I = ( x+x^3/1-x^2 – x-x^3/ 1+x^2 ) : ( 1+x / 1-x – 1 -x / 1+x ) Rút gọn biểu thức trên

0 bình luận về “I = ( x+x^3/1-x^2 – x-x^3/ 1+x^2 ) : ( 1+x / 1-x – 1 -x / 1+x ) Rút gọn biểu thức trên”

1. `I = (( x+x^3)/(1-x^2 )- (x-x^3)/( 1+x^2) ) : ( (1+x )/ (1-x) -(1 -x) /( 1+x) )(text(ĐKXĐ:)xne{0;+-1}) `

   `I = ( x(1+x^2)(1+x^2)-x(1-x^2)(1-x^2))/((1-x^2 )( 1+x^2)) : ( (1+x )(1+x)-(1-x)(1-x))/( (1-x) ( 1+x) ) `

   `I = ( x(1+2x^2+x^4)-x(1-2x^2+x^4))/((1-x^2 )( 1+x^2)) : ( 1+2x+x^2-(1-2x+x^2))/( 1-x^2)  `

   `I = ( x+2x^3+x^5-x+2x^3-x^5)/((1-x^2 )( 1+x^2)) *(1-x^2)/ ( 1+2x+x^2-1+2x-x^2) `

   `I = (4x^3)/((1-x^2 )( 1+x^2)) *(1-x^2)/ ( 4x) `

   `I = (x^2)/( 1+x^2) `

2. Giải thích các bước giải:

    ĐKXĐ: $x \ne \left\{ {0; \pm 1} \right\}$

   Ta có

   $\begin{array}{l}
   I = \left( {\dfrac{{x + {x^3}}}{{1 – {x^2}}} – \dfrac{{x – {x^3}}}{{1 + {x^2}}}} \right):\left( {\dfrac{{1 + x}}{{1 – x}} – \dfrac{{1 – x}}{{1 + x}}} \right)\\
    = \dfrac{{\left( {x + {x^3}} \right)\left( {1 + {x^2}} \right) – \left( {x – {x^3}} \right)\left( {1 – {x^2}} \right)}}{{\left( {1 – {x^2}} \right)\left( {1 + {x^2}} \right)}}:\dfrac{{{{\left( {1 + x} \right)}^2} – {{\left( {1 – x} \right)}^2}}}{{\left( {1 – x} \right)\left( {1 + x} \right)}}\\
    = \dfrac{{x\left( {{{\left( {1 + {x^2}} \right)}^2} – {{\left( {1 – {x^2}} \right)}^2}} \right)}}{{\left( {1 – {x^2}} \right)\left( {1 + {x^2}} \right)}}.\dfrac{{\left( {1 – x} \right)\left( {1 + x} \right)}}{{4x}}\\
    = \dfrac{{x.4{x^2}}}{{\left( {1 – x} \right)\left( {1 + x} \right)\left( {1 + {x^2}} \right)}}.\dfrac{{\left( {1 – x} \right)\left( {1 + x} \right)}}{{4x}}\\
    = \dfrac{{{x^2}}}{{1 + {x^2}}}
   \end{array}$

   Vậy $I = \dfrac{{{x^2}}}{{1 + {x^2}}}$ với $x \ne \left\{ {0; \pm 1} \right\}$