Câu 3:
a: Thay x=1 vào (d1), ta được:
\(y=2\cdot1-5=2-5=-3=y_A\)
=>A(1;-3) thuộc (d1)
Thay x=-1 vào (d1), ta được:
\(y=2\cdot\left(-1\right)-5=-2-5=-7\)
=>B(-1;0) không thuộc (d1)
b: Phương trình hoành độ giao điểm là:
3x-4=2x-5
=>3x-2x=-5+4
=>x=-1
Khi x=-1 thì \(y=3\cdot\left(-1\right)-4=-7\)
vậy: Tọa độ giao điểm là C(-1;-7)
Câu 1:
a: \(A=\dfrac{6x+2}{8}+\dfrac{3-3x}{12}\)
\(=\dfrac{18x+6}{24}+\dfrac{6-6x}{24}=\dfrac{18x+6+6-6x}{24}=\dfrac{12x+12}{24}=\dfrac{x+1}{2}\)
b: \(B=\dfrac{3x+4y}{x-y}-\dfrac{x+6y}{x-y}\)
\(=\dfrac{3x+4y-x-6y}{x-y}\)
\(=\dfrac{2x-2y}{x-y}=2\)
c: \(C=\left(\dfrac{x+1}{1-x}-\dfrac{1-x}{1+x}+\dfrac{4x^2}{1-x^2}\right):\dfrac{4-4x^2}{1+2x+x^2}\)
\(=\left(\dfrac{-\left(x+1\right)}{x-1}+\dfrac{x-1}{x+1}-\dfrac{4x^2}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{-4\left(x^2-1\right)}{\left(x+1\right)^2}\)
\(=\dfrac{-\left(x+1\right)^2+\left(x-1\right)^2-4x^2}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x+1\right)}{-4\left(x-1\right)}\)
\(=\dfrac{-x^2-2x-1+x^2-2x+1-4x^2}{\left(x-1\right)}\cdot\dfrac{1}{-4\left(x-1\right)}\)
\(=\dfrac{-4x^2-4x}{-4\left(x-1\right)^2}=\dfrac{x^2+x}{\left(x-1\right)^2}\)
