b) `5x^2+4x-3=0`
`⇔5(x^2+\frac{4}{5}x-\frac{3}{5})=0`
`⇔5[(x+\frac{2}{5})^2-\frac{19}{25}]=0`
`⇔5(x+\frac{2}{5})^2-\frac{19}{5}=0`
`⇔`\(\left[ \begin{array}{l}x+\dfrac{2}{5}-\dfrac{\sqrt{19}}{5}=0\\x+\dfrac{2}{5}+\dfrac{\sqrt{19}}{5}=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=\dfrac{\sqrt{19}-2}{5}\\x=-\dfrac{\sqrt{19}-2}{5}\end{array} \right.\)
Vậy `S={\frac{\sqrt{19}-2}{5};-\frac{\sqrt{19}-2}{5}}`
c) `2x^2+3x-5=0`
`⇔2x^2+5x-2x-5=0`
`⇔x(2x+5)-(2x+5)=0`
`⇔(2x+5)(x-1)=0`
`⇔`\(\left[ \begin{array}{l}2x+5=0\\x-1=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}2x=-5\\x=1\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=-\dfrac{5}{2}\\x=1\end{array} \right.\)
Vậy `S={-\frac{5}{2};1}`
d) `4x^2-4x+1=0`
`⇔(2x-1)^2=0`
`⇔2x-1=0`
`⇔2x=1`
`⇔x=\frac{1}{2}`
Vậy `S={\frac{1}{2}}`
e) `x^2+2x+4=0`
`⇔x^2+2x+1+3=0`
`⇔(x+1)^2+3=0`
`⇔(x+1)^2=-3`
Vì `(x+1)^2≥0∀x`
Mà `(x+1)^2=-3` (vô lí)
Vậy phương trình vô nghiệm.
