Bài 20 trang 29 SBT toán 8 tập 1

Đề bài

Cộng các phân thức: 

a. \(\dfrac{1}{{\left( {x - y} \right)\left( {y - z} \right)}}\) + \(\dfrac{1 }{ {\left( {y - z} \right)\left( {z - x} \right)}}\) + \(\dfrac{1}{ {\left( {z - x} \right)\left( {x - y} \right)}}\) 

b. \(\dfrac{4}{{\left( {y - x} \right)\left( {z - x} \right)}} + \dfrac{3}{{\left( {y - x} \right)\left( {y - z} \right)}}\) + \(\dfrac{3 }{{\left( {y - z} \right)\left( {x - z} \right)}}\)

c. \(\dfrac{1}{ {x\left( {x - y} \right)\left( {x - z} \right)}} + \dfrac{1}{{y\left( {y - z} \right)\left( {y - x} \right)}}\) + \(\dfrac{1}{{z\left( {z - x} \right)\left( {z - y} \right)}}\)

Lời giải

a. \(\displaystyle{1 \over {\left( {x - y} \right)\left( {y - z} \right)}} + {1 \over {\left( {y - z} \right)\left( {z - x} \right)}}\) + \(\displaystyle {1 \over {\left( {z - x} \right)\left( {x - y} \right)}}\) 

 \(\displaystyle = {{z - x} \over {\left( {x - y} \right)\left( {y - z} \right)\left( {z - x} \right)}} \)\(\displaystyle + {{x - y} \over {\left( {x - y} \right)\left( {y - z} \right)\left( {z - x} \right)}}\)\(\displaystyle + {{y - z} \over {\left( {x - y} \right)\left( {y - z} \right)\left( {z - x} \right)}}\)\(\displaystyle  = {{z - x + x - y + y - z} \over {\left( {x - y} \right)\left( {y - z} \right)\left( {z - x} \right)}} = 0 \)

b. \(\displaystyle{4 \over {\left( {y - x} \right)\left( {z - x} \right)}}\)\(\displaystyle + {3 \over {\left( {y - x} \right)\left( {y - z} \right)}}\)\(\displaystyle + {3 \over {\left( {y - z} \right)\left( {x - z} \right)}}\)

\(\displaystyle = {{ - 4} \over {\left( {y - x} \right)\left( {x - z} \right)}}\)\(\displaystyle + {3 \over {\left( {y - x} \right)\left( {y - z} \right)}}\)\(\displaystyle + {3 \over {\left( {y - z} \right)\left( {x - z} \right)}}\)

\(\displaystyle = {{ - 4\left( {y - z} \right)} \over {\left( {x - z} \right)\left( {y - z} \right)\left( {y - x} \right)}}\)\(\displaystyle + {{3\left( {x - z} \right)} \over {\left( {x - z} \right)\left( {y - z} \right)\left( {y - x} \right)}}\)\(\displaystyle + {{3\left( {y - x} \right)} \over {\left( {x - z} \right)\left( {y - z} \right)\left( {y - x} \right)}}\)

\(\displaystyle= {{ - 4y + 4z + 3x - 3z + 3y - 3x} \over {\left( {x - z} \right)\left( {y - z} \right)\left( {y - x} \right)}}\)\(\displaystyle = {{z - y} \over {\left( {x - z} \right)\left( {y - z} \right)\left( {y - x} \right)}}\) 

\(\displaystyle= {{ - \left( {y - z} \right)} \over {\left( {x - z} \right)\left( {y - z} \right)\left( {y - x} \right)}}\) \(\displaystyle = {{ - 1} \over {\left( {x - z} \right)\left( {y - x} \right)}} = {1 \over {\left( {x - z} \right)\left( {x - y} \right)}}\)

c. \(\displaystyle{1 \over {x\left( {x - y} \right)\left( {x - z} \right)}}\)\(\displaystyle + {1 \over {y\left( {y - z} \right)\left( {y - x} \right)}}\)\(\displaystyle + {1 \over {z\left( {z - x} \right)\left( {z - y} \right)}}\)

\(\displaystyle = {1 \over {x\left( {x - y} \right)\left( {x - z} \right)}}\)\(\displaystyle - {1 \over {y\left( {x - y} \right)\left( {y - z} \right)}}\)\(\displaystyle + {1 \over {z\left( {x - z} \right)\left( {y - z} \right)}}\)

\(\displaystyle = {{yz\left( {y - z} \right)} \over {xyz\left( {x - y} \right)\left( {x - z} \right)\left( {y - z} \right)}}\)\(\displaystyle + {{ - xz\left( {x - z} \right)} \over {xyz\left( {x - y} \right)\left( {x - z} \right)\left( {y - z} \right)}}\)\(\displaystyle + {{xy\left( {x - y} \right)} \over {xyz\left( {x - y} \right)\left( {x - z} \right)\left( {y - z} \right)}}\)

\(\displaystyle = {{{y^2}z - y{z^2} - {x^2}z + x{z^2} + {x^2}y - x{y^2}} \over {xyz\left( {x - y} \right)\left( {x - z} \right)\left( {y - z} \right)}}\)\(\displaystyle = {{{z^2}\left( {x - y} \right) + xy\left( {x - y} \right) - z\left( {x - y} \right)\left( {x + y} \right)} \over {xyz\left( {x - y} \right)\left( {x - z} \right)\left( {y - z} \right)}}\)\(\displaystyle  = {{\left( {x - y} \right)\left( {{z^2} + xy - xz - yz} \right)} \over {xyz\left( {x - y} \right)\left( {x - z} \right)\left( {y - z} \right)}}\)\(\displaystyle = {{\left( {x - y} \right)\left[ {x\left( {y - z} \right) - z\left( {y - z} \right)} \right]} \over {xyz\left( {x - y} \right)\left( {x - z} \right)\left( {y - z} \right)}}\)\(\displaystyle  = {{\left( {x - y} \right)\left( {y - z} \right)\left( {x - z} \right)} \over {xyz\left( {x - y} \right)\left( {x - z} \right)\left( {y - z} \right)}} = {1 \over {xyz}}\)