Câu 20 trang 29 Sách bài tập (SBT) Toán 8 tập 1

Cộng các phân thức:

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

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

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

Giải:

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

\(\eqalign{  &  = {{z - x} \over {\left( {x - y} \right)\left( {y - z} \right)\left( {z - x} \right)}} + {{x - y} \over {\left( {x - y} \right)\left( {y - z} \right)\left( {z - x} \right)}} + {{y - z} \over {\left( {x - y} \right)\left( {y - z} \right)\left( {z - x} \right)}}  \cr  &  = {{z - x + x - y + y - z} \over {\left( {x - y} \right)\left( {y - z} \right)\left( {z - x} \right)}} = 0 \cr} \)

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

\(\eqalign{  &  = {{ - 4} \over {\left( {y - x} \right)\left( {x - z} \right)}} + {3 \over {\left( {y - x} \right)\left( {y - z} \right)}} + {3 \over {\left( {y - z} \right)\left( {x - z} \right)}}  \cr  &  = {{ - 4\left( {y - z} \right)} \over {\left( {x - z} \right)\left( {y - z} \right)\left( {y - x} \right)}} + {{3\left( {x - z} \right)} \over {\left( {x - z} \right)\left( {y - z} \right)\left( {y - x} \right)}} + {{3\left( {y - x} \right)} \over {\left( {x - z} \right)\left( {y - z} \right)\left( {y - x} \right)}}  \cr  &  = {{ - 4y + 4z + 3x - 3z + 3y - 3x} \over {\left( {x - z} \right)\left( {y - z} \right)\left( {y - x} \right)}} = {{z - y} \over {\left( {x - z} \right)\left( {y - z} \right)\left( {y - x} \right)}}  \cr  &  = {{ - \left( {y - z} \right)} \over {\left( {x - z} \right)\left( {y - z} \right)\left( {y - x} \right)}} = {{ - 1} \over {\left( {x - z} \right)\left( {y - x} \right)}} = {1 \over {\left( {x - z} \right)\left( {x - y} \right)}} \cr} \)

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

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

Sachbaitap.com