a)
\(\Rightarrow\left[\left(x^3+3^3\right)+\left(x+3\right)\right].\left(x-9\right)=0\)
\(\Rightarrow\left[\left(x+3\right)\left(x^2-3x+3^2\right)+\left(x+3\right)\right].\left(x-9\right)=0\)
\(\Rightarrow\left[\left(x+3\right).\left(x^2-3x+9+1\right)\right].\left(x-9\right)=0\)
\(\Rightarrow\left[\left(x+3\right).\left(x^2-3x+10\right)\right].\left(x-9\right)=0\)
\(\Rightarrow\left(x+3\right)=0\) hoặc x-9=0 hoặc \(x^2-3x+10=0\)
1) x-9=0 => x=9
2) x+3 = 0 => x= -3
3) \(x^2-3x+10=0\Rightarrow x^2-2.x.\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{31}{4}=0\)
\(\Rightarrow\left(x-\dfrac{3}{2}\right)^2+\dfrac{31}{4}=0\Rightarrow\left(x-\dfrac{3}{2}\right)^2=-\dfrac{31}{4}\left(ktm\right)\)
Vậy x=9 ; x=-3
b)
\(\left(16x^2-9\right).\left(x+1\right)^2=0\)
\(\Rightarrow\left[\left(4x\right)^2-3^2\right].\left(x+1\right)^2=0\)
=> \(\left(4x\right)^2-3^2=0\) hoặc \(\left(x+1\right)^2=0\)
1) \(\left(x+1\right)^2=0\Rightarrow x+1=0\Rightarrow x=-1\)
2) \(\left(4x\right)^2-3^2=0\Rightarrow\left(4x+3\right).\left(4x-3\right)=0\)
=> 4x+3 =0 hoặc 4x-3=0
* 4x+3=0 => 4x=-3 => x= \(\dfrac{-3}{4}\)
* 4x-3=0 => 4x= 3 => x= \(\dfrac{3}{4}\)
Vậy x=-1 ; x=0 ; x=\(\dfrac{-3}{4}\) ; x=\(\dfrac{3}{4}\)
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