bien doi cac tich sau thanh tich cac da thuc
a.x3+8
b.27-8y3
c.y6+1
d.64×3-1/8y3
e.125×6-27y9
f.-x6/125-y3/64
bien doi cac tich sau thanh tich cac da thuc
a.x3+8
b.27-8y3
c.y6+1
d.64×3-1/8y3
e.125×6-27y9
f.-x6/125-y3/64
Đáp án:
Giải thích các bước giải:
\[\begin{array}{l}
a){x^3} + 8 = \left( {x + 2} \right)\left( {{x^2} – 2x + 4} \right)\\
b)27 – 8{y^3} = \left( {3 – 2y} \right)\left( {9 + 6y + 4{y^2}} \right)\\
c){y^6} + 1 = \left( {{y^2} + 1} \right)\left( {{y^4} – {y^2} + 1} \right)\\
d)64{x^3} – \frac{1}{8}{y^3} = \left( {4x – \frac{1}{2}y} \right)\left( {16{x^2} + 2xy + \frac{1}{4}{y^2}} \right)\\
e)125{x^6} – 27{y^9} = \left( {5{x^2} – 3{y^3}} \right)\left( {25{x^4} + 15{x^2}{y^3} + 9{y^6}} \right)\\
f)\,\frac{{ – {x^6}}}{{125}} – \frac{{{y^3}}}{{64}} = – \left( {\frac{{{x^6}}}{{125}} + \frac{{{y^3}}}{{64}}} \right) = – \left( {\frac{{{x^2}}}{5} + \frac{y}{4}} \right)\left( {\frac{{{x^4}}}{{25}} – \frac{{{x^2}y}}{{20}} + \frac{{{y^2}}}{{16}}} \right)
\end{array}\]