Question

Tìm x biết: x³ + 27 + (x + 3)(x - 9) = 0

Answer 1

\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-2=0\\x+3=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=2\\x=-3\end{array}\right.\)

Answer 2

<=> (x+3)(x²+3x+9)+(x+3)(x - 9)

<=> (x+3)(x²-3x+9+x - 9)=0

<=> (x+3)(x²-2x)=0

<=> (x+3)x(x-2)=0

\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=-3\\x=2\end{array}\right.\)

Answer 3

\(x^3+27+\left(x+3\right)\left(x-9\right)\)=0
\(\Leftrightarrow x^3+3^3+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x+3=0\\x-2=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=-3\\x=2\end{array}\right.\)
Vậy x=0 hoặc x=-3 hoặc x=2

Answer 4

\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(x^3+27+x^2-9x+3x-27=0\)

\(x^3+x^2-6x=0\)

Mk làm đến đây thôi tại vì chưa học

Answer 5

\(x^3+27+\left(x+3\right)\left(x-9\right)=\left(x+3\right)\left(x^2+3x+9\right)+\left(x+3\right)\left(x-9\right)\)=0

<=>\(\left(x+3\right)\left(x^2+4x\right)=0\)

<=>(x+3)x(x+4)=0

<=> \(\left[\begin{array}{nghiempt}x+3=0\\x=0\\x+4=0\end{array}\right.\)

<=> \(\left[\begin{array}{nghiempt}x=-3\\x=0\\x=-4\end{array}\right.\)

vậy tập nghiệm là S={0,-3,-4}

 

Answer 6

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

=> _x=0

    |_x+3=0=>x=-3

    |_ x-2=0=>x=2