bien doi cac tich sau thanh tich cac da thuc a.x3+8 b.27-8y3 c.y6+1 d.64x3-1/8y3 e.125x6-27y9 f.-x6/125-y3/64

Đáp án:

Giải thích các bước giải: \[\begin{array}{l} a){x^3} + 8 = \left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)\\ b)27 - 8{y^3} = \left( {3 - 2y} \right)\left( {9 + 6y + 4{y^2}} \right)\\ c){y^6} + 1 = \left( {{y^2} + 1} \right)\left( {{y^4} - {y^2} + 1} \right)\\ d)64{x^3} - \frac{1}{8}{y^3} = \left( {4x - \frac{1}{2}y} \right)\left( {16{x^2} + 2xy + \frac{1}{4}{y^2}} \right)\\ e)125{x^6} - 27{y^9} = \left( {5{x^2} - 3{y^3}} \right)\left( {25{x^4} + 15{x^2}{y^3} + 9{y^6}} \right)\\ f)\,\frac{{ - {x^6}}}{{125}} - \frac{{{y^3}}}{{64}} = - \left( {\frac{{{x^6}}}{{125}} + \frac{{{y^3}}}{{64}}} \right) = - \left( {\frac{{{x^2}}}{5} + \frac{y}{4}} \right)\left( {\frac{{{x^4}}}{{25}} - \frac{{{x^2}y}}{{20}} + \frac{{{y^2}}}{{16}}} \right) \end{array}\]

\[\begin{array}{l}
a){x^3} + 8 = (x + 2)({x^2} - 2x + 4)\\
b)27 - 8{y^3} = (3 - 2y)(9 + 6y + 4{y^2})\\
c){y^6} + 1 = ({y^2} + 1)({y^4} - {y^2} + 1)\\
d)64{x^3} - 18{y^3} = (4x - \frac{1}{2}y)(16{x^2} + 2xy + \frac{1}{4}{y^2})\\
e)125{x^6} - 27{y^9} = (5{x^2} - 3{y^3})(25{x^4} + 15{x^2}{y^3} + 9{y^6})\\
f) - \frac{{{x^6}}}{{125}} - \frac{{{y^3}}}{{64}} =  - (\frac{{{x^6}}}{{125}} + \frac{{{y^3}}}{{64}}) =  - (\frac{{{x^2}}}{5} + \frac{y}{4})(\frac{{{x^4}}}{{25}} - \frac{{{x^2}y}}{{20}} + \frac{{{y^2}}}{{16}})
\end{array}\]