Question

Giải phương trình \(x^2-1=2x\sqrt{x^2-2x}\)

Answer 1

​ĐK: \left\{{}\begin{matrix}\left(x+2\right)\left(x+3\right)\ge0\\x^2+5x-36\ge0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge4\\x\le-9\end{matrix}\right.{(x+2)(x+3)≥0x2+5x−36≥0​⇔[x≥4x≤−9​

pt\Leftrightarrow\sqrt{x^2+5x+6}=x^2+5x-36pt⇔x2+5x+6​=x2+5x−36

Đặt \sqrt{x^2+5x+6}=t\left(t\ge0\right)x2+5x+6​=t(t≥0) , phương trình trở thành:

t=t^2-42\Leftrightarrow t^2-t-42=0\Leftrightarrow\left(t+6\right)\left(t-7\right)=0t=t2−42⇔t2−t−42=0⇔(t+6)(t−7)=0

\Leftrightarrow\left[{}\begin{matrix}t=-6\left(ktmđk\right)\\t=7\end{matrix}\right.⇔[t=−6(ktmđk)t=7​

Với t=7\Rightarrow\sqrt{x^2+5x+6}=7\Rightarrow x^2+5x+6=49t=7⇒x2+5x+6​=7⇒x2+5x+6=49

\Rightarrow x^2+5x-43=0⇒x2+5x−43=0

\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5+\sqrt{197}}{2}\\x=\dfrac{-5-\sqrt{197}}{2}\end{matrix}\right.\left(tmđk\right)⇔⎣⎢⎢⎡​x=2−5+197​​x=2−5−197​​​(tmđk)

Vậy phương trình có tập nghiệm S=\left\{\dfrac{-5+\sqrt{197}}{2};\dfrac{-5-\sqrt{197}}{2}\right\}S={2−5+197​​;2−5−197​​}

​ĐK: \left\{{}\begin{matrix}\left(x+2\right)\left(x+3\right)\ge0\\x^2+5x-36\ge0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge4\\x\le-9\end{matrix}\right.{(x+2)(x+3)≥0x2+5x−36≥0​⇔[x≥4x≤−9​

pt\Leftrightarrow\sqrt{x^2+5x+6}=x^2+5x-36pt⇔x2+5x+6​=x2+5x−36

Đặt \sqrt{x^2+5x+6}=t\left(t\ge0\right)x2+5x+6​=t(t≥0) , phương trình trở thành:

t=t^2-42\Leftrightarrow t^2-t-42=0\Leftrightarrow\left(t+6\right)\left(t-7\right)=0t=t2−42⇔t2−t−42=0⇔(t+6)(t−7)=0

\Leftrightarrow\left[{}\begin{matrix}t=-6\left(ktmđk\right)\\t=7\end{matrix}\right.⇔[t=−6(ktmđk)t=7​

Với t=7\Rightarrow\sqrt{x^2+5x+6}=7\Rightarrow x^2+5x+6=49t=7⇒x2+5x+6​=7⇒x2+5x+6=49

\Rightarrow x^2+5x-43=0⇒x2+5x−43=0

\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5+\sqrt{197}}{2}\\x=\dfrac{-5-\sqrt{197}}{2}\end{matrix}\right.\left(tmđk\right)⇔⎣⎢⎢⎡​x=2−5+197​​x=2−5−197​​​(tmđk)

Vậy phương trình có tập nghiệm S=\left\{\dfrac{-5+\sqrt{197}}{2};\dfrac{-5-\sqrt{197}}{2}\right\}S={2−5+197​​;2−5−197​​}

​ĐK: \left\{{}\begin{matrix}\left(x+2\right)\left(x+3\right)\ge0\\x^2+5x-36\ge0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge4\\x\le-9\end{matrix}\right.{(x+2)(x+3)≥0x2+5x−36≥0​⇔[x≥4x≤−9​

pt\Leftrightarrow\sqrt{x^2+5x+6}=x^2+5x-36pt⇔x2+5x+6​=x2+5x−36

Đặt \sqrt{x^2+5x+6}=t\left(t\ge0\right)x2+5x+6​=t(t≥0) , phương trình trở thành:

t=t^2-42\Leftrightarrow t^2-t-42=0\Leftrightarrow\left(t+6\right)\left(t-7\right)=0t=t2−42⇔t2−t−42=0⇔(t+6)(t−7)=0

\Leftrightarrow\left[{}\begin{matrix}t=-6\left(ktmđk\right)\\t=7\end{matrix}\right.⇔[t=−6(ktmđk)t=7​

Với t=7\Rightarrow\sqrt{x^2+5x+6}=7\Rightarrow x^2+5x+6=49t=7⇒x2+5x+6​=7⇒x2+5x+6=49

\Rightarrow x^2+5x-43=0⇒x2+5x−43=0

\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5+\sqrt{197}}{2}\\x=\dfrac{-5-\sqrt{197}}{2}\end{matrix}\right.\left(tmđk\right)⇔⎣⎢⎢⎡​x=2−5+197​​x=2−5−197​​​(tmđk)

Vậy phương trình có tập nghiệm S=\left\{\dfrac{-5+\sqrt{197}}{2};\dfrac{-5-\sqrt{197}}{2}\right\}S={2−5+197​​;2−5−197​​}

​ĐK: \left\{{}\begin{matrix}\left(x+2\right)\left(x+3\right)\ge0\\x^2+5x-36\ge0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge4\\x\le-9\end{matrix}\right.{(x+2)(x+3)≥0x2+5x−36≥0​⇔[x≥4x≤−9​

pt\Leftrightarrow\sqrt{x^2+5x+6}=x^2+5x-36pt⇔x2+5x+6​=x2+5x−36

Đặt \sqrt{x^2+5x+6}=t\left(t\ge0\right)x2+5x+6​=t(t≥0) , phương trình trở thành:

t=t^2-42\Leftrightarrow t^2-t-42=0\Leftrightarrow\left(t+6\right)\left(t-7\right)=0t=t2−42⇔t2−t−42=0⇔(t+6)(t−7)=0

\Leftrightarrow\left[{}\begin{matrix}t=-6\left(ktmđk\right)\\t=7\end{matrix}\right.⇔[t=−6(ktmđk)t=7​

Với t=7\Rightarrow\sqrt{x^2+5x+6}=7\Rightarrow x^2+5x+6=49t=7⇒x2+5x+6​=7⇒x2+5x+6=49

\Rightarrow x^2+5x-43=0⇒x2+5x−43=0

\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5+\sqrt{197}}{2}\\x=\dfrac{-5-\sqrt{197}}{2}\end{matrix}\right.\left(tmđk\right)⇔⎣⎢⎢⎡​x=2−5+197​​x=2−5−197​​​(tmđk)

Vậy phương trình có tập nghiệm S=\left\{\dfrac{-5+\sqrt{197}}{2};\dfrac{-5-\sqrt{197}}{2}\right\}S={2−5+197​​;2−5−197​​}

​ĐK: \left\{{}\begin{matrix}\left(x+2\right)\left(x+3\right)\ge0\\x^2+5x-36\ge0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge4\\x\le-9\end{matrix}\right.{(x+2)(x+3)≥0x2+5x−36≥0​⇔[x≥4x≤−9​

pt\Leftrightarrow\sqrt{x^2+5x+6}=x^2+5x-36pt⇔x2+5x+6​=x2+5x−36

Đặt \sqrt{x^2+5x+6}=t\left(t\ge0\right)x2+5x+6​=t(t≥0) , phương trình trở thành:

t=t^2-42\Leftrightarrow t^2-t-42=0\Leftrightarrow\left(t+6\right)\left(t-7\right)=0t=t2−42⇔t2−t−42=0⇔(t+6)(t−7)=0

\Leftrightarrow\left[{}\begin{matrix}t=-6\left(ktmđk\right)\\t=7\end{matrix}\right.⇔[t=−6(ktmđk)t=7​

Với t=7\Rightarrow\sqrt{x^2+5x+6}=7\Rightarrow x^2+5x+6=49t=7⇒x2+5x+6​=7⇒x2+5x+6=49

\Rightarrow x^2+5x-43=0⇒x2+5x−43=0

\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5+\sqrt{197}}{2}\\x=\dfrac{-5-\sqrt{197}}{2}\end{matrix}\right.\left(tmđk\right)⇔⎣⎢⎢⎡​x=2−5+197​​x=2−5−197​​​(tmđk)

Vậy phương trình có tập nghiệm S=\left\{\dfrac{-5+\sqrt{197}}{2};\dfrac{-5-\sqrt{197}}{2}\right\}S={2−5+197​​;2−5−197​​}

​ĐK: \left\{{}\begin{matrix}\left(x+2\right)\left(x+3\right)\ge0\\x^2+5x-36\ge0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge4\\x\le-9\end{matrix}\right.{(x+2)(x+3)≥0x2+5x−36≥0​⇔[x≥4x≤−9​

pt\Leftrightarrow\sqrt{x^2+5x+6}=x^2+5x-36pt⇔x2+5x+6​=x2+5x−36

Đặt \sqrt{x^2+5x+6}=t\left(t\ge0\right)x2+5x+6​=t(t≥0) , phương trình trở thành:

t=t^2-42\Leftrightarrow t^2-t-42=0\Leftrightarrow\left(t+6\right)\left(t-7\right)=0t=t2−42⇔t2−t−42=0⇔(t+6)(t−7)=0

\Leftrightarrow\left[{}\begin{matrix}t=-6\left(ktmđk\right)\\t=7\end{matrix}\right.⇔[t=−6(ktmđk)t=7​

Với t=7\Rightarrow\sqrt{x^2+5x+6}=7\Rightarrow x^2+5x+6=49t=7⇒x2+5x+6​=7⇒x2+5x+6=49

\Rightarrow x^2+5x-43=0⇒x2+5x−43=0

\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5+\sqrt{197}}{2}\\x=\dfrac{-5-\sqrt{197}}{2}\end{matrix}\right.\left(tmđk\right)⇔⎣⎢⎢⎡​x=2−5+197​​x=2−5−197​​​(tmđk)

Vậy phương trình có tập nghiệm S=\left\{\dfrac{-5+\sqrt{197}}{2};\dfrac{-5-\sqrt{197}}{2}\right\}S={2−5+197​​;2−5−197​​}

​ĐK: \left\{{}\begin{matrix}\left(x+2\right)\left(x+3\right)\ge0\\x^2+5x-36\ge0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge4\\x\le-9\end{matrix}\right.{(x+2)(x+3)≥0x2+5x−36≥0​⇔[x≥4x≤−9​

pt\Leftrightarrow\sqrt{x^2+5x+6}=x^2+5x-36pt⇔x2+5x+6​=x2+5x−36

Đặt \sqrt{x^2+5x+6}=t\left(t\ge0\right)x2+5x+6​=t(t≥0) , phương trình trở thành:

t=t^2-42\Leftrightarrow t^2-t-42=0\Leftrightarrow\left(t+6\right)\left(t-7\right)=0t=t2−42⇔t2−t−42=0⇔(t+6)(t−7)=0

\Leftrightarrow\left[{}\begin{matrix}t=-6\left(ktmđk\right)\\t=7\end{matrix}\right.⇔[t=−6(ktmđk)t=7​

Với t=7\Rightarrow\sqrt{x^2+5x+6}=7\Rightarrow x^2+5x+6=49t=7⇒x2+5x+6​=7⇒x2+5x+6=49

\Rightarrow x^2+5x-43=0⇒x2+5x−43=0

\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5+\sqrt{197}}{2}\\x=\dfrac{-5-\sqrt{197}}{2}\end{matrix}\right.\left(tmđk\right)⇔⎣⎢⎢⎡​x=2−5+197​​x=2−5−197​​​(tmđk)

Vậy phương trình có tập nghiệm S=\left\{\dfrac{-5+\sqrt{197}}{2};\dfrac{-5-\sqrt{197}}{2}\right\}S={2−5+197​​;2−5−197​​}

​ĐK: \left\{{}\begin{matrix}\left(x+2\right)\left(x+3\right)\ge0\\x^2+5x-36\ge0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge4\\x\le-9\end{matrix}\right.{(x+2)(x+3)≥0x2+5x−36≥0​⇔[x≥4x≤−9​

pt\Leftrightarrow\sqrt{x^2+5x+6}=x^2+5x-36pt⇔x2+5x+6​=x2+5x−36

Đặt \sqrt{x^2+5x+6}=t\left(t\ge0\right)x2+5x+6​=t(t≥0) , phương trình trở thành:

t=t^2-42\Leftrightarrow t^2-t-42=0\Leftrightarrow\left(t+6\right)\left(t-7\right)=0t=t2−42⇔t2−t−42=0⇔(t+6)(t−7)=0

\Leftrightarrow\left[{}\begin{matrix}t=-6\left(ktmđk\right)\\t=7\end{matrix}\right.⇔[t=−6(ktmđk)t=7​

Với t=7\Rightarrow\sqrt{x^2+5x+6}=7\Rightarrow x^2+5x+6=49t=7⇒x2+5x+6​=7⇒x2+5x+6=49

\Rightarrow x^2+5x-43=0⇒x2+5x−43=0

\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5+\sqrt{197}}{2}\\x=\dfrac{-5-\sqrt{197}}{2}\end{matrix}\right.\left(tmđk\right)⇔⎣⎢⎢⎡​x=2−5+197​​x=2−5−197​​​(tmđk)

Vậy phương trình có tập nghiệm S=\left\{\dfrac{-5+\sqrt{197}}{2};\dfrac{-5-\sqrt{197}}{2}\right\}S={2−5+197​​;2−5−197​​}

​ĐK: \left\{{}\begin{matrix}\left(x+2\right)\left(x+3\right)\ge0\\x^2+5x-36\ge0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge4\\x\le-9\end{matrix}\right.{(x+2)(x+3)≥0x2+5x−36≥0​⇔[x≥4x≤−9​

pt\Leftrightarrow\sqrt{x^2+5x+6}=x^2+5x-36pt⇔x2+5x+6​=x2+5x−36

Đặt \sqrt{x^2+5x+6}=t\left(t\ge0\right)x2+5x+6​=t(t≥0) , phương trình trở thành:

t=t^2-42\Leftrightarrow t^2-t-42=0\Leftrightarrow\left(t+6\right)\left(t-7\right)=0t=t2−42⇔t2−t−42=0⇔(t+6)(t−7)=0

\Leftrightarrow\left[{}\begin{matrix}t=-6\left(ktmđk\right)\\t=7\end{matrix}\right.⇔[t=−6(ktmđk)t=7​

Với t=7\Rightarrow\sqrt{x^2+5x+6}=7\Rightarrow x^2+5x+6=49t=7⇒x2+5x+6​=7⇒x2+5x+6=49

\Rightarrow x^2+5x-43=0⇒x2+5x−43=0

\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5+\sqrt{197}}{2}\\x=\dfrac{-5-\sqrt{197}}{2}\end{matrix}\right.\left(tmđk\right)⇔⎣⎢⎢⎡​x=2−5+197​​x=2−5−197​​​(tmđk)

Vậy phương trình có tập nghiệm S=\left\{\dfrac{-5+\sqrt{197}}{2};\dfrac{-5-\sqrt{197}}{2}\right\}S={2−5+197​​;2−5−197​​}

​ĐK: \left\{{}\begin{matrix}\left(x+2\right)\left(x+3\right)\ge0\\x^2+5x-36\ge0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge4\\x\le-9\end{matrix}\right.{(x+2)(x+3)≥0x2+5x−36≥0​⇔[x≥4x≤−9​

pt\Leftrightarrow\sqrt{x^2+5x+6}=x^2+5x-36pt⇔x2+5x+6​=x2+5x−36

Đặt \sqrt{x^2+5x+6}=t\left(t\ge0\right)x2+5x+6​=t(t≥0) , phương trình trở thành:

t=t^2-42\Leftrightarrow t^2-t-42=0\Leftrightarrow\left(t+6\right)\left(t-7\right)=0t=t2−42⇔t2−t−42=0⇔(t+6)(t−7)=0

\Leftrightarrow\left[{}\begin{matrix}t=-6\left(ktmđk\right)\\t=7\end{matrix}\right.⇔[t=−6(ktmđk)t=7​

Với t=7\Rightarrow\sqrt{x^2+5x+6}=7\Rightarrow x^2+5x+6=49t=7⇒x2+5x+6​=7⇒x2+5x+6=49

\Rightarrow x^2+5x-43=0⇒x2+5x−43=0

\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5+\sqrt{197}}{2}\\x=\dfrac{-5-\sqrt{197}}{2}\end{matrix}\right.\left(tmđk\right)⇔⎣⎢⎢⎡​x=2−5+197​​x=2−5−197​​​(tmđk)

Vậy phương trình có tập nghiệm S=\left\{\dfrac{-5+\sqrt{197}}{2};\dfrac{-5-\sqrt{197}}{2}\right\}S={2−5+197​​;2−5−197​​}​ĐK: 

\left\{{}\begin{matrix}\left(x+2\right)\left(x+3\right)\ge0\\x^2+5x-36\ge0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge4\\x\le-9\end{matrix}\right.{(x+2)(x+3)≥0x2+5x−36≥0​⇔[x≥4x≤−9​

pt\Leftrightarrow\sqrt{x^2+5x+6}=x^2+5x-36pt⇔x2+5x+6​=x2+5x−36

Đặt \sqrt{x^2+5x+6}=t\left(t\ge0\right)x2+5x+6​=t(t≥0) , phương trình trở thành:

t=t^2-42\Leftrightarrow t^2-t-42=0\Leftrightarrow\left(t+6\right)\left(t-7\right)=0t=t2−42⇔t2−t−42=0⇔(t+6)(t−7)=0

\Leftrightarrow\left[{}\begin{matrix}t=-6\left(ktmđk\right)\\t=7\end{matrix}\right.⇔[t=−6(ktmđk)t=7​

Với t=7\Rightarrow\sqrt{x^2+5x+6}=7\Rightarrow x^2+5x+6=49t=7⇒x2+5x+6​=7⇒x2+5x+6=49

\Rightarrow x^2+5x-43=0⇒x2+5x−43=0

\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5+\sqrt{197}}{2}\\x=\dfrac{-5-\sqrt{197}}{2}\end{matrix}\right.\left(tmđk\right)⇔⎣⎢⎢⎡​x=2−5+197​​x=2−5−197​​​(tmđk)

Vậy phương trình có tập nghiệm S=\left\{\dfrac{-5+\sqrt{197}}{2};\dfrac{-5-\sqrt{197}}{2}\right\}S={2−5+197​​;2−5−197​​}