Bài 2:
2x(x-5)+9x-6=0
=>\(2x^2-10x+9x-6=0\)
=>\(2x^2-x-6=0\)
=>\(2x^2-4x+3x-6=0\)
=>2x(x-2)+3(x-2)=0
=>(x-2)(2x+3)=0
=>\(\left[{}\begin{matrix}x-2=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\)
Bài 4:
a:\(\text{Δ}=\left(-2\right)^2-4\cdot3\cdot\left(-7\right)=4+12\cdot7=4+84=88>0\)
=>Phương trình luôn có hai nghiệm phân biệt
b: Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{2}{3}\\x_1x_2=\dfrac{c}{a}=-\dfrac{7}{3}\end{matrix}\right.\)
\(A=\dfrac{x_1}{x_2-2}+\dfrac{x_2}{x_1-2}=\dfrac{x_1\left(x_1-2\right)+x_2\left(x_2-2\right)}{\left(x_1-2\right)\left(x_2-2\right)}\)
\(=\dfrac{\left(x_1^2+x_2^2\right)-2\left(x_1+x_2\right)}{x_1x_2-2\left(x_1+x_2\right)+4}=\dfrac{\left(x_1+x_2\right)^2-2x_1x_2-2\left(x_1+x_2\right)}{x_1x_2-2\left(x_1+x_2\right)+4}\)
\(=\dfrac{\left(\dfrac{2}{3}\right)^2-2\cdot\dfrac{-7}{3}-2\cdot\dfrac{2}{3}}{-\dfrac{7}{3}-2\cdot\dfrac{2}{3}+4}=\dfrac{\dfrac{4}{9}+\dfrac{14}{3}-\dfrac{4}{3}}{-\dfrac{7}{3}-\dfrac{4}{3}+4}\)
\(=\dfrac{\dfrac{4}{9}+\dfrac{10}{3}}{\dfrac{1}{3}}=\dfrac{34}{9}\cdot3=\dfrac{34}{3}\)
