Advertisements (Quảng cáo)
Thực hiện phép chia:
a) \(2{x^2}:x\)
b) \(5{x^4}:2{x^2}\) ;
c) \(( – 8{x^2}):4x\) ;
d) \(x{y^3}{z^4}:( – 3xyz)\) ;
Advertisements (Quảng cáo)
e) \({2 \over 3}{x^3}{y^4}:\left( {{{ – 4} \over 9}{x^2}{y^3}} \right)\) .
\(\eqalign{ & a)\,\,2{x^2}:x = {{2{x^2}} \over x} = 2{x^{2 – 1}} = 2x \cr & b)\,\,5{x^4}:\left( {2{x^2}} \right) = {{5{x^4}} \over {2{x^2}}} = {5 \over 2}.{{{x^4}} \over {{x^2}}} = {5 \over 2}{x^{4 – 2}} = {5 \over 2}{x^2} \cr & c)\,\,\left( { – 8{x^2}} \right):\left( {4x} \right) = {{ – 8{x^2}} \over {4x}} = {{ – 8} \over 4}.{{{x^2}} \over x} = – 2{x^{2 – 1}} = – 2x \cr & d)\,\,x{y^3}{z^4}:\left( { – 3xyz} \right) = {{x{y^3}{z^4}} \over { – 3xyz}} = {1 \over { – 3}}.{{x{y^3}{z^4}} \over {xyz}} = – {1 \over 3}{y^2}{z^3} \cr & e)\,\,{2 \over 3}{x^3}{y^4}:\left( { – {4 \over 9}{x^2}{y^3}} \right) = {{{2 \over 3}{x^3}{y^4}} \over { – {4 \over 9}{x^2}{y^3}}} = {{{2 \over 3}} \over { – {4 \over 9}}}.{{{x^3}{y^4}} \over {{x^2}{y^3}}} = {{{2 \over 3}} \over { – {{\left( {{2 \over 3}} \right)}^2}}}xy = {1 \over { – {2 \over 3}}}xy = – {3 \over 2}xy \cr} \)