a) \(\sqrt{\dfrac{x-2}{x-1}+\dfrac{x+2}{\left(\sqrt{x+2}+\sqrt{x-2}\right)^2}}=1\)
\(\Leftrightarrow\dfrac{x-2}{x-1}+\dfrac{x+2}{\left(\sqrt{x+2}+\sqrt{x-2}\right)^2}=1\)
\(\Leftrightarrow\dfrac{\left(x-2\right)\left(\sqrt{x+2}+\sqrt{x-2}\right)^2+\left(x+2\right)\left(x-1\right)}{\left(x-1\right)\left(\sqrt{x+2}+\sqrt{x-2}\right)^2}=1\)
\(\Leftrightarrow\left(x-2\right)\left(2x+2\sqrt{x^2-4}\right)+\left(x+2\right)\left(x-1\right)=\left(x-1\right)\left(2x+2\sqrt{x^2-4}\right)\)
\(\Leftrightarrow2\sqrt{x^2-4}=x^2-x-2\)
\(\Leftrightarrow2\sqrt{\left(x+2\right)\left(x-2\right)}=\left(x-2\right)\left(x+1\right)\)
Đến đây tự giải nha
b) \(\sqrt{\left(x^2+2x\right)^2+4\left(x+1\right)^2}-\sqrt{x^2+\left(x+1\right)^2+\left(x^2+x\right)^2}=2017\)
\(\Leftrightarrow\sqrt{x^4+4x^3+8x^2+8x+4}+\sqrt{x^4+2x^3+3x^2+2x+1}=2017\)
\(\Leftrightarrow\sqrt{\left(x^2+2x+2\right)^2}+\sqrt{\left(x^2+x+1\right)^2}=2017\)
Vì \(x^2+2x+2=\left(x+1\right)^2+1>0\forall x\)
\(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\)
pt\(\Leftrightarrow x^2+2x+2+x^2+x+1=2017\)
\(\Leftrightarrow2x^2+3x-2014=0\)
Đến đây tự giải nha
