~ Câu 1: \(\left(4x-1\right)\sqrt{x^2+1}=2x^2-2x+2\left(1\right)\)
\(\Leftrightarrow2\left(x^2+1\right)-\left(4x-1\right)\sqrt{x^2+1}-2x=0\)
Đặt \(x^2+1=a\left(a\ge1\right)\)
\(\left(1\right)\Rightarrow2a^2-\left(4x-1\right)a-2x=0\left(1'\right)\)
\(\Delta=\left[-\left(4x-1\right)\right]^2-4\times2\times\left(-2x\right)\)
\(=\left(16x^2-8x+1\right)+16x\)
\(=\left(4x+1\right)^2>0\)
\(\Rightarrow\left(1'\right)\) có 2 no phân biệt:
\(a_1=\dfrac{-\left[-\left(4x-1\right)\right]+\sqrt{\left(4x+1\right)^2}}{2\times2}=2x\)
\(a_2=\dfrac{-\left[-\left(4x-1\right)\right]-\sqrt{\left(4x+1\right)^2}}{2\times2}=-\dfrac{1}{2}\left(l\right)\)
\(\Rightarrow\sqrt{x^2+1}=2x\)
\(\Leftrightarrow x^2+1=4x^2\)
\(\Leftrightarrow3x^2-1=0\) (vô lý)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{1}{3}}\left(n\right)\\x=-\sqrt{\dfrac{1}{3}}\left(l\right)\end{matrix}\right.\)
Vậy (1) có tập nghiệm \(S=\left\{\sqrt{\dfrac{1}{3}}\right\}\)
._._._._._.
Đkxđ: \(x>\dfrac{1}{4}\)
Giải thích: \(VT=2x^2-2x+2\ge\dfrac{3}{2}>0\forall x\)
mà \(\sqrt{x^2+1}>0\forall x\)
\(\Rightarrow4x-1>0\)
\(\Leftrightarrow x>\dfrac{1}{4}\)
