Đáp án:

\(\begin{array}{l}
a,\\
\dfrac{{{x^2} + 1}}{{{x^2}}}\\
b,\\
\dfrac{{x - 3}}{{x\left( {x + 4} \right)}}\\
c,\\
\dfrac{{8 - y}}{{y - 2}}
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
a,\\
\dfrac{{7{x^2} + 7}}{{7{x^2}}} = \dfrac{{7\left( {{x^2} + 1} \right)}}{{7{x^2}}} = \dfrac{{{x^2} + 1}}{{{x^2}}}\\
b,\\
\dfrac{{{x^2} - 7x + 12}}{{{x^3} - 16x}} = \dfrac{{\left( {{x^2} - 3x} \right) + \left( { - 4x + 12} \right)}}{{x\left( {{x^2} - 16} \right)}}\\
 = \dfrac{{\left( {{x^2} - 3x} \right) - \left( {4x - 12} \right)}}{{x\left( {{x^2} - {4^2}} \right)}} = \dfrac{{x.\left( {x - 3} \right) - 4.\left( {x - 3} \right)}}{{x\left( {x - 4} \right)\left( {x + 4} \right)}}\\
 = \dfrac{{\left( {x - 3} \right)\left( {x - 4} \right)}}{{x\left( {x - 4} \right)\left( {x + 4} \right)}} = \dfrac{{x - 3}}{{x\left( {x + 4} \right)}}\\
c,\\
\dfrac{{9 - {{\left( {y - 5} \right)}^2}}}{{{y^2} - 4y + 4}} = \dfrac{{{3^2} - {{\left( {y - 5} \right)}^2}}}{{{y^2} - 2.y.2 + {2^2}}}\\
 = \dfrac{{\left[ {3 - \left( {y - 5} \right)} \right].\left[ {3 + \left( {y - 5} \right)} \right]}}{{{{\left( {y - 2} \right)}^2}}}\\
 = \dfrac{{\left( {3 - y + 5} \right)\left( {3 + y - 5} \right)}}{{{{\left( {y - 2} \right)}^2}}}\\
 = \dfrac{{\left( {8 - y} \right).\left( {y - 2} \right)}}{{{{\left( {y - 2} \right)}^2}}}\\
 = \dfrac{{8 - y}}{{y - 2}}
\end{array}\)

Đáp án `+` Giải thích các bước giải `!`

`a)`

`(7x^2+7)/(7x^2)` `(x \ne 0)`

`= (7(x^2+1))/(7x^2)`

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

`b)`

`(x^2-7x+12)/(x^3-16x)` `(x \ne 0; x\ne +-4)`

`= (x^2-3x-4x+12)/(x(x^2-16))`

`= ((x^2-3x)-(4x-12))/(x(x-4)(x+4))`

`= (x(x-3)-4(x-3))/(x(x-4)(x+4))`

`= ((x-4)(x-3))/(x(x-4)(x+4))`

`= (x-3)/(x(x+4)`

`c)`

`(9-(y-5)^2)/(y^2-4y+4)` `(y \ne 2)`

`= ((3-y+5)(3+y-5))/((y-2)^2)`

`= ((8-y)(y-2))/((y-2)^2)`

`= (8-y)/(y-2)`