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\(\left(y-2\right)\left(y-3\right)+\left(y-2\right)-1=0\)
\(\Leftrightarrow\left(y-2\right)\left(y-3\right)+\left(y-3\right)=0\)
\(\Leftrightarrow\left(y-3\right)^2=0\)
\(\Leftrightarrow y=3\)
\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow\left(x+3\right)x\left(x-2\right)=0\)
\(\Leftrightarrow x\in\left\{0;-3;2\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x^3+27+(x+3)(x-9)=0
(x+3)(x^2-3x+9)+(x+3)(x-9)=0
(x+3)(x^2-3x+9+x-9)=0
(x+3)(x^2-2x)=0
(x+3)x(x-2)=0
suy ra x+3=0 hoặc x=0 hoặc x-2=0
x=-3 hoặc x=0 hoặc x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
=x3+33+(x+3)(x-9)
=(x+3)(x2-3x+9)+(x+3)(x-9)
=(x+3)(x2-3x+9+x-9)
=(x+3)(x2-2x)
=(x+3)(x-2)x
![](https://rs.olm.vn/images/avt/0.png?1311)
=>\(\left(x^3+3^3\right)+\left(x+3\right)\left(x-9\right)=0\)
=>\(\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
=>\(\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
=>\(\left(x+3\right)\left(x^2-2x\right)=0\)
=>\(\left(x+3\right)x\left(x-2\right)=0\)
.........
tu tinh nhe
Vậy x=-3,x=0,x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow x=0;x=-3;x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x3 + 27 + (x + 3)(x - 9) = 0
(x + 3)(x2 - 3x + 9) + (x + 3)(x - 9) = 0
(x + 3)(x2 - 3x + 9 + x - 9) = 0
(x + 3)(x2 - 2x) = 0
x(x + 3)(x - 2) = 0
\(\left[\begin{array}{nghiempt}x=0\\x+3=0\\x-2=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=0\\x=-3\\x=2\end{array}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(x+3)(x^2-3x+9)+(x+3)(x-9)=0
(x+3)(x^2-3x+9+x-9)=0
(x+3)(x^2-2x)=0
rồi bạn tự làm đi
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