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\(a,\left(3x+x\right)\left(x^2-9\right)-\left(x-3\right)\left(x^2+3x+9\right)\)
\(=4x\left(x^2-9\right)-x^3+27\)
\(=4x^3-36x-x^3+27\)
\(=3x^3-36x+27\)
\(\left(x+6\right)^2-2x.\left(x+6\right)+\left(x-6\right).\left(x+6\right)\)
\(=\left(x+6\right).\left(x+6-2x+x-6\right)\)
\(=\left(x+6\right).0\)
\(=0\)
c, viết thiếu kìa :))
\(a,\left(2x-1\right)-\left(5x+1\right)=\left(x+3\right)-\left(x-2\right)\)
\(\Leftrightarrow2x-1-5x-1=x+3-x+2\)
\(\Leftrightarrow-3x-2=5\Leftrightarrow-3x=7\Leftrightarrow x=\frac{7}{3}\)
\(b,\left(2x-3\right)-\left(x-5\right)=\left(x+2\right)-\left(x-1\right)\)
\(\Leftrightarrow2x-3-x+5=x+2-x+1\Leftrightarrow x+2=3\Leftrightarrow x=1\)
\(c,2\left(x-1\right)-5\left(x+2\right)=0\)
\(\Leftrightarrow2x-2-5x-10=0\Leftrightarrow-3x-12=0\Leftrightarrow-3x=12\Leftrightarrow x=-4\)
(x+1)3+(x−2)3−2x2(x−1,5)=3
⇔(x3+3x2+3x+1)+(x3−6x2+12x−8)−(2x3−3x2)=3
⇔x3+3x2+3x+1+x3−6x2+12x−8−2x3+3x2= 3
⇔15x−12=0
⇔15x=10
⇔x= 2/3
\(2x\left(x+3\right)-3\left(x^2+1\right)=x+1-x\left(x-2\right)\)
\(\Leftrightarrow2x^2+6x-3x^2-3=x+1-x^2+2x\)
\(\Leftrightarrow-x^2+6x-3=-x^2+3x+1\)
\(\Leftrightarrow3x=4\)
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 x=43
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