Cho biểu thức A = 2x/x^2+2x – 8/x^2-4 + 2/x-2 (x khác 0 , x khác _+ 2) giúp mik vs 29/07/2021 Bởi Harper Cho biểu thức A = 2x/x^2+2x – 8/x^2-4 + 2/x-2 (x khác 0 , x khác _+ 2) giúp mik vs
Giải thích các bước giải: $\begin{array}{l} \frac{{2x}}{{{x^2} + 2x}} – \frac{8}{{{x^2} – 4}} + \frac{2}{{x – 2}}\\ = \frac{{2x}}{{x(x + 2)}} – \frac{8}{{{x^2} – 4}} + \frac{2}{{x – 2}}\\ = \frac{2}{{x + 2}} – \frac{8}{{{x^2} – 4}} + \frac{2}{{x – 2}}\\ = \frac{{2(x – 2) – 8 + 2(x + 2)}}{{(x + 2)(x – 2)}}\\ = \frac{{2x – 4 – 8 + 2x + 4}}{{(x – 2)(x + 2)}}\\ = \frac{{4x – 8}}{{(x – 2)(x + 2)}}\\ = \frac{{4(x – 2)}}{{(x – 2)(x + 2)}} = \frac{4}{{x + 2}} \end{array}$ Bình luận
` A = ( 2x)/(x^2+2x) -8/(x^2-4) +2/(x-2)` ` = (2x)/(x(x+2) – 8/((x-2)(x+2)) +2/(x-2)` ` = ( 2x(x-2))/(x(x-2)(x+2)) – (8x)/(x(x-2)(x+2)) + (2x(x+2))/(x(x-2)(x+2))` ` = ( 2x^2 – 4x – 8x + 2x^2+4x)/(x(x-2)(x+2))` ` = ( 4x^2 -8x)/(x(x-2)(x+2))` ` = ( 4x(x-2))//(x(x-2)(x+2))` ` = (4x)/(x(x+2))` `= 4/(x+2)` Bình luận
Giải thích các bước giải:
$\begin{array}{l} \frac{{2x}}{{{x^2} + 2x}} – \frac{8}{{{x^2} – 4}} + \frac{2}{{x – 2}}\\ = \frac{{2x}}{{x(x + 2)}} – \frac{8}{{{x^2} – 4}} + \frac{2}{{x – 2}}\\ = \frac{2}{{x + 2}} – \frac{8}{{{x^2} – 4}} + \frac{2}{{x – 2}}\\ = \frac{{2(x – 2) – 8 + 2(x + 2)}}{{(x + 2)(x – 2)}}\\ = \frac{{2x – 4 – 8 + 2x + 4}}{{(x – 2)(x + 2)}}\\ = \frac{{4x – 8}}{{(x – 2)(x + 2)}}\\ = \frac{{4(x – 2)}}{{(x – 2)(x + 2)}} = \frac{4}{{x + 2}} \end{array}$
` A = ( 2x)/(x^2+2x) -8/(x^2-4) +2/(x-2)`
` = (2x)/(x(x+2) – 8/((x-2)(x+2)) +2/(x-2)`
` = ( 2x(x-2))/(x(x-2)(x+2)) – (8x)/(x(x-2)(x+2)) + (2x(x+2))/(x(x-2)(x+2))`
` = ( 2x^2 – 4x – 8x + 2x^2+4x)/(x(x-2)(x+2))`
` = ( 4x^2 -8x)/(x(x-2)(x+2))`
` = ( 4x(x-2))//(x(x-2)(x+2))`
` = (4x)/(x(x+2))`
`= 4/(x+2)`