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c: \(\Leftrightarrow x^3-9x^2+27x-27-x^3+9x^2=0\)
hay x=1
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b) 4x(2-x)+(2x+1)^2=2
8x-4x^2+4x^2+4x+1-2=0
(8x+4x)+(-4x^2+4x^2)+(1-2)=0
12x + 0 -1 =0
12x=1
x=1/12
Vậy x= 1/2
c) (x-3)^3-x^2(x-9)=0
x^3-9x^2+27x-x^3+9x^2=0
(x^3-x^3)+(-9x^2+9x^2)+27x=0
0 + 0 + 27x=0
x= 0
Vậy x=0
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\(a,=12-3x+4x-x^2+x^2-2x=12-x\\ b,=x^2-2x+1-x^2+4=-2x+5\)
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\(a,=x^2+4x+4\\ b,=x^3+3x^2+3x+1\\ c,=\left(x-3\right)\left(x+3\right)\)
a,\(\left(x+2\right)^2=x^2+2.x.2+2^2=x^2+4x+4\)
b, \(\left(x+1\right)^3=x^3+3.x^2.1+3.x.1^2+1^3=x^3+3x^2+3x+1\)
c,\(x^2-3^2=\left(x-3\right).\left(x+3\right)\)
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\(\Rightarrow8x-4x^2+4x^2+4x+1=2\\ \Rightarrow12x=1\Rightarrow x=\dfrac{1}{12}\)
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\(\Leftrightarrow8x-4x^2+4x^2+4x+1=2\)
\(\Leftrightarrow x=\dfrac{1}{12}\)
\(\Rightarrow x^2+3x-x^2=6\\ \Rightarrow3x=6\Rightarrow x=2\)