Giải bpt x2 - x +1 >0
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\(a,\frac{x+5}{x^2-2x+1}>0\)
\(\Leftrightarrow\frac{x+5}{\left(x-1\right)^2}>0\)
\(\Leftrightarrow x>-5\)
\(b,x^2+x+1>0\)
\(\Leftrightarrow\left(x^2+x+\frac{1}{4}\right)+\frac{3}{4}>0\)
\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\) ( luôn đúng)
\(x^2-2^2>0\)
\(\Rightarrow\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\x+2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\x+2< 0\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x>-2\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x< -2\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x>2\\x< -2\end{matrix}\right.\)
\(x^2-4>0\Leftrightarrow\left(x+2\right)\left(x-2\right)>0\)
\(TH1:\left\{{}\begin{matrix}x-2>0\\x+2>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>2\\x>-2\end{matrix}\right.\Rightarrow x>2\)
\(TH2:\left\{{}\begin{matrix}x-2< 0\\x+2< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< 2\\x< -2\end{matrix}\right.\Rightarrow x< -2\)
Từ 2 trường hợp: \(\Rightarrow\left[{}\begin{matrix}x>2\\x< -2\end{matrix}\right.\)
x2 - x +1 > 0
<=> x2 - 2.1/2.x + 1/4 +3/4 >0
<=> (x-1/2)2 + 3/4 > 0
<=> (x-1/2)2 > 3/4
tự tính tiếp ạ
Trả lời
Ta có \(x^2-x+1=x^2-2\times x\times\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)
\(=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\)
Mà \(\left(x-\frac{1}{2}\right)^2\ge0\) Dấu "=" xảy ra khi \(x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\) dấu "=" khi x=1/2
Mà \(\frac{3}{4}>0\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\)
\(\Rightarrow x^2-x+1>0\)