Đề bài
Thực hiện phép chia:
a) \(2{x^2}:x\)
b) \(5{x^4}:2{x^2}\) ;
c) \(( - 8{x^2}):4x\) ;
d) \(x{y^3}{z^4}:( - 3xyz)\) ;
e) \({2 \over 3}{x^3}{y^4}:\left( {{{ - 4} \over 9}{x^2}{y^3}} \right)\) .
Lời giải chi tiết
\(\eqalign{ & a)\,\,2{x^2}:x = {{2{x^2}} \over x} = 2{x^{2 - 1}} = 2x \cr & b)\,\,5{x^4}:\left( {2{x^2}} \right) = {{5{x^4}} \over {2{x^2}}} = {5 \over 2}.{{{x^4}} \over {{x^2}}} = {5 \over 2}{x^{4 - 2}} = {5 \over 2}{x^2} \cr & c)\,\,\left( { - 8{x^2}} \right):\left( {4x} \right) = {{ - 8{x^2}} \over {4x}} = {{ - 8} \over 4}.{{{x^2}} \over x} = - 2{x^{2 - 1}} = - 2x \cr & d)\,\,x{y^3}{z^4}:\left( { - 3xyz} \right) = {{x{y^3}{z^4}} \over { - 3xyz}} = {1 \over { - 3}}.{{x{y^3}{z^4}} \over {xyz}} = - {1 \over 3}{y^2}{z^3} \cr & e)\,\,{2 \over 3}{x^3}{y^4}:\left( { - {4 \over 9}{x^2}{y^3}} \right) = {{{2 \over 3}{x^3}{y^4}} \over { - {4 \over 9}{x^2}{y^3}}} = {{{2 \over 3}} \over { - {4 \over 9}}}.{{{x^3}{y^4}} \over {{x^2}{y^3}}} = {{{2 \over 3}} \over { - {{\left( {{2 \over 3}} \right)}^2}}}xy = {1 \over { - {2 \over 3}}}xy = - {3 \over 2}xy \cr} \)
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