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Tìm x.
a) 9x^2 – 6x – 3 = 0
b) x^3 + 9x^2 + 27x + 19 = 0
c) x(x + 5)(x – 5) – (x + 2)(x^2 – 2x + 4) = 3
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(9x^2-6x-3=0\)
\(\Leftrightarrow3\left(x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
b) \(x^3+9x^2+27x+19=0\)
\(\Leftrightarrow x^2\left(x+1\right)+8x\left(x+1\right)+19\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+8x+19\right)=0\)
\(\Leftrightarrow x=-1\)( do \(x^2+8x+19=\left(x+4\right)^2+3>0\))
c) \(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\)
\(\Leftrightarrow x\left(x^2-25\right)-x^3-8=3\)
\(\Leftrightarrow x^3-25x-x^3=8\Leftrightarrow-25x=11\Leftrightarrow x=-\dfrac{11}{25}\)
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27x3-27x2+9x-1=1
<=> (3x)3-3.9x2.1+3.3.12-13 =1 (hằng đẳng thức thứ năm)
<=> (3x-1)3=1
<=> 3x-1=1
<=> 3x=2
<=> x=2/3
Nhớ k đúng cho mik nha
27x3 - 27x2 + 9x -1 =1
<=>(3x-1)3=1 (hằng đẳng thức thứ 5)
<=> 3x-1=1
<=> x=2/3
(mk tóm tắt bn tự hiểu nha)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^3+9x^2+27x+26=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+7x+13\right)=0\Rightarrow x=-2\)
\(x^3+9x^2+27x+26=0\)
\(\Leftrightarrow x^3+9x^2+27x+27=1\)
\(\Leftrightarrow\left(x+3\right)^3=1^3\)
\(\Leftrightarrow x+3=1\Leftrightarrow x=-2\)
Ta có : \(\left(27x^3+27x^2+9x\right)=26\)
\(\Leftrightarrow\left(27x^3+27x^2+9x+1\right)=27\)
\(\Leftrightarrow\left(3x+1\right)^3=27\) \(\Leftrightarrow3x+1=3\Leftrightarrow3x=2\Leftrightarrow x=\frac{2}{3}\)
Vậy ....
2x^3 + 27x^2 + 9x = 26
<=> 2x^3 + 27x^2 + 9x - 26 = 0
<=> (3x - 2)(9x^2 + 15x + 13) = 0
vì 9x^2 + 15x + 13 >= 0 nên:
<=> 3x - 2 = 0
<=> 3x = 2
<=> x = 2/3