\(\left(3x+1\right)^2-\left(3x-2\right)^2=0\\ \left(3x+1-3x+2\right)\left(3x+1+3x-2\right)=0\)
\(3\left(6x-1\right)=0\\ =>6x-1=0\\ 6x=0+1\\ 6x=1\\ x=1:6\\ x=\dfrac{1}{6}\)
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