Rút gọn:$\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right)\div \left(x-2+\dfrac{10-x^2}{x+2}\right)$

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1. Đáp án:

    

   Giải thích các bước giải:

    `( x/( x^2 – 4 ) + 2/( 2 – x ) + 1/( x + 2 )) : ( x – 2 + ( 10 – x^2 )/( x + 2 ) )`

   `= ( x/(( x + 2 )( x – 2 )) – 2/( x – 2 ) + 1/( x + 2 )) : ( (( x – 2 )( x + 2 ))/( x + 2 ) + ( 10 – x^2 )/( x + 2 ) )`

   ( Đk: x $\neq$ ± 2 )

   `= ( x/(( x + 2 )( x – 2 )) – (2( x + 2 ))/(( x – 2 )( x + 2 )) + (1( x – 2 ))/(( x + 2 )( x – 2 )) ) : ( ( x^2 – 4 )/( x` `+ 2 ) + ( 10 – x^2 )/( x + 2 ) )`

   `= ( x/(( x + 2 )( x – 2 )) – (2x + 4 )/(( x – 2 )( x + 2 )) + ( x – 2 )/(( x + 2 )( x – 2 )) ) :  6/( x + 2 )`

   `= -6/(( x + 2 )( x – 2 )) . ( x + 2 )/6`

   `= -1/( x – 2 )` 

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2. Đáp án:

    

   Giải thích các bước giải:

    `(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}):(x-2+\frac{10-x^2}{x+2})`

   ĐK: `x \ne +-2`

   `=[\frac{x}{(x-2)(x+2)}-\frac{2(x+2)}{(x-2)(x+2)}+\frac{x-2}{(x-2)(x+2)}]:[\frac{(x-2)(x+2)}{x+2}+\frac{10-x^2}{x+2}]`

   `=[\frac{x-2x-4+x-2}{(x-2)(x+2)}]:[\frac{x^2-4+10-x^2}{x+2}]`

   `=[\frac{-6}{(x-2)(x+2)}].[\frac{x+2}{6}]`

   `=\frac{-1}{x-2}`

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