\(2x\left(x+3\right)-3\left(x^2+1\right)=x+1-x\left(x-2\right)\)
\(\Leftrightarrow2x^2+6x-3x^2-3=x+1-x^2+2x\)
\(\Leftrightarrow-x^2+6x-3=-x^2+3x+1\)
\(\Leftrightarrow3x=4\)
hay \(x=\dfrac{4}{3}\)
\(2x\left(x+3\right)-3\left(x^2+1\right)=x+1-x\left(x-2\right)\)
\(\Leftrightarrow2x^2+6x-3x^2-3=x+1-x^2+2x\)
\(\Leftrightarrow3x=4\Leftrightarrow x=\dfrac{4}{3}\)
B=2x^2+6x-3x^2-3=x+1-x(x-2)=0
=-x^2+6x-3=x+1-x^2+x=0
=4x-3=0
x=3/4
2x(x + 3) - 3(x2 + 1) = x + 1 - x(x - 2)
<=> 2x2 + 6x - 3x2 - 3 = x + 1 - x2 + 2x
<=> 2x2 - 3x2 - x2 + 6x - 2x - x = 1 + 3
<=> 6x2 + 3x - 4 = 0
<=> Số ko đẹp: x = 0,603....
2x(x+3)−3(x2+1)=x+1−x(x−2)2�(�+3)−3(�2+1)=�+1−�(�−2)
⇔2x2+6x−3x2−3=x+1−x2+2x⇔2�2+6�−3�2−3=�+1−�2+2�
⇔−x2+6x−3=−x2+3x+1⇔−�2+6�−3=−�2+3�+1
⇔3x=4⇔3�=4
hay
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