a/ \(\sqrt{4\left(x-1\right)^2}=6\)
\(\Leftrightarrow\left|2\left(x-1\right)\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2\left(x-1\right)=6\\2\left(x-1\right)=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)
vậy.......
b/ \(\sqrt{x^2-4x+9}=3\)
\(\Leftrightarrow x^2-4x+9=9\)
\(\Leftrightarrow x\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\Rightarrow x=4\end{matrix}\right.\)
Vậy.............
c/ đk: x ≤ 3 - √5
\(\sqrt{x^2-6x+4}=\sqrt{4-x}\)
\(\Leftrightarrow x^2-6x+4=4-x\)
\(\Leftrightarrow x^2-5x=0\)
\(\Leftrightarrow x\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x-5=0\Rightarrow x=5\left(KTM\right)\end{matrix}\right.\)
vậy.......
a. \(\sqrt{4\left(x-1\right)^2}\) =6
Với x ≥1, bình phương 2 vế ta có:
=> 4(x-1)2 =36
=> 4(x2 -2x+1) = 36
=> 4x2 - 8x +4 =36
=> 4x2 -8x -32=0
=> 4x2 -16x + 8x -32=0
=> 4x(x-4) +8(x-4)=0
=> (x-4)(4x+8)=0
=> x=4(TM)
a)\(\sqrt{4\left(x-1\right)^2}=6\Leftrightarrow2\left|x-1\right|=6\Leftrightarrow\left|x-1\right|=3\)
\(\Leftrightarrow x\in\left\{-2;4\right\}\)
b)\(\sqrt{x^2-4x+9}=3\Leftrightarrow x^2-4x+9=9\)
\(\Leftrightarrow x^2-4x=0\Leftrightarrow x\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
c)\(\sqrt{x^2-6x+4}=\sqrt{4-x}\)
ĐKXĐ: \(x\le4;x\ge3+\sqrt{5};x\le3-\sqrt{5}\)
\(\Leftrightarrow x^2-6x+4=4-x\Leftrightarrow x^2-5x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\left(loại\right)\end{matrix}\right.\)