|x2+|3x-1||=x2+7
Vì |x2+|3x-1|| \(\ge\) 0;
x2+7 \(\ge\) 0
nên VT=VP
<=>x2+|3x-1|=x2+7
<=>|3x-1|=x2+7-x2=7
<=>\(\int^{3x-1=7\Rightarrow3x=8\Rightarrow x=\frac{8}{3}}_{3x-1=-7\Rightarrow3x=-6\Rightarrow x=-2}\)
Vậy x \(\in\) (-2;8/3}
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