Tìm x biết: x2+x+1=0
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b: \(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) x = 1; x = - 1 3 b) x = 2.
c) x = 3; x = -2. d) x = -3; x = 0; x = 2.
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\Rightarrow x\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
b) \(\Rightarrow x\left(x^2-4\right)=0\Rightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
c) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
d) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\Rightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
e) \(\Rightarrow2x^2-10x-3x-2x^2=26\)
\(\Rightarrow-13x=26\Rightarrow x=-2\)
f) \(\Rightarrow\left(x-2012\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2012\\x=\dfrac{1}{5}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) x = -1. b) x = 4 hoặc x = 5.
c) x = ± 2 . d) x = 1 hoặc x = 2.
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(x-1)2-1+x2-(1-x)(x+3)=0
⇒x2-2x+1-1+x2-x(1-x)+3(1-x)=0
⇒x2-2x+1-1+x2-x+x2+3-3x=0
⇒3x2-6x+3=0
⇒3(x2-2x+1)=0
⇒x2-2x+1=0
⇒(x-1)2=0
⇒x-1=0
⇒x=1
Lời giải:
$(x-1)^2-1+x^2-(1-x)(x+3)=0$
$\Leftrightarrow (x^2-2x+1)-1+x^2-(3-x^2-2x)=0$
$\Leftrightarrow x^2-2x+1-1+x^2-3+x^2+2x=0$
$\Leftrightarrow 3x^2-3=0$
$\Leftrightarrow x^2-1=0$
$\Leftrightarrow (x-1)(x+1)=0$
$\Leftrightarrow x=1$ hoặc $x=-1$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(b,3x+x^2=0\\ \Rightarrow x\left(3+x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\\ c,\left(x-1\right)\left(x-3\right)< 0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1< 0\\x-3>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1>0\\x-3< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 1\\x>3\left(vô.lí\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x>1\\x< 3\end{matrix}\right.\end{matrix}\right.\)
Vậy 1<x<3
![](https://rs.olm.vn/images/avt/0.png?1311)
1. a) \(7x^2\left(2x^3+3x^5\right)=7x^2\cdot2x^3+7x^2\cdot3x^5=14x^5+21x^7\)
b) \(\left(x^3-x^2+x-1\right):\left(x-1\right)=\dfrac{x^3-x^2+x-1}{x-1}\)
\(=\dfrac{x^2\left(x-1\right)+\left(x-1\right)}{x-1}=\dfrac{\left(x-1\right)\left(x^2+1\right)}{x-1}=x^2+1\)
2: \(x^2-8x+7=0\)
=>\(x^2-x-7x+7=0\)
=>\(x\left(x-1\right)-7\left(x-1\right)=0\)
=>\(\left(x-1\right)\left(x-7\right)=0\)
=>\(\left[{}\begin{matrix}x-1=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=7\end{matrix}\right.\)
1:
a: \(7x^2\left(2x^3+3x^5\right)=7x^2\cdot2x^3+7x^2\cdot3x^5=21x^7+14x^5\)
b: \(\dfrac{x^3-x^2+x-1}{x-1}=\dfrac{x^2\left(x-1\right)+\left(x-1\right)}{\left(x-1\right)}\)
\(=x^2+1\)
Ta có
\(x^2+x+1=x^2+2.\frac{1}{2}x+\frac{1}{4}+1-\frac{1}{4}\)
\(=\left(x^2+2.\frac{1}{2}x+\frac{1}{4}\right)+\frac{3}{4}\)
\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Vì \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
\(\frac{3}{4}>0\)
Nên \(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)hay \(x^2+x+1>0\)
mà theo đề bài \(x^2+x+1=0\)
Vậy \(x\in\varnothing\)