Question

Tìm x:

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

Answer 1

Bài làm

x³ + 27 + ( x + 3 )( x + 9 ) = 0

( 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 +18 ) = 0

~ Đến đây làm tiếp, áp dụng A . B = 0 => A = 0 hoặc B = 0 ~
# Học tôtd #

Answer 2

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

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x+9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x+18\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2-2x+1+17=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\\left(x-1\right)^2+17=0\end{matrix}\right.\)
Vì \(\left(x-1\right)^2\ge0\forall x\) \(\Rightarrow\left(x-1\right)^2+17\ge17>0\forall x\)
\(\Rightarrow\left(x-1\right)^2+17\ne0\forall x\) \(\Rightarrow\) Pt vô nghiệm
Vậy \(x=-3\) thì \(x^3+27+\left(x+3\right)\left(x+9\right)=0\)