3/(x+2)(x+5)+5/(x+5)(x+10)+7/(x+10)(x+17)=x/(x+2)(x+17) 01/10/2021 Bởi Kennedy 3/(x+2)(x+5)+5/(x+5)(x+10)+7/(x+10)(x+17)=x/(x+2)(x+17)
$$\eqalign{ & {3 \over {\left( {x + 2} \right)\left( {x + 5} \right)}} + {5 \over {\left( {x + 5} \right)\left( {x + 10} \right)}} + {7 \over {\left( {x + 10} \right)\left( {x + 17} \right)}} = {x \over {\left( {x + 2} \right)\left( {x + 17} \right)}}\,\,\left( {x \ne – 2;\,\,x \ne – 5;\,\,x \ne – 10;\,\,x \ne – 17} \right) \cr & \Leftrightarrow {{\left( {x + 5} \right) – \left( {x + 2} \right)} \over {\left( {x + 2} \right)\left( {x + 5} \right)}} + {{\left( {x + 10} \right) – \left( {x + 5} \right)} \over {\left( {x + 5} \right)\left( {x + 10} \right)}} + {{\left( {x + 17} \right) – \left( {x + 10} \right)} \over {\left( {x + 10} \right)\left( {x + 17} \right)}} = {x \over {\left( {x + 2} \right)\left( {x + 17} \right)}} \cr & \Leftrightarrow {1 \over {x + 2}} – {1 \over {x + 5}} + {1 \over {x + 5}} – {1 \over {x + 10}} + {1 \over {x + 10}} – {1 \over {x + 17}} = {x \over {\left( {x + 2} \right)\left( {x + 17} \right)}} \cr & \Leftrightarrow {1 \over {x + 2}} – {1 \over {x + 17}} = {x \over {\left( {x + 2} \right)\left( {x + 17} \right)}} \cr & \Leftrightarrow {{x + 17 – x – 2} \over {\left( {x + 2} \right)\left( {x + 17} \right)}} = {x \over {\left( {x + 2} \right)\left( {x + 17} \right)}} \cr & \Leftrightarrow {{15} \over {\left( {x + 2} \right)\left( {x + 17} \right)}} = {x \over {\left( {x + 2} \right)\left( {x + 17} \right)}} \cr & \Leftrightarrow x = 15\,\,\left( {tm} \right) \cr} $$ Bình luận
$$\eqalign{
& {3 \over {\left( {x + 2} \right)\left( {x + 5} \right)}} + {5 \over {\left( {x + 5} \right)\left( {x + 10} \right)}} + {7 \over {\left( {x + 10} \right)\left( {x + 17} \right)}} = {x \over {\left( {x + 2} \right)\left( {x + 17} \right)}}\,\,\left( {x \ne – 2;\,\,x \ne – 5;\,\,x \ne – 10;\,\,x \ne – 17} \right) \cr
& \Leftrightarrow {{\left( {x + 5} \right) – \left( {x + 2} \right)} \over {\left( {x + 2} \right)\left( {x + 5} \right)}} + {{\left( {x + 10} \right) – \left( {x + 5} \right)} \over {\left( {x + 5} \right)\left( {x + 10} \right)}} + {{\left( {x + 17} \right) – \left( {x + 10} \right)} \over {\left( {x + 10} \right)\left( {x + 17} \right)}} = {x \over {\left( {x + 2} \right)\left( {x + 17} \right)}} \cr
& \Leftrightarrow {1 \over {x + 2}} – {1 \over {x + 5}} + {1 \over {x + 5}} – {1 \over {x + 10}} + {1 \over {x + 10}} – {1 \over {x + 17}} = {x \over {\left( {x + 2} \right)\left( {x + 17} \right)}} \cr
& \Leftrightarrow {1 \over {x + 2}} – {1 \over {x + 17}} = {x \over {\left( {x + 2} \right)\left( {x + 17} \right)}} \cr
& \Leftrightarrow {{x + 17 – x – 2} \over {\left( {x + 2} \right)\left( {x + 17} \right)}} = {x \over {\left( {x + 2} \right)\left( {x + 17} \right)}} \cr
& \Leftrightarrow {{15} \over {\left( {x + 2} \right)\left( {x + 17} \right)}} = {x \over {\left( {x + 2} \right)\left( {x + 17} \right)}} \cr
& \Leftrightarrow x = 15\,\,\left( {tm} \right) \cr} $$
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