Đề bài - bài tập 1 trang 83 tài liệu dạy – học toán 8 tập 1
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\(\eqalign{ & a)\,\,{{3{a^2}b} \over {4c}}.{{{c^2}} \over {ab}} = {{3{a^2}b{c^2}} \over {4cab}} = {{3{a^2}b{c^2}} \over {4abc}} = {{3ac} \over 4} \cr & b)\,\,{9 \over {d - 3f}}:{6 \over {9f - 3d}} = {9 \over {d - 3f}}.{{9f - 3d} \over 6} \cr & \,\,\,\,\, = {9 \over {d - 3f}}.{{ - 3\left( {d - 3f} \right)} \over 6} = {{ - 9} \over 2} \cr & c)\,\,{{10k} \over {k - 5}}.{{3{k^2} - 15k} \over {5{k^2}}} = {{10k\left( {3{k^2} - 15k} \right)} \over {\left( {k - 5} \right)5{k^2}}} \cr & \,\,\,\,\, = {{10k.3k.\left( {k - 5} \right)} \over {\left( {k - 5} \right)5{k^2}}} = {{30{k^2}} \over {5{k^2}}} = 6 \cr & d)\,\,{{4x - 1} \over {4y - 4x}}:{3 \over {9x - 9y}} = {{ - \left( {4x - 1} \right)} \over {4x - 4y}}.{{9x - 9y} \over 3} \cr & \,\,\,\,\, = {{ - \left( {4x - 1} \right)9\left( {x - y} \right)} \over {4\left( {x - y} \right).3}} = {{ - \left( {4x - 1} \right).3} \over 4} = - {{3\left( {4x - 1} \right)} \over 4} \cr & e)\,\,{{mn - 4m + 2{n^2} - 8n} \over {m + 2n}}:{{4 - n} \over n} = {{mn - 4m + 2{n^2} - 8n} \over {m + 2n}}.{n \over {4 - n}} \cr & \,\,\,\,\,\, = {{m\left( {n - 4} \right) + 2n\left( {n - 4} \right)} \over {m + 2n}}.{n \over {4 - n}} \cr & \,\,\,\,\, = {{\left( {n - 4} \right)\left( {m + 2n} \right).n} \over {\left( {m + 2n} \right)\left( {4 - n} \right)}} = {{ - \left( {4 - n} \right)n} \over {4 - n}} = - n \cr & f)\,\,{{7{{\left( {p - q} \right)}^2}} \over {pq}}.{{2{p^3}{q^2}} \over {p - {q^2}}} = {{7{{\left( {p - q} \right)}^2}.2{p^3}{q^2}} \over {pq\left( {p - {q^2}} \right)}} = {{14{{\left( {p - q} \right)}^2}{p^2}q} \over {p - {q^2}}} \cr} \) Đề bài Rút gọn biểu thức : a) \({{3{a^2}b} \over {4c}}.{{{c^2}} \over {ab}}\) ; b) \({9 \over {d - 3f}}:{6 \over {9f - 3d}}\) ; c) \({{10k} \over {k - 5}}.{{3{k^2} - 15k} \over {5{k^2}}}\) ; d) \({{4x - 1} \over {4y - 4x}}:{3 \over {9x - 9y}}\) ; e) \({{mn - 4m + 2{n^2} - 8n} \over {m + 2n}}:{{4 - n} \over n}\) ; f) \({{7{{(p - q)}^2}} \over {pq}}.{{2{p^3}{q^2}} \over {p - {q^2}}}\) Lời giải chi tiết \(\eqalign{ & a)\,\,{{3{a^2}b} \over {4c}}.{{{c^2}} \over {ab}} = {{3{a^2}b{c^2}} \over {4cab}} = {{3{a^2}b{c^2}} \over {4abc}} = {{3ac} \over 4} \cr & b)\,\,{9 \over {d - 3f}}:{6 \over {9f - 3d}} = {9 \over {d - 3f}}.{{9f - 3d} \over 6} \cr & \,\,\,\,\, = {9 \over {d - 3f}}.{{ - 3\left( {d - 3f} \right)} \over 6} = {{ - 9} \over 2} \cr & c)\,\,{{10k} \over {k - 5}}.{{3{k^2} - 15k} \over {5{k^2}}} = {{10k\left( {3{k^2} - 15k} \right)} \over {\left( {k - 5} \right)5{k^2}}} \cr & \,\,\,\,\, = {{10k.3k.\left( {k - 5} \right)} \over {\left( {k - 5} \right)5{k^2}}} = {{30{k^2}} \over {5{k^2}}} = 6 \cr & d)\,\,{{4x - 1} \over {4y - 4x}}:{3 \over {9x - 9y}} = {{ - \left( {4x - 1} \right)} \over {4x - 4y}}.{{9x - 9y} \over 3} \cr & \,\,\,\,\, = {{ - \left( {4x - 1} \right)9\left( {x - y} \right)} \over {4\left( {x - y} \right).3}} = {{ - \left( {4x - 1} \right).3} \over 4} = - {{3\left( {4x - 1} \right)} \over 4} \cr & e)\,\,{{mn - 4m + 2{n^2} - 8n} \over {m + 2n}}:{{4 - n} \over n} = {{mn - 4m + 2{n^2} - 8n} \over {m + 2n}}.{n \over {4 - n}} \cr & \,\,\,\,\,\, = {{m\left( {n - 4} \right) + 2n\left( {n - 4} \right)} \over {m + 2n}}.{n \over {4 - n}} \cr & \,\,\,\,\, = {{\left( {n - 4} \right)\left( {m + 2n} \right).n} \over {\left( {m + 2n} \right)\left( {4 - n} \right)}} = {{ - \left( {4 - n} \right)n} \over {4 - n}} = - n \cr & f)\,\,{{7{{\left( {p - q} \right)}^2}} \over {pq}}.{{2{p^3}{q^2}} \over {p - {q^2}}} = {{7{{\left( {p - q} \right)}^2}.2{p^3}{q^2}} \over {pq\left( {p - {q^2}} \right)}} = {{14{{\left( {p - q} \right)}^2}{p^2}q} \over {p - {q^2}}} \cr} \)
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