giup mik bai 2 vs may ban eiiiiiiiiiii

Đáp án:

a. \({x^3} – 9{x^2} + 27x – 27\)

Giải thích các bước giải:

\(\begin{array}{l}
a.{\left( {x – 3} \right)^3}\\
 = {x^3} – 3{x^2}.3 + 3.x.9 – 27\\
 = {x^3} – 9{x^2} + 27x – 27\\
b.{\left( {2x – 3} \right)^3}\\
 = 8{x^3} – 3.4{x^2}.3 + 3.2x.9 – 27\\
 = 8{x^3} – 36{x^2} + 54x – 27\\
d.{\left( {{x^3} – 2} \right)^3}\\
 = {x^9} – 3{x^6}.2 + 3.{x^3}.4 – 8\\
 = {x^9} – 18{x^6} + 12{x^3} – 8\\
e.{\left( {2x – 3y} \right)^3}\\
 = 8{x^3} – 3.4{x^2}.3y + 3.2x.9{y^2} – 27{y^3}\\
 = 8{x^3} – 36{x^2}y + 54x{y^2} – 27{y^3}\\
c.{\left( {x – \dfrac{1}{2}} \right)^3}\\
 = {x^3} – 3{x^2}.\dfrac{1}{2} + 3x.\dfrac{1}{4} – \dfrac{1}{8}\\
 = {x^3} – \dfrac{3}{2}{x^2} + \dfrac{3}{4}x – \dfrac{1}{8}\\
f.{\left( {\dfrac{1}{2}x – {y^2}} \right)^3}\\
 = \dfrac{1}{8}{x^3} – 3.\dfrac{1}{4}{x^2}.{y^2} + 3.\dfrac{1}{2}x.{y^4} – {y^6}\\
 = \dfrac{1}{8}{x^3} – \dfrac{3}{4}{x^2}.{y^2} + \dfrac{3}{2}x.{y^4} – {y^6}
\end{array}\)