Question

giải phương trình

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

\(\sqrt{2x^2-2x+1}=2x-1\)

Answer 1

Vì \(\sqrt{x^2-2x+4} \)≥ 0 ( đúng với ∀ x )
→ \(2x - 2\) ≥ 0 
→x ≥ 1
Ta có : \(\sqrt{x^2-2x+4} \) = \(2x - 2\)
⇔ \(x^2-2x+4 \) = \((2x - 2)^2\)
⇔ \(x^2-2x+4 \) = \(4x^2 - 8x + 4 \)
⇔ \(0 = 3x^2 - 6x \)
⇔ 0 = \(3x(x-1)\)
⇔\(\begin{cases} x=0\\ x-1=0 \end{cases} \)
Mà x ≥ 1
Vậy x ∈ { 1}

Answer 2

Xin lỗi mình lm sai chút :)))
Vì \(\sqrt{x^2-2x+4} \)≥ 0 ( đúng với ∀ x )
→  ≥ 0 
→x ≥ 1
Ta có : \(\sqrt{x^2-2x+4} \) = 
⇔ \(x^2 - 2x + 4\)= 
⇔ 
⇔ 0 = 
⇔\(\left[\begin{array}{} x=0\\ x=2 \end{array} \right.\)
Mà x ≥ 1
→ x ∈ {2}

Answer 3

a.

\(\Leftrightarrow\left\{{}\begin{matrix}2x-2\ge0\\x^2-2x+4=\left(2x-2\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x^2-2x+4=4x^2-8x+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\3x^2-6x=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow x=2\)

Answer 4

b.

\(\Leftrightarrow\left\{{}\begin{matrix}2x-1\ge0\\2x^2-2x+1=\left(2x-1\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\2x^2-2x+1=4x^2-4x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\2x^2-2x=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow x=1\)