Question

Answer 1

\(5+\left|x+3\right|=9\)

\(\Leftrightarrow\left|x+3\right|=4\)

\(\Leftrightarrow x+3=\pm4\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)

Answer 2

đề bài : tìm x bt

Answer 3

5 + Ix + 3I = 9

      Ix + 3 I = 9 -5

      Ix + 3 I = 4

=> x + 3 = -+4

     x        = 4 - 3

     x         = 1

x + 3 = -4 

x       = -4 -3

x=  -7

Vậy x={ .....}

Answer 4

\(5+|x+3|=9\\ \Leftrightarrow|x+3|=4\\ \Leftrightarrow\left[{}\begin{matrix}x+3=4\\x+3=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)

\(\left[{}\begin{matrix}\\\end{matrix}\right.\)là hoặc nha bn

tik mik nha

Answer 5

Ta có 2 TH:

Nếu x \(\ge\) -3 thì \(|x+3|=x+3\)

Nếu x < -3 thì \(|x+3|=-\left(x+3\right)\)

TH1:

Nếu x \(\ge\) -3:

5 + x + 3 = 9

<=> x = 9 - 5 - 3

<=> x = 1 (thỏa mãn)

Vậy x = 1

TH2: Nếu x < -3:

5 + \(\left[-\left(x+3\right)\right]=9\)

<=> 5 + (x - 3) = 9

<=> 5 + x - 3 = 9

<=> x = 9 - 5 + 3

<=> x = 7 (loại)

Vậy PT có nghiệm duy nhất S = \(\left\{1\right\}\)

Answer 6

g: Ta có: \(5+\left|x+3\right|=9\)

\(\Leftrightarrow\left|x+3\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=4\\x+3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)