Question

[Joy]

Giải phương trình:

\(4x^2-11x+6=\left(x-1\right).\sqrt{2x^2-6x+6}\)

Answer 1

[Đặng Ngọc Quỳnh]

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

Đặt \(a=2x-3;b=\sqrt{\left(x-1\right)\left(2x-3\right)-\left(x-3\right)}\)

Ta có hpt \(\hept{\begin{cases}a^2+x-3=\left(x-1\right)b\\b^2+x-3=\left(x-1\right)a\end{cases}}\)

Trừ 2 pt trên ta được: \(a^2-b^2=\left(x-1\right)\left(b-a\right)\Rightarrow\left(a-b\right)\left(a+b+x-1\right)=0\)

+) Nếu \(a=b\Leftrightarrow2x-3=\sqrt{2x^2-6x+6}\Leftrightarrow\hept{\begin{cases}x\ge\frac{3}{2}\\2x^2-6x+3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{3-\sqrt{3}}{2}\left(loại\right)\\x=\frac{3+\sqrt{3}}{2}\left(tm\right)\end{cases}}}\)

+) Nếu \(2x-3+\sqrt{2x^2-6x+6}+x-3=0\Leftrightarrow\sqrt{2x^2-6x}=6-3x\)\(\Leftrightarrow\hept{\begin{cases}x\le2\\7x^2-30x+36=0\end{cases}\left(VN\right)}\)

Vậy pt có nghiệm duy nhất: \(x=\frac{3+\sqrt{3}}{2}\)

Answer 2

[Victorique de Blois]

\(4x^2-11x+6=\left(x-1\right)\sqrt{2x^2-6x+6}\)

\(\Leftrightarrow\left(4x^2-12x+9\right)+x-3=\left(x-1\right)\sqrt{2x^2-5x+3-x+3}\)

\(\Leftrightarrow\left(2x-3\right)^2+x+3=\left(x-1\right)\sqrt{\left(x-1\right)\left(2x-3\right)-\left(x-3\right)}\)

đặt \(\hept{\begin{cases}t=2x-3\\y=\sqrt{\left(x-1\right)\left(2x-3\right)-\left(x-3\right)}\left(y\ge0\right)\end{cases}}\)

ta có hệ : \(\hept{\begin{cases}t^2+x-3=\left(x-1\right)y\\y^2-\left(x-1\right)t+x-3=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}t^2-\left(x-1\right)y+\left(x-3\right)=0\\y^2-\left(x-1\right)t+\left(x-3\right)=0\end{cases}}\)

\(\Rightarrow t^2-y^2-\left(x-1\right)y+\left(x-1\right)t=0\)

\(\Leftrightarrow\left(t-y\right)\left(t+y\right)+\left(x-1\right)\left(t-y\right)=0\)

\(\Leftrightarrow\left(t-y\right)\left(t+y+x-1\right)=0\)

th1 : \(t-y=0\Leftrightarrow t=y\Leftrightarrow2x-3=\sqrt{\left(x-1\right)\left(2x-3\right)-\left(x-3\right)}\) ĐK : \(x\ge\frac{3}{2}\)

\(\Leftrightarrow4x^2-12x+9=2x^2-6x+6\)

\(\Leftrightarrow2x^2-6x+3=0\)

\(\Delta=b^2-4ac=\left(-6\right)^2-4\cdot2\cdot3=12\)

\(\Rightarrow\orbr{\begin{cases}x=3+\sqrt{3}\left(tm\right)\\x=3-\sqrt{3}\left(loai\right)\end{cases}}\)

th2 : \(x+y+t-1=0\Leftrightarrow y=1-x-t\)

\(\Leftrightarrow\sqrt{\left(x-1\right)\left(2x-3\right)-\left(x-3\right)}=1-x-2x+3\)

\(\Leftrightarrow\sqrt{2x^2-6x+6}=4-3x\left(đk:x\le\frac{4}{3}\right)\)

\(\Leftrightarrow2x^2-6x+6=16-24x+9x^2\)

\(\Leftrightarrow7x^2-18x+10=0\)

\(\Delta=b^2-4ac=\left(-18\right)^2-4\cdot7\cdot10=44\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{18+\sqrt{44}}{2}=9+\sqrt{11}\left(loai\right)\\x=\frac{18-\sqrt{44}}{2}=9-\sqrt{11}\left(loai\right)\end{cases}}\)