- Ta có : \(5\left(x+2\right)^3+7=2\)
=> \(5\left(x^3+6x^2+12x+8\right)+7=2\)
=> \(5x^3+30x^2+60x+40+7=2\)
=> \(5x^3+30x^2+60x+40+7-2=0\)
=> \(5x^3+30x^2+60x+45=0\)
=> \(5x^3+15x^2+15x^2+45x+15x+45=0\)
=> \(5x^2\left(x+3\right)+15x\left(x+3\right)+15\left(x+3\right)=0\)
=> \(\left(5x^2+15x+15\right)\left(x+3\right)=0\)
=> \(\left[{}\begin{matrix}x+3=0\\5x^2+15x+15=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-3\\x^2+3x+3=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-3\\x^2+2x.\frac{3}{2}+\frac{9}{4}+\frac{3}{4}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-3\\\left(x+\frac{3}{2}\right)^2+\frac{3}{4}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-3\\\left(x+\frac{3}{2}\right)^2=-\frac{3}{4}\left(VL\right)\end{matrix}\right.\)
=> \(x=-3\)
Vậy phương trình có tập nghiệm là \(S=\left\{-3\right\}\)
5(x + 2)3 + 7 = 2
(x + 2)3 = -1
=> x + 2 = -1
<=> x = -3
\(5.\left(x+2\right)^3+7=2\)
\(\Rightarrow5.\left(x+2\right)^3=2-7\)
\(\Rightarrow5.\left(x+2\right)^3=-5\)
\(\Rightarrow\left(x+2\right)^3=\left(-5\right):5\)
\(\Rightarrow\left(x+2\right)^3=-1\)
\(\Rightarrow\left(x+2\right)^3=\left(-1\right)^3\)
\(\Rightarrow x+2=-1\)
\(\Rightarrow x=\left(-1\right)-2\)
\(\Rightarrow x=-3\)
Vậy \(x=-3.\)
Chúc bạn học tốt!
