§2. Phương trình quy về phương trình bậc nhất, bậc hai

ĐK: \(x\ge0;x\le-1\)

\(pt\Leftrightarrow x^2+x-6=\sqrt{x^2+x}-4\)

\(\Leftrightarrow x^2+x-\sqrt{x^2+x}-2=0\)

\(\Leftrightarrow\left(\sqrt{x^2+x}+1\right)\left(\sqrt{x^2+x}-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+x}=-1\left(l\right)\\\sqrt{x^2+x}=2\end{matrix}\right.\)

\(\sqrt{x^2+x}=2\)

\(\Leftrightarrow x^2+x-4=0\)

\(\Leftrightarrow x=\dfrac{-1\pm\sqrt{17}}{2}\left(tm\right)\)

ĐKXĐ: \(x\ge\dfrac{1}{2}\)

\(pt\Leftrightarrow3x+2+2\sqrt{\left(x+3\right)\left(2x-1\right)}=9\left(3x-2\right)\)

\(\Leftrightarrow2\sqrt{\left(x+3\right)\left(2x-1\right)}=24x-20\)

\(\Leftrightarrow\sqrt{2x^2+5x-3}=12x-10\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x^2+5x-3=144x^2+100-240x\\12x-10\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(142x-103\right)=0\\x\ge\dfrac{5}{6}\end{matrix}\right.\)

\(\Leftrightarrow x=1\left(tm\right)\)

Ta có: \(\left|A-B\right|=\left|A\right|-\left|B\right|\Leftrightarrow\left(A-B\right)^2=\left(\left|A\right|-\left|B\right|\right)^2\)\(\Leftrightarrow A^2-2AB+B^2=A^2-2\left|A\right|\left|B\right|+B^2\)\(\Leftrightarrow\left|AB\right|=AB\)\(\Leftrightarrow AB\ge0\)