\(\left|4x+3\right|-\left|x-1\right|=7\)
+) Xét \(x\ge1\) ta có:
\(4x+3-\left(x-1\right)=7\)
\(\Rightarrow4x-x+1=4\)
\(\Rightarrow x=1\) ( t/m )
+) Xét \(\dfrac{-3}{4}\le x< 1\) ta có:
\(4x+3-\left(1-x\right)=7\)
\(\Rightarrow4x+3-1+x=7\)
\(\Rightarrow x=1\) ( không t/m )
+) Xét \(x< \dfrac{-3}{4}\) ta có:
\(\left(-4x-3\right)-\left(1-x\right)=7\)
\(\Rightarrow-4x-3-1+x=7\)
\(\Rightarrow-3x-4=7\)
\(\Rightarrow x=\dfrac{-11}{3}\) ( t/m )
Vậy x = 1 hoặc \(x=\dfrac{-11}{3}\)
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