Môn Toán Lớp 7: |x – 1| = 2x – 5 ||x + 5 | – 4 | = 3

Trả Lời

1. a)

   ĐKXĐ:2x-5≥0=>2x≥5=>x≥5/2

   |x-1|=2x-5

   => \(\left[ \begin{array}{l}x-1=2x-5\\x-1=-2x+5\end{array} \right.\) 

   => \(\left[ \begin{array}{l}x-2x=-5+1\\x+2x=5+1\end{array} \right.\) 

   => \(\left[ \begin{array}{l}-x=-4\\3x=6\end{array} \right.\) 

   => \(\left[ \begin{array}{l}x=4(tm)\\x=2(l)\end{array} \right.\) 

   b)||x+5|-4|=3

   => \(\left[ \begin{array}{l}|x+5|-4=3\\|x+5|-4=-3\end{array} \right.\) 

   => \(\left[ \begin{array}{l}|x+5|=7\\|x+5|=1\end{array} \right.\) 

   => \(\left[ \begin{array}{l}x+5=-7\\x+5=7\\x+5=-1\\x+5=1\end{array} \right.\) 

   => \(\left[ \begin{array}{l}x=-12\\x=2\\x=-6\\x=-4\end{array} \right.\) 

   => x in { -12;2;-6;-4}

   Trả lời
2. | x – 1 | = 2x – 5

   +) TH1 : x – 1 >= 0 hay x >= 1 thì | x – 1 | = 2x – 5 => x – 1 = 2x – 5

   => x – 2x – 1 + 5 = 0

   => – x + 4 = 0

   => – x = -4

   => x = 4 ( tm )

   +) TH2 : x – 1 < 0 hay x < 1 thì | x – 1 | = 2x – 5 => 1 – x = 2x – 5

   => – x – 2x + 1 + 5 = 0

   => – 3x + 6 = 0

   => – 3x = -6

   => x = -6 : ( – 3 )

   => x = 2 ( loại )

   Vậy x = 4

   ||x + 5 | – 4 | = 3

   => {(| x + 5 | – 4 = – 3),(| x + 5 | – 4 = 3):}

   => {(| x + 5 | = 1),(| x + 5 | = 7):} => {({(x+5=-1),(x+5=1):}),({(x+5=-7),(x+5=7):})}

   => {({(x=-6),(x=-4):}),({(x=-12),(x=2):}):}

   Vậy x in { -6 ; -4 ; -12 ; 2 } 

   Trả lời