2} }=6-2x[/latex]  или  [latex]x+ \sqrt{4- x^{2} }=2x-6[/latex]    [latex]\sqrt{4- x^{2} }=6-3x[/latex]         или   [latex]\sqrt{4- x^{2} }=x-6[/latex] [latex] \left \{ {{4- x^{2} =(6-3x)^2} \atop {6-3x \geq 0}} \right. [/latex]                 или   [latex] \left \{ {{4- x^{2} =(x-6)^2} \atop {x-6 \geq 0}} \right. [/latex] [latex] \left \{ {{4- x^{2} =36+9x^2-36x} \atop {x \leq 2}} \right. [/latex]         или   [latex] \left \{ {{4- x^{2} = x^{2} -12x+36} \atop {x \geq 6}} \right. [/latex] [latex] \left \{ {10x^2-36x+32=0} \atop {x \leq 2}} \right. [/latex]               или   [latex] \left \{ {2x^2-12x+32=0} \atop {x \geq 6}} \right. [/latex] [latex] \left \{ {5x^2-18x+16=0} \atop {x \leq 2}} \right. [/latex]             или   [latex] \left \{ {x^2-6x+16=0} \atop {x \geq 6}} \right. [/latex] [latex]D=(-18)^2-4*5*16=4[/latex]  или   [latex]D=(-6)^2-4*1*16=-28\ \textless \ 0[/latex] [latex]x_1= \frac{18+2}{10} =2[/latex]      или                               ∅ [latex]x_2= \frac{18-2}{10}=1.6 [/latex] Ответ: [latex]1.6;[/latex]   [latex]2[/latex]