a: \(4x^2-9=0\)
=>\(4x^2=9\)
=>\(x^2=\dfrac{9}{4}\)
=>\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
d: \(x^2-7x+10=0\)
=>(x-2)(x-5)=0
=>\(\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
g: \(4x^2+28x+49=0\)
=>\(\left(2x\right)^2+2\cdot2x\cdot7+7^2=0\)
=>\(\left(2x+7\right)^2=0\)
=>2x+7=0
=>2x=-7
=>\(x=-\dfrac{7}{2}\)
j: \(3x^2-5x-1=0\)
\(\text{Δ}=\left(-5\right)^2-4\cdot3\cdot\left(-1\right)=25+12=37>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left[{}\begin{matrix}x=\dfrac{5-\sqrt{37}}{2\cdot3}=\dfrac{5-\sqrt{37}}{6}\\x=\dfrac{5+\sqrt{37}}{6}\end{matrix}\right.\)
b: \(2x^2+5=0\)
=>\(2x^2=-5\)
=>\(x^2=-\dfrac{5}{2}\)(vô lý)
=>\(x\in\varnothing\)
e: \(2x^2-5x-13=0\)
\(\text{Δ}=\left(-5\right)^2-4\cdot2\cdot\left(-13\right)=25+104=129>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left[{}\begin{matrix}x=\dfrac{5-\sqrt{129}}{2\cdot2}=\dfrac{5-\sqrt{129}}{4}\\x=\dfrac{5+\sqrt{129}}{4}\end{matrix}\right.\)
h: \(x^2-4x-21=0\)
=>\(x^2-7x+3x-21=0\)
=>(x-7)(x+3)=0
=>\(\left[{}\begin{matrix}x-7=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
k: \(-3x^2+5x-3=0\)
\(\text{Δ}=5^2-4\cdot\left(-3\right)\cdot\left(-3\right)=25-36=-11< 0\)
=>Phương trình vô nghiệm
