`C = a + ( 2a + x )/( 2 – x ) – ( 2a – x )/( 2 + x ) + 4x( x^2 – 4 )` với `x = a( a + 1 )` Tính giá trị của `C`

2 bình luận về “`C = a + ( 2a + x )/( 2 – x ) – ( 2a – x )/( 2 + x ) + 4x( x^2 – 4 )` với `x = a( a + 1 )` Tính giá trị của `C`”

1. Giải đáp + Lời giải và giải thích chi tiết:

   $C=a+{\displaystyle \frac{2a+x}{2–x}}–{\displaystyle \frac{2a–x}{2+x}}+{\displaystyle \frac{4a}{x^2–4}}$

   $=a+{\displaystyle \frac{2a+x}{2–x}}–{\displaystyle \frac{2a–x}{2+x}}–{\displaystyle \frac{4a}{4–x^2}}$

   $=a+{\displaystyle \frac{2a+x}{2–x}}–{\displaystyle \frac{2a–x}{2+x}}–{\displaystyle \frac{4a}{(2+x)(2–x)}}$

   $=a+{\displaystyle \frac{(2a+x)(x+2)}{(2+x)(2–x)}}–{\displaystyle \frac{(2a–x)(2–x)}{(2+x)(2–x)}}–{\displaystyle \frac{4a}{(2+x)(2–x)}}$

   $=a+{\displaystyle \frac{2ax+x^2+4a+2x}{(2+x)(2–x)}}–{\displaystyle \frac{4a–2x–2ax+x^2}{(2+x)(2–x)}}–{\displaystyle \frac{4a}{(2+x)(2–x)}}$

   $=a+{\displaystyle \frac{2ax+x^2+4a+2x–4a+2x+2ax–x^2–4a}{(2+x)(2–x)}}$

   $=a+{\displaystyle \frac{4ax+4x–4a}{(2+x)(2–x)}}$

   $=a+{\displaystyle \frac{4ax+4x–4a}{(2+x)(2–x)}}$

   Thay $x={\displaystyle \frac{a}{a+1}}$, ta có:

   $C=a+{\displaystyle \frac{4a({\displaystyle \frac{a}{a+1}})+4({\displaystyle \frac{a}{a+1}})–4a}{(2+{\displaystyle \frac{a}{a+1}})(2–{\displaystyle \frac{a}{a+1}})}}$

   $=a+{\displaystyle \frac{{\displaystyle \frac{a(4a+4)}{a+1}}–{\displaystyle \frac{4a(a+1)}{a+1}}}{(2+{\displaystyle \frac{a}{a+1}})(2–{\displaystyle \frac{a}{a+1}})}}$

   $=a+{\displaystyle \frac{{\displaystyle \frac{4a(a+1)–4a(a+1)}{a+1}}}{(2+{\displaystyle \frac{a}{a+1}})(2–{\displaystyle \frac{a}{a+1}})}}$

   $=a+{\displaystyle \frac{0}{(2+{\displaystyle \frac{a}{a+1}})(2–{\displaystyle \frac{a}{a+1}})}}$

   $=a$
   Vậy $C=a$
   $\text{\#Vexi’s Return}$

   Trả lời
2. Giải đáp + Lời giải và giải thích chi tiết:

   Ta có:C = a + \frac{2a+x}{2-x} – \frac{2a-x}{2+x} + \frac{4a}{x^2-4}

   =a + \frac{(2a+x)(2+x) – (2a-x)(2-x)}{(2-x)(2+x)} + \frac{4a}{x^2-4}

   =a + \frac{4a+2ax+2x+x^2 – (4a – 2ax – 2x + x^2)}{(2-x)(2+x)} + \frac{4a}{x^2-4}

   =a + \frac{4ax + 4x}{4-a^2} – \frac{4a}{4-x^2}

   =a + \frac{4ax + 4x – 4a}{4-x^2}

   =a + \frac{4x(a + 1) – 4a}{4-x^2}

   Thay x = \frac{a}{a+1} vào ta được:

   a + \frac{4.\frac{a}{a+1}.(a + 1) – 4a}{4-(\frac{a}{a+1})^2}

   =a + \frac{4a – 4a}{4-(\frac{a}{a+1})^2}

   =a

   Vậy C=a

    

   Trả lời