Ta có:\(\left|x+1\right|\ge0;\left|x-2\right|\ge0;\left|x+7\right|\ge0\)
\(\Rightarrow\left|x+1\right|+\left|x-2\right|+\left|x+7\right|\ge0\)
\(\Rightarrow5x-10\ge0\)
\(\Rightarrow5x\ge10\)
\(\Rightarrow x\ge2\)
\(\Rightarrow\left|x+1\right|=x+1\)
\(\left|x-2\right|=x-2\)
\(\left|x+7\right|=x+7\)
Ta có:\(\left|x+1\right|+\left|x-2\right|+\left|x+7\right|=5x-10\)
\(\Rightarrow x+1+x-2+x+7=5x-10\)
\(\Rightarrow\)\(3x+6=5x-10\)
\(\Rightarrow6+10=5x-3x\)
\(\Rightarrow2x=16\)
\(\Rightarrow x=8\)
Vậy x=8 thỏa mãn
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