Câu 20 trang 29 SBT Toán 8 tập 1: Cộng các phân thức:...

Cộng các phân thức:

a. ({1 over {left( {x – y} ight)left( {y – z} ight)}} + {1 over {left( {y – z} ight)left( {z – x} ight)}} + {1 over {left( {z – x} ight)left( {x – y} ight)}})

b. ({4 over {left( {y – x} ight)left( {z – x} ight)}} + {3 over {left( {y – x} ight)left( {y – z} ight)}} + {3 over {left( {y – z} ight)left( {x – z} ight)}})

c. ({1 over {xleft( {x – y} ight)left( {x – z} ight)}} + {1 over {yleft( {y – z} ight)left( {y – x} ight)}} + {1 over {zleft( {z – x} ight)left( {z – y} ight)}})

Giải:

a. ({1 over {left( {x – y} ight)left( {y – z} ight)}} + {1 over {left( {y – z} ight)left( {z – x} ight)}} + {1 over {left( {z – x} ight)left( {x – y} ight)}})

(eqalign{  &  = {{z – x} over {left( {x – y} ight)left( {y – z} ight)left( {z – x} ight)}} + {{x – y} over {left( {x – y} ight)left( {y – z} ight)left( {z – x} ight)}} + {{y – z} over {left( {x – y} ight)left( {y – z} ight)left( {z – x} ight)}}  cr  &  = {{z – x + x – y + y – z} over {left( {x – y} ight)left( {y – z} ight)left( {z – x} ight)}} = 0 cr} )

b. ({4 over {left( {y – x} ight)left( {z – x} ight)}} + {3 over {left( {y – x} ight)left( {y – z} ight)}} + {3 over {left( {y – z} ight)left( {x – z} ight)}})

(eqalign{  &  = {{ – 4} over {left( {y – x} ight)left( {x – z} ight)}} + {3 over {left( {y – x} ight)left( {y – z} ight)}} + {3 over {left( {y – z} ight)left( {x – z} ight)}}  cr  &  = {{ – 4left( {y – z} ight)} over {left( {x – z} ight)left( {y – z} ight)left( {y – x} ight)}} + {{3left( {x – z} ight)} over {left( {x – z} ight)left( {y – z} ight)left( {y – x} ight)}} + {{3left( {y – x} ight)} over {left( {x – z} ight)left( {y – z} ight)left( {y – x} ight)}}  cr  &  = {{ – 4y + 4z + 3x – 3z + 3y – 3x} over {left( {x – z} ight)left( {y – z} ight)left( {y – x} ight)}} = {{z – y} over {left( {x – z} ight)left( {y – z} ight)left( {y – x} ight)}}  cr  &  = {{ – left( {y – z} ight)} over {left( {x – z} ight)left( {y – z} ight)left( {y – x} ight)}} = {{ – 1} over {left( {x – z} ight)left( {y – x} ight)}} = {1 over {left( {x – z} ight)left( {x – y} ight)}} cr} )

c. ({1 over {xleft( {x – y} ight)left( {x – z} ight)}} + {1 over {yleft( {y – z} ight)left( {y – x} ight)}} + {1 over {zleft( {z – x} ight)left( {z – y} ight)}})

(eqalign{  &  = {1 over {xleft( {x – y} ight)left( {x – z} ight)}} + {1 over {yleft( {x – y} ight)left( {y – z} ight)}} + {1 over {zleft( {x – z} ight)left( {y – z} ight)}}  cr  &  = {{yzleft( {y – z} ight)} over {xyzleft( {x – y} ight)left( {x – z} ight)left( {y – z} ight)}} + {{ – xzleft( {x – z} ight)} over {xyzleft( {x – y} ight)left( {x – z} ight)left( {y – z} ight)}} + {{xyleft( {x – y} ight)} over {xyzleft( {x – y} ight)left( {x – z} ight)left( {y – z} ight)}}  cr  &  = {{{y^2}z – y{z^2} – {x^2}z + x{z^2} + {x^2}y – x{y^2}} over {xyzleft( {x – y} ight)left( {x – z} ight)left( {y – z} ight)}} = {{{z^2}left( {x – y} ight) + xyleft( {x – y} ight) – zleft( {x – y} ight)left( {x + y} ight)} over {xyzleft( {x – y} ight)left( {x – z} ight)left( {y – z} ight)}}  cr  &  = {{left( {x – y} ight)left( {{z^2} + xy – xz – yz} ight)} over {xyzleft( {x – y} ight)left( {x – z} ight)left( {y – z} ight)}} = {{left( {x – y} ight)left[ {xleft( {y – z} ight) – zleft( {y – z} ight)} ight]} over {xyzleft( {x – y} ight)left( {x – z} ight)left( {y – z} ight)}}  cr  &  = {{left( {x – y} ight)left( {y – z} ight)left( {x – z} ight)} over {xyzleft( {x – y} ight)left( {x – z} ight)left( {y – z} ight)}} = {1 over {xyz}} cr} )