Question

Solve irrational equation pls

Answer 1

$\hbox{Domain:}\\ x^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\\ x^2-x+2x-2\geq0 \wedge x^2-x-3x+3\geq0 \wedge x^2\geq1\\ x(x-1)+2(x-1)\geq 0 \wedge x(x-1)-3(x-1)\geq0 \wedge (x\geq 1 \vee x\leq-1)\\ (x+2)(x-1)\geq0 \wedge (x-3)(x-1)\geq0\wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\ x\in(-\infty,-2\rangle\cup\langle1,\infty) \wedge x\in(-\infty,1\rangle \cup\langle3,\infty) \wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\ x\in(-\infty,-2\rangle\cup\langle3,\infty)$


$\sqrt{x^2+x-2}+\sqrt{x^2-4x+3}=\sqrt{x^2-1}\\ x^2-1=x^2+x-2+2\sqrt{(x^2+x-2)(x^2-4x+3)}+x^2-4x+3\\ 2\sqrt{(x^2+x-2)(x^2-4x+3)}=-x^2+3x-2\\ \sqrt{(x^2+x-2)(x^2-4x+3)}=\dfrac{-x^2+3x-2}{2}\\ (x^2+x-2)(x^2-4x+3)=\left(\dfrac{-x^2+3x-2}{2}\right)^2\\ (x+2)(x-1)(x-3)(x-1)=\left(\dfrac{-x^2+x+2x-2}{2}\right)^2\\ (x+2)(x-3)(x-1)^2=\left(\dfrac{-x(x-1)+2(x-1)}{2}\right)^2\\ (x+2)(x-3)(x-1)^2=\left(\dfrac{-(x-2)(x-1)}{2}\right)^2\\ (x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\ 4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2$
$4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\\ (x-1)^2(4(x+2)(x-3)-(x-2)^2)=0\\ (x-1)^2(4(x^2-3x+2x-6)-(x^2-4x+4))=0\\ (x-1)^2(4x^2-4x-24-x^2+4x-4)=0\\ (x-1)^2(3x^2-28)=0\\ x-1=0 \vee 3x^2-28=0\\ x=1 \vee 3x^2=28\\ x=1 \vee x^2=\dfrac{28}{3}\\ x=1 \vee x=\sqrt{\dfrac{28}{3}} \vee x=-\sqrt{\dfrac{28}{3}}$

There's one more condition I forgot about
$-(x-2)(x-1)\geq0\\ x\in\langle1,2\rangle$

Finally
$x\in(-\infty,-2\rangle\cup\langle3,\infty) \wedge x\in\langle1,2\rangle \wedge x=\{1,\sqrt{\dfrac{28}{3}}, -\sqrt{\dfrac{28}{3}}\}\\ \boxed{\boxed{x=1}}$