\(\sqrt{x-1}+x-1=2\sqrt{2\left(x-3\right)^2+2x-2}\)
\(\Leftrightarrow\sqrt{x-1}+x-1=\sqrt{8x^2-40x+64}\)
\(\Leftrightarrow\sqrt{x-1}+x-1=\sqrt{8\left(x-1\right)^2-24\left(x-1\right)+32}\)
Đặt \(\sqrt{x-1}=a\ge0\) thì ta có:
\(a^2+a=\sqrt{8a^4-40a^2+32}\)
\(\Leftrightarrow7a^4-2a^3-25a^2+32=0\)
\(\Leftrightarrow\left(a^4-2a^3+a^2\right)+\left(6a^4-26a^2+\dfrac{169}{6}\right)+\dfrac{23}{6}=0\)
\(\Leftrightarrow\left(a^2-a\right)^2+\left(\sqrt{6}a^2-\dfrac{13\sqrt{6}}{6}\right)^2+\dfrac{23}{6}>0\)
Vậy PT vô nghiệm
\(\sqrt{x-1}+x-1=2\sqrt{2\left(x-3\right)^2+2x-2}\)
\(\sqrt{x-1}+x-1=\sqrt{8}.\sqrt{\left(x-1-2\right)^2+x-1}\)
\(\sqrt{x-1}+x-1=\sqrt{8}.\sqrt{\left(x-1\right)^2-4\left(x-1\right)+4+x-1}\)
\(\sqrt{x-1}+x-1=\sqrt{8}.\sqrt{\left(x-1\right)^2-3\left(x-1\right)+4}\)
Điều kiện: \(\left(x-1\right)\ge0\Rightarrow x\ge1\)
Có \(\left(x-1\right)^2-3\left(x-1\right)+4>0\) mọi x
Điều kiện: \(x\ge1\)
Đặt: \(\sqrt{x-1}=a\) cho gọn \(a\ge0\)
\(\Leftrightarrow a^2+a=\sqrt{8}.\sqrt{a^4-3a^2+4}\)
BP: \(\Leftrightarrow a^4+2a^3+a^2=8a^4-24a^2+32\)
\(\Leftrightarrow a^4-2a^3+a^2+6a^4-26a^2+32=0\)
\(\Leftrightarrow\left(a^2-a\right)^2+6a^4-26a^2+32=0\)
\(\Leftrightarrow\left(a^2-a\right)^2+6\left(a^2-\dfrac{13}{6}\right)^2+\dfrac{23}{6}>0\)
\(\Rightarrow\) PT vô nghiệm
KL: PT vô nghiệm
