Bài 1:
a) \(\Delta=b^2-4ac=\left(-5\right)^2-4\cdot2\cdot1=25-8=17\)
Vì Δ>0 nên phương trình \(2x^2-5x+1=0\) có hai nghiệm là:
\(\left\{{}\begin{matrix}x_1=\frac{-b-\sqrt{\Delta}}{2a}\\x_2=\frac{-b+\sqrt{\Delta}}{2a}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=\frac{5-\sqrt{17}}{2\cdot2}=\frac{5-\sqrt{17}}{4}\\x_2=\frac{5+\sqrt{17}}{2\cdot2}=\frac{5+\sqrt{17}}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{5-\sqrt{17}}{4};\frac{5+\sqrt{17}}{4}\right\}\)
b) Ta có: \(4x^2+4x+1=0\)
\(\Leftrightarrow\left(2x+1\right)^2=0\)
\(\Leftrightarrow2x+1=0\)
\(\Leftrightarrow2x=-1\)
hay \(x=-\frac{1}{2}\)
Vậy: \(S=\left\{\frac{-1}{2}\right\}\)
c) Ta có: \(-3x^2+2x+8=0\)
\(\Leftrightarrow-3x^2+6x-4x+8=0\)
\(\Leftrightarrow-3x\left(x-2\right)-4\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(-3x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\-3x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\-3x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{-4}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{2;\frac{-4}{3}\right\}\)
d) Ta có: \(5x^2-6x-1=0\)
\(\Delta=b^2-4\cdot a\cdot c=\left(-6\right)^2-4\cdot5\cdot\left(-1\right)=56\)
Vì Δ>0 nên phương trình \(5x^2-6x-1=0\) có hai nghiệm là:
\(\left\{{}\begin{matrix}x_1=\frac{-b-\sqrt{\Delta}}{2a}\\x_2=\frac{-b+\sqrt{\Delta}}{2a}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=\frac{6-\sqrt{56}}{2\cdot5}=\frac{3-\sqrt{14}}{5}\\x_2=\frac{6+\sqrt{56}}{2\cdot5}=\frac{3+\sqrt{14}}{5}\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{3-\sqrt{14}}{5};\frac{3+\sqrt{14}}{5}\right\}\)
e) Ta có: \(-3x^2+14x-8=0\)
\(\Leftrightarrow-3x^2+12x+2x-8=0\)
\(\Leftrightarrow-3x\left(x-4\right)+2\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(-3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\-3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\-3x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\frac{2}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{4;\frac{2}{3}\right\}\)
g) Ta có: \(-7x^2+4x-3=0\)
\(\Delta=b^2-4ac=4^2-4\cdot\left(-7\right)\cdot\left(-3\right)=-68\)
Vì Δ<0 nên phương trình \(-7x^2+4x-3=0\) không có nghiệm
Vậy: S=∅
