Помогите решить уравнение:√2x+1=5-2x

√ 2Х + 1 = 5 - 2Х  ( √ 2Х + 1)^2 = ( 5 - 2X)^2  2X + 1 > 0 ; 2X > - 1 ; X > - 0,5  2X + 1 = 25 - 20X + 4X^2  4X^2 - 20X - 2X + 25 - 1 = 0  4X^2 - 22X + 24 = 0  2 * ( 2X^2 - 11X + 12 ) = 0 D = 121 - 96 = 25 ; √ D = 5  X1 = ( 11 + 5 ) : 4 = 4 ( > - 0,5 ) X2 = ( 11 - 5 ) : 4 = 1,5 ( > - 0,5)  Ответ 4 и 1,5 

[latex] \sqrt{2x+1} =5-2x \\ 2x+1 \geq 0 \\ 2x \geq -1 \\ x \geq - \frac{1}{2} \\ x \geq -0.5[/latex] x∈ [latex][-0.5;+ \infty )[/latex] [latex] \sqrt{2x+1} =5-2x \\ ( \sqrt{2x+1} ) ^{2} =(5-2x ) ^{2} \\ 2x+1=25+4 x^{2} -20x \\ 2x+1-25-4 x^{2} +20x=0 \\ -4 x^{2} +22x-24=0/*(-1)\\4 x^{2} -22x+24=0 /:2\\ 2 x^{2} -11x+12=0 \\ D=121-4*2*12=121-96=25 \\ \sqrt{D} =5 \\ x_{1} = \frac{11+5}{4} = \frac{16}{4} =4 \\ x_{2} = \frac{11-5}{4} = \frac{6}{4} =1.5[/latex]