Bài 6: Phép trừ các phân thức đại số

\(=\dfrac{5(x+4)}{(x+4)(x-4)}+\dfrac{4(x-4)}{(x+4)(x-4)}-\dfrac{2x-24}{(x+4)(x-4)} \\=\dfrac{7x+28}{(x+4)(x-4)} \\=\dfrac{7(x+4)}{(x+4)(x-4)} \\=\dfrac{7}{x-4}\)

b: \(=\dfrac{5x+20+4x-16-2x+24}{\left(x-4\right)\left(x+4\right)}=\dfrac{7x+28}{\left(x-4\right)\left(x+4\right)}=\dfrac{7}{x-4}\)

\(\dfrac{5}{x-4}+\dfrac{4}{x+4}-\dfrac{2x-24}{x^2-16}=\dfrac{5}{x-4}+\dfrac{4}{x+4}-\dfrac{2x-24}{\left(x-4\right)\left(x+4\right)}=\dfrac{5\text{(x+4)+4(x-4)-2(x-12)}}{\left(x-4\right)\left(x+4\right)}=\dfrac{5x+20+4x-16-\left(2x-24\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{7x+28}{\left(x-4\right)\left(x+4\right)}=\dfrac{7\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{7}{\left(x-4\right)}\)

Bài nào ạ

Giúp tui với mn ơi

\(=\dfrac{8x+9-5x+6}{3x\left(x+5\right)}=\dfrac{3x+15}{3x\left(x+5\right)}=\dfrac{3\left(x+5\right)}{3x\left(x+5\right)}=\dfrac{1}{x}\)

\(=\dfrac{1}{8}\cdot24^3-\dfrac{3}{2}\cdot24^2-2\cdot24=816\)

Giúp tôi với mn

b: \(=\dfrac{x^3+6x^2-25}{x\left(x+5\right)\left(x-2\right)}+\dfrac{x+5}{x\left(x-2\right)}\)

\(=\dfrac{x^3+6x^2-25+x^2+10x+25}{x\left(x+5\right)\left(x-2\right)}=\dfrac{x^3+7x^2+10x}{x\left(x+5\right)\left(x-2\right)}=\dfrac{x+2}{x-2}\)

\(VP=\dfrac{1}{2}\cdot\dfrac{a+2-a}{a\left(a+2\right)}=\dfrac{2}{2a\left(a+2\right)}=\dfrac{1}{a\left(a+2\right)}=VT\)

 = \(\dfrac{x}{x\left(x-2\right)}-\dfrac{x-2}{x\left(x-2\right)}\) = \(\dfrac{x-x+2}{x\left(x-2\right)}=\dfrac{2}{x\left(x-2\right)}\)

Mình giải ra luôn chứ không ghi đề nữa nha bn! :D

\(=\dfrac{x}{x^2-2x}-\dfrac{x-2}{x^2-2x}=\dfrac{x-x+2}{x^2-2x}=\dfrac{2}{x\left(x-2\right)}\)