Tìm x:
X^3 + 27 + (x+3)×(x-9)=0
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=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
a) \(x+546=46\\ x=46-546\\ x=-500\)
b) \(2x-19\times3=27\\ 2x-57=27\\ 2x=27+57\\ 2x=84\\ x=84:2\\ x=42\)
c) \(x+12=23+3\times3^4\\ x+12=23+3\times81\\ x=23+243-12\\ x=254\)
d) \(x-12=3-3\times2^4\\ x-12=3-3\times16\\ x=3-48+12\\ x=-33\)
e) \(\left(27-x\right)\left(x+9\right)=0\\ \Rightarrow\left[{}\begin{matrix}27-x=0\\x+9=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=27\\x=-9\end{matrix}\right.\)
f) \(\left(-x\right)\left(x-43\right)=0\\ \Rightarrow\left[{}\begin{matrix}-x=0\\x-43=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=43\end{matrix}\right.\)
\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-2=0\\x+3=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=2\\x=-3\end{array}\right.\)
<=> (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
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=-3\\x=2\end{array}\right.\)
Đáp án của bạn Nguyễn Tiến Hải còn thiếu trường hợp:
(x + 1/2)2 = 25/4
TH1: x + 1/2 = 5/2 và giải như bạn Hải
TH2: x + 1/2 = -5/2
x = -3
=> (x+3)(x2-3x+9) + (x+3)(x-9) =0
=> (x+3)(x2-2x)=0 => (x+3)(x-2)x=0
=> x=-3 hoặc x=2 hoặc x=0
\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)
\(a,\Leftrightarrow\left(x-2\right)^3-3x\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-2-3x\right)=0\\ \Leftrightarrow\left(x-2\right)\left(-2x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\\ b,\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\\ \Leftrightarrow x\left(x-2\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)
\(\left(y-2\right)\left(y-3\right)+\left(y-2\right)-1=0\)
\(\Leftrightarrow\left(y-2\right)\left(y-3\right)+\left(y-3\right)=0\)
\(\Leftrightarrow\left(y-3\right)^2=0\)
\(\Leftrightarrow y=3\)
\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow\left(x+3\right)x\left(x-2\right)=0\)
\(\Leftrightarrow x\in\left\{0;-3;2\right\}\)
(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
(x+3)x(x-2)=0
=> x=-3;0;2
OK. NHỚ K CHO MK NHÉ