a) \(5x^2-4x+1=0\)
Ta có: \(5x^2-4x+1=5\left(x^2-\dfrac{4}{5}x+\dfrac{1}{5}\right)\)
= \(5\left(x^2-2x.\dfrac{2}{5}+\dfrac{4}{25}+\dfrac{1}{25}\right)\)
= \(5\left[\left(x-\dfrac{2}{5}\right)^2+\dfrac{1}{25}\right]\)
= \(5\left(x-\dfrac{2}{5}\right)^2+\dfrac{1}{5}>0\forall x\)
Do đó phương trình trên vô nghiệm.
b) \(x^2-x-6=0\)
\(\Leftrightarrow\) \(x^2-3x+2x-6=0\)
\(\Leftrightarrow\) \(x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\) \(\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy x = 3 hoặc x = -2
c) \(2x^2+x-1=0\)
\(\Leftrightarrow\) \(2x^2+2x-x-1=0\)
\(\Leftrightarrow\) \(2x\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\) \(\left(x+1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy x = -1 hoặc x = \(\dfrac{1}{2}\)
a)
5x2 - 4x + 1 = 0
<=> \(5\left(x^2-\dfrac{4}{5}x+\dfrac{4}{25}\right)+\dfrac{1}{5}=0\)
\(\Leftrightarrow5\left(x-\dfrac{2}{5}\right)^2+\dfrac{1}{5}=0\)
mà \(5\left(x-\dfrac{2}{5}\right)^2+\dfrac{1}{5}\ge\dfrac{1}{5}>0\)
Vậy pt vô nghiệm.
b)
x2 - x - 6 = 0
<=> (x - 3)(x + 2) = 0
<=> \(\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy S = {3 ; - 2}
c)
2x2 + x - 1 = 0
<=> (2x - 1)(x + 1) = 0
<=> \(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-1\end{matrix}\right.\)
Vậy S = {- 1 ; 0,5}
a, 5x^2 - 4x+1 = 0
Ta có: \(5x^2-4x+1=0\)
\(\Rightarrow\left(x-\dfrac{2}{5}\right)^2+\dfrac{1}{5}=0\) ( vô lý )
Vậy pt vô nghiệm.
b, x^2 - x - 6=0
Ta có: \(x^2-x-6=0\)
\(\Leftrightarrow x^2-\left(3x-2x\right)-6=0\)
\(\Leftrightarrow x^2-3x+2x-6=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy ...
c, 2x^2 +x-1=0
Ta có: \(2x^2+x-1=0\)
\(\Leftrightarrow2x^2+2x-x+1=0\)
\(\Leftrightarrow2x\left(x+1\right)-x-1=0\)
\(\Leftrightarrow2x\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\2x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy...
a) đề có vấn đề
b) \(x^2-x-6=0\Leftrightarrow x^2+2x-3x-6=0\)
\(\Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\) vậy \(x=3;x=-2\)
c) \(2x^2+x-1=0\Leftrightarrow2x^2+2x-x-1=0\)
\(\Leftrightarrow2x\left(x+1\right)-\left(x+1\right)=0\Leftrightarrow\left(2x-1\right)\left(x+1\right)=0\)
\(\left\{{}\begin{matrix}2x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=1\\x=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\x=-1\end{matrix}\right.\) vậy \(x=\dfrac{1}{2};x=-1\)
