`Q = [(2x – x^2) / (2x^2 + 8) – (2x^2)/(8+ 2x^2 – 4x -x^3)][1-1/x -2/x^2]` 30/10/2021 Bởi Valerie `Q = [(2x – x^2) / (2x^2 + 8) – (2x^2)/(8+ 2x^2 – 4x -x^3)][1-1/x -2/x^2]`
Giải thích các bước giải : `↓↓↓` Rút gọn Q : Ta có : `Q = [ (2x-x^2)/(2x^2 +8) – (2x^2)/{8+2x^2 -4x-x^3} ][ 1 – 1/x – 2/{x^2} ]` `→= [ {x(2-x)}/{2(x^2 +4)} – (2x^2)/{(x^2 +4)(x-2)} ][ 2/{x^2} + {x-x^2}/(x^2) ]` `→=[{-x(x-2)^2 – 4x^2}/{2(x^2 + 4)(x-2)]][ {2x+x-x^2}/{x^2} ]` `→=[{-x^3 + 4x^2 -4x-4x^2}/{2(x^2+4)(x-2)} ][ {2+x^2-x^2}/{x^2} ]` `→=[{-x(x^2 +4)}/{2(x^2 +4)(x-2)} ][ {(x-2)(x+1)}/{x^2} ]` `→=[{x.4(x+1)}/{2.x^2}` `→=(2x+1)/{2x}` Bình luận
$Q=[\frac{2x-x^2}{2x^2+8}-\frac{2x^2}{8+2x^2-4x-x^3}](1-\frac{1}x-\frac{2}{x^2})$ $=[\frac{x(2-x)}{2(x^2+4)}-\frac{2x^2}{(2-x)(x^2+4)}].\frac{x^2-x-2}{x^2}$ $=\frac{x(2-x)^2-2.2x^2}{2(2-x)(x^2+4)}.\frac{(x+1)(x-2)}{x^2}$ $=\frac{4+x^2+4x-4x}{2(x^2+4)}.\frac{x+1}x=\frac{x+1}{2x}$ Bình luận
Giải thích các bước giải :
`↓↓↓`
Rút gọn Q :
Ta có :
`Q = [ (2x-x^2)/(2x^2 +8) – (2x^2)/{8+2x^2 -4x-x^3} ][ 1 – 1/x – 2/{x^2} ]`
`→= [ {x(2-x)}/{2(x^2 +4)} – (2x^2)/{(x^2 +4)(x-2)} ][ 2/{x^2} + {x-x^2}/(x^2) ]` `→=[{-x(x-2)^2 – 4x^2}/{2(x^2 + 4)(x-2)]][ {2x+x-x^2}/{x^2} ]`
`→=[{-x^3 + 4x^2 -4x-4x^2}/{2(x^2+4)(x-2)} ][ {2+x^2-x^2}/{x^2} ]`
`→=[{-x(x^2 +4)}/{2(x^2 +4)(x-2)} ][ {(x-2)(x+1)}/{x^2} ]`
`→=[{x.4(x+1)}/{2.x^2}`
`→=(2x+1)/{2x}`
$Q=[\frac{2x-x^2}{2x^2+8}-\frac{2x^2}{8+2x^2-4x-x^3}](1-\frac{1}x-\frac{2}{x^2})$
$=[\frac{x(2-x)}{2(x^2+4)}-\frac{2x^2}{(2-x)(x^2+4)}].\frac{x^2-x-2}{x^2}$
$=\frac{x(2-x)^2-2.2x^2}{2(2-x)(x^2+4)}.\frac{(x+1)(x-2)}{x^2}$
$=\frac{4+x^2+4x-4x}{2(x^2+4)}.\frac{x+1}x=\frac{x+1}{2x}$