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\(x\left(x-1\right)>0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x-1< 0\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x>1\\x< 0\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2-1>0\Rightarrow x^2>1\Rightarrow\left|x\right|>1\Rightarrow\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2-4x-21>0\)
\(\Leftrightarrow\) \(x^2-4x+4>25\)
\(\Leftrightarrow\) \(\left(x-2\right)^2>25\)
\(\Leftrightarrow\) \(\left|x-2\right|>5\)
\(\Leftrightarrow\orbr{\begin{cases}x-2>5\\x-2>-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x>7\\x>-3\end{cases}}}\)
\(x^2-4x-21>0\)
\(x^2-4x+4-25>0\)
\(\left(x-2\right)^2>25\)
Ta có: \(25=5^2=\left(-5\right)^2\)
TH1: \(\left(x-2\right)^2>5^2\)
\(x-2>5\)
\(x>7\)
TH2: \(\left(x-2\right)^2>\left(-5\right)^2\)
\(x-2>-5\)
\(x>-3\)
Kết hợp cả 2 TH ta đc x>-3
=.= hok tốt!!
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1-3x<0
<=> -3x<-1
<=> x<\(\frac{-1}{-3}\)
<=> x<\(\frac{1}{3}\)
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\(\sqrt{x+6}-2\sqrt{x}>0\)
\(\Leftrightarrow\sqrt{x+6}>2\sqrt{x}\)
\(\Leftrightarrow x+6>4x\)
\(\Leftrightarrow-3x>-6\)
\(\Leftrightarrow x
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(2x^2-5x+5=2\left(x^2-2.\dfrac{5}{4}x+\dfrac{25}{16}\right)+\dfrac{15}{8}=2\left(x-\dfrac{5}{4}\right)^2+\dfrac{15}{8}>0\)
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 \(\frac{3}{4}>0\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\)
\(\Rightarrow x^2-x+1>0\)