Question

Tìm x biết:   \(\sqrt{\left(2x-5\right)^2}=3\)

Các bn giúp mk vs. Mk đang cần gấp. Cảm ơn các bn nhiều!!!

Answer 1

\(\sqrt{\left(2x-5\right)^2}=3\)

\(\Rightarrow\left(2x-5\right)^2=9\)

\(\Rightarrow\left[\begin{array}{nghiempt}2x-5=3\\2x-5=-3\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=4\\x=1\end{array}\right.\)

Vậy x=4 ; x=1

Answer 2

\(\sqrt{\left(2x-5\right)^2}=3\)

\(\Leftrightarrow\left|2x-5\right|=3\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-5=3\\2x-5=-3\end{array}\right.\) 

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=4\\x=1\end{array}\right.\)

Answer 3

\(\sqrt{\left(2x-5\right)^2}=3\)

\(\Leftrightarrow\left|2x-5\right|=3\)                 (1)

+)TH1: \(2x-5\ge0\Leftrightarrow x\ge\frac{5}{2}\) thì:

(1)<=> \(2x-5=3\Leftrightarrow2x=8\Leftrightarrow x=4\left(tm\right)\)

+) TH2: \(2x-5< 0\Leftrightarrow x< \frac{5}{2}\) thì

(1)<=> \(5-2x=3\Leftrightarrow-2x=-2\Leftrightarrow x=1\left(tm\right)\)

Vậy x={1;4}

Answer 4

dìm hàng ak

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