Hãy chứng tỏ các phân thức sau bằng nhau
a/ \(\dfrac{x+3}{2x-5}=\dfrac{x^2+3x}{2x^2-5x}\)
b/ \(\dfrac{3-x}{x+3}=\dfrac{x^2-6x+9}{9-x^{ }}\)
c/ \(\dfrac{x^3+64}{\left(3-x\right)\left(x^2-4x+16\right)}\)\(=\dfrac{x-4}{x-3}\)
d/ \(\dfrac{x^3+6x^2-x-30}{x^3+3x^2-25x-75}=\dfrac{x-2}{x-5}\)
AI GIÚP MK VS Ạ AI NHANH MK SẼ VOTE Ạ
\(a,VP=\dfrac{x\left(x+3\right)}{x\left(2x-5\right)}=\dfrac{x+3}{2x-5}=VT\\ b,VP=\dfrac{\left(3-x\right)^2}{\left(3-x\right)\left(3+x\right)}=\dfrac{3-x}{x+3}=VT\\ c,VP=\dfrac{\left(x+4\right)\left(x^2-4x+16\right)}{\left(3-x\right)\left(x^2-4x+16\right)}=\dfrac{x+4}{3-x}=VP\left(bạn.sửa.lại.đề.đi\right)\\ d,VT=\dfrac{x^3-2x^2+8x^2-16x+15x-30}{x^3-5x^2+8x^2-40x+15x-75}\\ =\dfrac{\left(x-2\right)\left(x^2+8x+15\right)}{\left(x-5\right)\left(x^2+8x+15\right)}=\dfrac{x-2}{x-5}=VP\)