Bài 14 trang 57 Tài liệu dạy – học toán 6 tập 2

Đề bài

Tìm x, biết :

a) $1{1 \over {30}}:\left( {24{1 \over 6} – 24{1 \over 5}} \right) – {{1{1 \over 2} – {3 \over 4}} \over {4x – {1 \over 2}}} = {{{1 \over {13.16}} + {1 \over {14.17}}} \over {{1 \over {13.15}} + {1 \over {14.16}} + {1 \over {15.17}}}}$

b) ${{\left( {27{5 \over {19}} – 26{4 \over {13}}} \right)\left( {{3 \over 4} + {{19} \over {59}} – {3 \over {118}}} \right)} \over {\left( {{3 \over 4} + x} \right){{27} \over {33}}}} = {{{1 \over {13.16}} + {1 \over {14.17}}} \over {{1 \over {13.15}} + {1 \over {14.16}} + {1 \over {15.17}}}}$

c) ${1 \over {1.3}} + {1 \over {3.5}} + … + {1 \over {x\left( {x + 2} \right)}} = {{20} \over {41}}.$

Lời giải chi tiết

$\eqalign{  & a)1{1 \over {30}}:\left( {24{1 \over 6} – 24{1 \over 5}} \right) – {{1{1 \over 2} – {3 \over 4}} \over {4x – {1 \over 2}}} = {{{1 \over {13.16}} + {1 \over {14.17}}} \over {{1 \over {13.15}} + {1 \over {14.16}} + {1 \over {15.17}}}}  \cr  & {{31} \over {30}}:\left( {24{5 \over {30}} – 24{6 \over {30}}} \right) – {{{3 \over 2} – {3 \over 4}} \over {4x – {1 \over 2}}} = {{{1 \over 3}.\left( {{3 \over {13.16}} + {3 \over {14.17}}} \right)} \over {{1 \over 2}.\left( {{2 \over {13.15}} + {2 \over {14.16}} + {2 \over {15.17}}} \right)}}  \cr  & {{31} \over {30}}:{{ – 1} \over {30}} – {{{6 \over 4} – {3 \over 4}} \over {4x – {1 \over 2}}} = {{{1 \over 3}.\left( {{1 \over {13}} – {1 \over {16}} + {1 \over {14}} – {1 \over {17}}} \right)} \over {{1 \over 2}.\left( {{1 \over {13}} – {1 \over {15}} + {1 \over {14}} – {1 \over {16}} + {1 \over {15}} – {1 \over {17}}} \right)}}  \cr  &  – 31 – {{{3 \over 4}} \over {4x – {1 \over 2}}} = {{{1 \over 3}} \over {{1 \over 2}}} \Leftrightarrow  – 31 – {{{3 \over 4}} \over {4x – {1 \over 2}}} = {2 \over 3} \Leftrightarrow  – {{{3 \over 4}} \over {4x – {1 \over 2}}} = {2 \over 3} + 31 \Leftrightarrow {{ – {3 \over 4}} \over {4x – {1 \over 2}}} = {{95} \over 3}  \cr  & 4x – {1 \over 2} =  – {3 \over 4}:{{95} \over 3} \Leftrightarrow 4x – {1 \over 2} = {{ – 9} \over {380}} \Leftrightarrow 4x = {{181} \over {380}} \Leftrightarrow x = {{181} \over {1520}}  \cr  & b){{\left( {27{5 \over {19}} – 26{4 \over {13}}} \right).\left( {{{177} \over {236}} + {{76} \over {236}} – {6 \over {236}}} \right)} \over {\left( {{3 \over 4} + x} \right).{{27} \over {33}}}} = {2 \over 3}(apdungketquacaua)  \cr  & {{\left( {26{{312} \over {247}} – 26{{76} \over {247}}} \right).{{247} \over {236}}} \over {\left( {{3 \over 4} + x} \right).{{27} \over {33}}}} = {2 \over 3} \Leftrightarrow {{236} \over {247}}.{{247} \over {236}} = {2 \over 3}.\left( {{3 \over 4} + x} \right).{{27} \over {33}}  \cr  & 1 = {6 \over {11}}.\left( {{3 \over 4} + x} \right) \Leftrightarrow {3 \over 4} + x = 1:{6 \over {11}} \Leftrightarrow {3 \over 4} + x = {{11} \over 6}  \cr  & x = {{11} \over 6} – {3 \over 4} \Leftrightarrow x = {{22} \over {12}} – {9 \over {12}} \Leftrightarrow x = {{13} \over {12}} \Leftrightarrow x = 1{1 \over {12}}  \cr  & c){1 \over {1.3}} + {1 \over {3.5}} + … + {1 \over {x(x + 2)}} = {{20} \over {41}}  \cr  & {1 \over 2}\left[ {{2 \over {1.3}} + {2 \over {3.5}} + {2 \over {5.7}} + … + {2 \over {(x – 2).x}} + {2 \over {x(x + 2)}}} \right] = {{20} \over {41}}  \cr  & {1 \over 2}\left( {1 – {1 \over 3} + {1 \over 3} – {1 \over 5} + {1 \over 5} – {1 \over 7} + … + {1 \over {x – 2}} – {1 \over x} + {1 \over x} – {1 \over {x + 2}}} \right) = {{20} \over {41}}  \cr  & {1 \over 2}\left( {1 – {1 \over {x + 2}}} \right) = {{20} \over {41}} \Leftrightarrow 1 – {1 \over {x + 2}} = {{20} \over {41}}:{1 \over 2} \Leftrightarrow 1 – {1 \over {x + 2}} = {{40} \over {41}} \Leftrightarrow {1 \over {x + 2}} = 1 – {{40} \over {41}}  \cr  & {1 \over {x + 2}} = {1 \over {41}} \Leftrightarrow x + 2 = 41 \Leftrightarrow x = 39. \cr}$

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