Question

ПОМОГИТЕ ПОЖАЛУЙСТА СРОЧНО!!!!!!!!!!!!!​

Answer 1

$1- \dfrac{ x }{ x+1 } = \dfrac{ x }{ x+2 } |*(x+1)(x+2) \\x\neq -1\\x\neq -2\\(x+1)(x+2)-(x+2)x=(x+1)x\\x^2+3x+2-x^2-2x=(x+1)x\\x+2=(x+1)x\\x+2=x^ 2+x\\x+2-x^ 2-x=0\\-x^2=-2\\x=\sqrt{2} \\x=-\sqrt{2}$

${ x }^{ 2 } + \dfrac{ 1-3x }{ x-4 } = 16- \dfrac{ 3x-1 }{ x-4 } |*x-4 \\x\neq 4\\(x-4)x ^2+1-3x=(x-4)16-(3x-1)\\x^3-4x^2+1-3x=16x-64-3x+1\\x^3-4x^2-16x+64=0 |/x-4\\x^{2} -16=0\\x^{2} =16\\x=-4\\x=4\\otvet :-4\\x\neq 4$

$\dfrac{ 36 }{ x ^ { 2 } -12x } - \dfrac{ 3 }{ x-12 } =3 |*x(x-12)\\\\x\neq 0\\x\neq 12\\36-x*3=3x^2-36x\\36+33x-3x ^2 =0 |/3\\12+11x-x^{2} =0\\\sqrt{D} =\sqrt{11^2-4*(-1)(12)} =13\\x_{1=} \dfrac{-11-13}{-2} =12\\x_{2=} \dfrac{-11+13}{-2} =-1\\\\otvet :x=-1\\x\neq 12$

$\dfrac{ 12 }{ x ^ { 2 } +2x } - \dfrac{ 3 }{ x ^ { 2 } +2x-2 } =1 \\x^2+2x=t\\ \dfrac{ 12 }{ t } - \dfrac{ 3 }{ t-2 } =1 |* t(t-2) \\12t-24-t*3=t ^2 -2t\\11t-24-t ^2 =0\\\sqrt{D} =\sqrt{11^2-4*-1*-24} =5\\t_{1} =\dfrac{-11+5}{-2} =3\\t_{2} =\dfrac{-11-5}{-2} =8\\x^2+2x=8\\x ^2 +2x-8=0\\\sqrt{D} =\sqrt{4-4*1*-8} =6\\x_{1} =\dfrac{-2+6}{2} =2\\x_{2} =\dfrac{-2-6}{2} =-4\\x^2+2x=3\\x^2+2x-3=0\\\sqrt{D} =\sqrt{4-4*1*-3} =4\\x_{1} =\dfrac{-2+4}{2} =1$

$x_{2} =\dfrac{-2-4}{2}=-3\\otvet:x=-3\\x=1\\x=2\\x=-4$