a) Ta có: \(x^2-8x+7=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=7\end{matrix}\right.\)
b) Ta có: \(x^2+x-20=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=4\end{matrix}\right.\)
c) Ta có: \(3x^2+4x-4=0\)
\(\Leftrightarrow3x^2+6x-2x-4=0\)
\(\Leftrightarrow3x\left(x+2\right)-2\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{2}{3}\end{matrix}\right.\)
d) Ta có: \(3x^2-4x-7=0\)
\(\Leftrightarrow3x^2-7x+3x-7=0\)
\(\Leftrightarrow\left(3x-7\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-1\end{matrix}\right.\)
e) Ta có: \(5x^2-16x+3=0\)
\(\Leftrightarrow5x^2-15x-x+3=0\)
\(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)
f) Ta có: \(x^2+3x-10=0\)
\(\Leftrightarrow x^2+5x-2x-10=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
a)
\(x^2-8x+7=0\text{⇔}\text{⇔}x^2-7x-x-7=\left(x-7\right)\left(x-1\right)=0\text{⇔}\left[{}\begin{matrix}x=1\\x=7\end{matrix}\right.\)
Vậy nghiệm của đa thức : \(S=\left\{1;7\right\}\)
c)
\(3x^2+4x-4=0\text{⇔}3x^2+6x-2x-4=\left(3x-2\right)\left(x+2\right)=0\text{⇔}\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)
Vậy nghiệm của đa thức : \(S=\left\{\dfrac{2}{3};-2\right\}\)
b)
\(x^2+x-20=0⇔\left(x-4\right)\left(x+5\right)=0\text{⇔}\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)
d)
\(3x^2-4x-7=0\text{⇔}\left(3x-7\right)\left(x+1\right)=0\text{⇔}\left[{}\begin{matrix}x=-1\\x=\dfrac{7}{3}\end{matrix}\right.\)
e)
\(5x^2-16x+3\text{⇔}\left(x-3\right)\left(5x-1\right)=0\text{⇔}\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)
f)
\(x^2+3x-10=0\text{⇔}\left(x-2\right)\left(x+5\right)=0\text{⇔}\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
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