Bài 2:
a: DKXĐ: x<>5; x<>-5
b: \(B=\dfrac{\left(x-5\right)^2}{\left(x-5\right)\left(x+5\right)}=\dfrac{x-5}{x+5}=\dfrac{-1-5}{-1+5}=\dfrac{-6}{4}=-\dfrac{3}{2}\)
Bài 1
Thay x = -2 và y = 3 vào A, ta có:
A = 2.(-2) - 3.(-2).32 + 1
= -4 + 54 + 1
= 51
Bài 3
\(C=\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
ĐKXĐ: \(x\ne0;x\ne3;x\ne-3\)
\(C=\left(\dfrac{9}{x\left(x+3\right)\left(x-3\right)}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x\left(x+3\right)}-\dfrac{x}{3\left(x+3\right)}\right)\)
\(=\left[\dfrac{9+x\left(x-3\right)}{x\left(x+3\right)\left(x-3\right)}\right]:\left[\dfrac{3\left(x-3\right)-x.x}{3x\left(x+3\right)}\right]\)
\(=\dfrac{x^2-3x+9}{x\left(x+3\right)\left(x-3\right)}:\dfrac{-x^2+3x-9}{3x\left(x-3\right)}\)
\(=\dfrac{x^2-3x+9}{x\left(x+3\right)\left(x-3\right)}.\dfrac{-3x\left(x-3\right)}{x^2-3x+9}\)
\(=\dfrac{-3}{x+3}\)
còn bài 4 nữa ah , vẽ hình luôn ah
