\(-2x^2-8x=0\)

\(\Leftrightarrow-2x\left(x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x=0\\x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-4\end{cases}}}\)

#H

1. B

2. A

3. A

4. B 

5. C

6. B

7. B

8. D

9. C

10. D

A d c b c

26. A

27. D

28. C

29. B

30. C

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

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

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

\(\Leftrightarrow\left(4x-3\right)\left(4x+3-x\right)=0\)

=>(4x-3)(3x+3)=0

=>x=3/4 hoặc x=-1

uầy bn cx bt biệt đội sinh tố à :) ? mik là fan zeros :)

và bn cx lên mạng tìm hiểu nha :)

thì ra là như zậy