\(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)
\(\left(x^2-4x+4\right)-\left(x^2-9\right)=6\)
\(x^2-4x+4-x^2+9=6\)
\(13-4x=6\)
4x=7
\(x=\dfrac{7}{4}\)
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\(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)
\(\Leftrightarrow\left(x^2-2.x.2+2^2\right)-\left(x^2-3^2\right)-6=0\)
\(\Leftrightarrow x^2-4x+4-x^2+9-6=0\)
\(\Leftrightarrow-4x+7=0\)
\(\Leftrightarrow-4x=-7\)
\(\Leftrightarrow x=\dfrac{-7}{-4}\)
\(\Leftrightarrow x=1,75\)
Vậy x=1,75
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