x3+6x2+12x+8
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a)
=(x-2)3
b)\(\left(2-x\right)^3\)
c)\(\left(x+\dfrac{1}{3}\right)^3\)
d)\(\left(\dfrac{x}{2}+y\right)^3\)
e)
\(=\left(x-1\right)^2\left(x-1-15\right)+25\left[3\left(x-1\right)-5\right]\)
\(=\left(x-1\right)^2\left(x-16\right)+25\left(3x-3-5\right)\)
\(=\left(x-1\right)^2\left(x-16\right)+25\left(3x-8\right)\)
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\(\Leftrightarrow\left(x-2\right)^3=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)
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\(8.\left(x-3\right)^3+x^3=6x^2-12x+8\)
\(\Leftrightarrow\left(2x-6\right)^3=-x^3+6x^2-12x+8\)
\(\Leftrightarrow\left(2x-6\right)^3=\left(2-x\right)^3\)
\(\Leftrightarrow2x-6=2-x\)
\(\Leftrightarrow3x=8\)
\(\Leftrightarrow x=\dfrac{8}{3}\)
Vậy pt có nghiệm x = \(\dfrac{8}{3}\)
\(8\left(x-3\right)^3+x^3=6x^2-12x+8\)
\(< =>8\left(x^3-9x^2+27x-27\right)+x^3-6x^2+12x-8=0\)
\(< =>8x^3-72x^2+216x-216+x^3-6x^2+12x-8=0\)
\(< =>9x^3-78x^2+228x-224=0\)
\(< =>\left(3x-8\right)\left(3x^2-18x+28\right)=0\)
đến đây dễ rồi bạn tự làm
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\left(x-2\right)^3=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\Rightarrow\left(x-1\right)^3=0\Rightarrow x=1\)
b) \(\Rightarrow\left(x^3-1\right)\left(x^3+1\right)=0\Rightarrow\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)(do \(\left\{{}\begin{matrix}x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\\x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\end{matrix}\right.\))
c) \(\Rightarrow4x\left(x^2-9\right)=0\Rightarrow4x\left(x-3\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
d) \(\Rightarrow\left(x-2\right)^3=0\Rightarrow x=2\)
a) \(x^3-3x^2+3x-1=0\Rightarrow\left(x-1\right)^3=0\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
b) \(x^6-1=0\Rightarrow\left(x^3\right)^2-1=0\Rightarrow\left(x^3-1\right)\left(x^3+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^3-1=0\\x^3+1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
c) \(4x^3-36x=0\Rightarrow4x\left(x^2-36\right)=0\Rightarrow4x\left(x-6\right)\left(x+6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}4x=0\\x-6=0\\x+6=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\\x=-6\end{matrix}\right.\)
d) \(x^3-6x^2+12x-8=0\) (đề bài như vậy mới làm đc, nếu là +8 thì mình xin bó tay nhé)
\(\Rightarrow x^3-3\cdot x^2\cdot2+3\cdot x\cdot2^2-2^3=0\)
\(\Rightarrow\left(x-2\right)^3=0\Rightarrow x-2=0\Rightarrow x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x3 – 6x2 + 12x – 8 = x3 – 3.x2.2 + 3.x.22 – 23 = (x – 2)3
Tại x = 22, giá trị biểu thức bằng (22 – 2)3 = 203 = 8000.
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có x 3 – 6 x 2 + 12 x – 8 = x 3 – 3 . x 2 . 2 + 3 . x . 2 2 – 2 3 = ( x – 2 ) 3
Đáp án cần chọn là: D
x\(^3\) + 6x\(^2\) + 12x + 8
= x\(^3\) + 3.2x\(^2\) + 3.2\(^2\)x + 2\(^3\)
= (x + 2)\(^3\)
Học tốt
Đề bài là gì? PTĐTTNT?