Đáp án:

f) \(\dfrac{{x + 3}}{{{{\left( {x - 1} \right)}^2}}}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\dfrac{{x + 1 - x}}{{x\left( {x + 1} \right)}} = \dfrac{1}{{x\left( {x + 1} \right)}}\\
c)\dfrac{{7x + 6}}{{2x\left( {x + 7} \right)}} - \dfrac{{3x + 6}}{{2x\left( {x + 7} \right)}}\\
 = \dfrac{{7x + 6 - 3x - 6}}{{2x\left( {x + 7} \right)}}\\
 = \dfrac{{4x}}{{2x\left( {x + 7} \right)}} = \dfrac{2}{{x + 7}}\\
e)\dfrac{{\left( {x + 1} \right)\left( {x + 3} \right) - \left( {1 - x} \right)\left( {x - 3} \right) + 2x\left( {1 - x} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
 = \dfrac{{{x^2} + 4x + 3 + {x^2} - 4x + 3 + 2x - 2{x^2}}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
 = \dfrac{{2x + 6}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = \dfrac{2}{{x - 3}}\\
f)\dfrac{{\left( {3x + 1} \right)\left( {x + 1} \right) - {{\left( {x - 1} \right)}^2} - \left( {x + 3} \right)\left( {x - 1} \right)}}{{{{\left( {x - 1} \right)}^2}\left( {x + 1} \right)}}\\
 = \dfrac{{3{x^2} + 4x + 1 - {x^2} + 2x - 1 - {x^2} - 2x + 3}}{{{{\left( {x - 1} \right)}^2}\left( {x + 1} \right)}}\\
 = \dfrac{{{x^2} + 4x + 3}}{{{{\left( {x - 1} \right)}^2}\left( {x + 1} \right)}} = \dfrac{{\left( {x + 1} \right)\left( {x + 3} \right)}}{{{{\left( {x - 1} \right)}^2}\left( {x + 1} \right)}}\\
 = \dfrac{{x + 3}}{{{{\left( {x - 1} \right)}^2}}}
\end{array}\)

\(\begin{array}{l}
d)\dfrac{1}{{x\left( {1 - 5x} \right)}} - \dfrac{{25x - 15}}{{\left( {5x - 1} \right)\left( {5x + 1} \right)}}\\
 = \dfrac{{ - 5x - 1 - 25{x^2} + 15x}}{{x\left( {5x - 1} \right)\left( {5x + 1} \right)}}\\
 = \dfrac{{ - 25{x^2} + 10x - 1}}{{x\left( {5x - 1} \right)\left( {5x + 1} \right)}}
\end{array}\)

Đáp án:

`↓↓` 

Giải thích các bước giải:

`a) 1/x-1/(x+1)`

`=(x+1)/(x(x+1)) - x/(x(x+1))=(x+1-x)/(x(x+1))=1/(x(x+1))`

`b) 1/(xy-x^2)-1/(y^2-xy)`

`=1/(x(y-x))-1/(y(y-x))=y/(xy(y-x))-x/(xy(y-x))=(y-x)/(xy(y-x))=1/(xy)`

`c) (7x+6)/(2x(x+7))-(3x+6)/(2x^2+14x)=(7x+6)/(2x(x+7))-(3x+6)/(2x(x+7))`

`=(7x+6-3x-6)/(2x(x+7))=(4x)/(2x(x+7))=2/(x+7)`

`d) 1/(x-5x^2)-(25x-15)/(25x^2-1)`

`=-1/(5x^2-x)-(25x-15)/((5x-1)(5x+1))=-1/(x(5x-1))-(25x-15)/((5x-1)(5x+1))`

`=(-1.(5x+1))/(x(5x-1)(5x+1))-((25x-15).x)/(x(5-1)(5x+1))=(-5x-1-25x^2+15x)/(x(5x-1)(5x+1))`

`=(-25x^2+10x-1)/(x(5x-1)(5x+1))`

`e) (x+1)/(x-3)-(1-x)/(x+3)-(2x(1-x))/(9-x^2)`

`=(x+1)/(x-3)-(1-x)/(x+3)-(2x(1-x))/(x^2-9)`

`=((x+1)(x+3))/((x-3)(x+3))-((1-x)(x-3))/((x-3)(x+3))-(2x(1-x))/((x-3)(x+3))`

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

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

`=(2x+6)/((x-3)(x+3))=(2(x+3))/((x-3)(x+3))=2/(x-3)`

`f) (3x+1)/(x-1)^2-1/(x+1)+(x+3)/(1-x^2)`

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

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

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

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