Bài 2:
a: \(3x^2-7x-10=0\)
=>\(3x^2-10x+3x-10=0\)
=>(3x-10)(x+1)=0
=>\(\left[{}\begin{matrix}3x-10=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{10}{3}\\x=-1\end{matrix}\right.\)
b: \(x^2-3x+2=0\)
=>\(x^2-x-2x+2=0\)
=>x(x-1)-2(x-1)=0
=>(x-1)(x-2)=0
=>\(\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
c: \(3x^2-2\sqrt{3}x-3=0\)
=>\(3x^2-3\sqrt{3}\cdot x+\sqrt{3}\cdot x-3=0\)
=>\(3x\left(x-\sqrt{3}\right)+\sqrt{3}\left(x-\sqrt{3}\right)=0\)
=>\(\left(x-\sqrt{3}\right)\left(3x+\sqrt{3}\right)=0\)
=>\(\left[{}\begin{matrix}x-\sqrt{3}=0\\3x+\sqrt{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}\\x=-\dfrac{\sqrt{3}}{3}\end{matrix}\right.\)
d: \(x^2-\left(\sqrt{2}+1\right)x+\sqrt{2}=0\)
=>\(x^2-x\sqrt{2}-x+\sqrt{2}=0\)
=>\(x\left(x-\sqrt{2}\right)-\left(x-\sqrt{2}\right)=0\)
=>\(\left(x-\sqrt{2}\right)\left(x-1\right)=0\)
=>\(\left[{}\begin{matrix}x-\sqrt{2}=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=1\end{matrix}\right.\)
e: \(\sqrt{3}\cdot x^2-\left(1-\sqrt{3}\right)x-1=0\)
=>\(x^2\cdot\sqrt{3}-x+\sqrt{3}\cdot x-1=0\)
=>\(x\left(x\sqrt{3}-1\right)+\left(x\sqrt{3}-1\right)=0\)
=>\(\left(x\sqrt{3}-1\right)\left(x+1\right)=0\)
=>\(\left[{}\begin{matrix}x\sqrt{3}-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{\sqrt{3}}\\x=-1\end{matrix}\right.\)
f: \(x^2-4x-5=0\)
=>\(x^2-5x+x-5=0\)
=>(x-5)(x+1)=0
=>\(\left[{}\begin{matrix}x-5=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
g: \(\left(2+\sqrt{3}\right)x^2-2\sqrt{3}x-2+\sqrt{3}=0\)
\(\text{Δ}=\left(-2\sqrt{3}\right)^2-4\left(2+\sqrt{3}\right)\left(-2+\sqrt{3}\right)\)
\(=12+4\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)=16>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{2\sqrt{3}-4}{2\left(2+\sqrt{3}\right)}=-7+4\sqrt{3}\\x_2=\dfrac{2\sqrt{3}+4}{2\left(2+\sqrt{3}\right)}=1\end{matrix}\right.\)
h: \(x^2-\left|x\right|-6=0\)
=>\(\left(\left|x\right|\right)^2-\left|x\right|-6=0\)
=>\(\left(\left|x\right|-3\right)\left(\left|x\right|+2\right)=0\)
=>\(\left|x\right|-3=0\)
=>|x|=3
=>\(\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
