Toán Lớp 8: Rút gọn biểu thức:
a) C = 6(c – d)$(c+d)^{2}$ + 12$(c-d)^{2}$ (c + d) + $(c+d)^{3}$ + 8$(c-d)^{3}$
b)D = $(m-n)^{3}$ – $(n+p)^{3}$ – 3$(n+p)^{2}$ (n – m) – 3(n + p) $(n-m)^{2}$
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Toán Lớp 8: Rút gọn biểu thức:
a) C = 6(c – d)$(c+d)^{2}$ + 12$(c-d)^{2}$ (c + d) + $(c+d)^{3}$ + 8$(c-d)^{3}$
b)D = $(m-n)^{3}$ – $(n+p)^{3}$ – 3$(n+p)^{2}$ (n – m) – 3(n + p) $(n-m)^{2}$
Comments ( 1 )
a)C = 6\left( {c – d} \right){\left( {c + d} \right)^2} + 12{\left( {c – d} \right)^2}\left( {c + d} \right)\\
+ {\left( {c + d} \right)^3} + 8{\left( {c – d} \right)^3}\\
= {\left( {c + d} \right)^3} + 3.{\left( {c + d} \right)^2}.2\left( {c – d} \right)\\
+ 3.\left( {c + d} \right).4{\left( {c – d} \right)^2} + {\left[ {2\left( {c – d} \right)} \right]^3}\\
= {\left( {c + d + 2\left( {c – d} \right)} \right)^3}\\
= {\left( {3c – d} \right)^3}\\
= 27{c^3} – 27{c^2}d + 9c{d^2} – {d^3}\\
d)\\
D = {\left( {m – n} \right)^3} – {\left( {n + p} \right)^3}\\
– 3{\left( {n + p} \right)^2}.\left( {n – m} \right) – 3\left( {n + p} \right){\left( {n – m} \right)^2}\\
= {\left( {m – n} \right)^3} – 3\left( {n + p} \right){\left( {m – n} \right)^2}\\
+ 3{\left( {n + p} \right)^2}.\left( {m – n} \right) – {\left( {n + p} \right)^3}\\
= {\left( {m – n – n – p} \right)^3}\\
= {\left( {m – 2n – p} \right)^3}
\end{array}$