Rút gọn biểu thức (a−b+1)[a2+b2+ab−(a+2b)+1]−(a3+1)\left( {a - b + 1} \right)\left[ {{a^2} + {b^2} + ab - (a + 2b) + 1} \right] - ({a^3} + 1)(a−b+1)[a2+b2+ab−(a+2b)+1]−(a3+1)
(1+b)3−1{(1 + b)^3} - 1(1+b)3−1.
(1+b)3+1{(1 + b)^3} + 1(1+b)3+1.
(1−b)3−1{(1 - b)^3} - 1(1−b)3−1.
(1−b)3+1{(1 - b)^3} + 1(1−b)3+1.
Đáp án: <p><span data-type="math-inline" data-latex="{(1 - b)^3} - 1">${(1 - b)^3} - 1$</span>.</p>
Phương pháp giải:
Áp dụng hằng đẳng thức: A3+B3=(A+B)(A2−AB+B2){A^3} + {B^3} = (A + B)({A^2} - AB + {B^2})A3+B3=(A+B)(A2−AB+B2)
Lời giải chi tiết:
(a−b+1)[a2+b2+ab−(a+2b)+1]−(a3+1)=[a+(1−b)][a2−(a−ab)+(b2−2b+1)]−(a3+1)=[a+(1−b)][a2−a(1−b)+(b−1)2]−(a3+1)=a3+(1−b)3−a3−1=(1−b)3−1\begin{array}{l}\left( {a - b + 1} \right)\left[ {{a^2} + {b^2} + ab - (a + 2b) + 1} \right] - ({a^3} + 1)\\ = \left[ {a + \left( {1 - b} \right)} \right]\left[ {{a^2} - (a - ab) + ({b^2} - 2b + 1)} \right] - \left( {{a^3} + 1} \right)\\ = \left[ {a + \left( {1 - b} \right)} \right]\left[ {{a^2} - a(1 - b) + {{\left( {b - 1} \right)}^2}} \right] - \left( {{a^3} + 1} \right)\\ = {a^3} + {(1 - b)^3} - {a^3} - 1\\ = {(1 - b)^3} - 1\end{array}(a−b+1)[a2+b2+ab−(a+2b)+1]−(a3+1)=[a+(1−b)][a2−(a−ab)+(b2−2b+1)]−(a3+1)=[a+(1−b)][a2−a(1−b)+(b−1)2]−(a3+1)=a3+(1−b)3−a3−1=(1−b)3−1
