Rút gọn các biểu thức sau:\(a)\;2x{\left( {2x - 1}...

A

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

\(\begin{array}{l}a)\;2x{\left( {2x - 1} \right)^2} - 3x\left( {x + 3} \right)\left( {x - 3} \right) - 4x{\left( {x + 1} \right)^2}\\ = 2x.\left( {{{\left( {2x} \right)}^2} - 2.2x.1 + {1^2}} \right) - 3x\left( {{x^2} - {3^2}} \right) - 4x.\left( {{x^2} + 2.x.1 + {1^2}} \right)\\ = 2x.\left( {4{x^2} - 4x + 1} \right) - 3x\left( {{x^2} - 9} \right) - 4x\left( {{x^2} + 2x + 1} \right)\\ = 2x.4{x^2} - 2x.4x + 2x.1 - 3x.{x^2} + 3x.9 - 4x.{x^2} - 4x.2x - 4x.1\\ = 8{x^3} - 8{x^2} + 2x - 3{x^3} + 27x - 4{x^3} - 8{x^2} - 4x\\ = {x^3} - 16{x^2} + 25x\end{array}\)

\(\begin{array}{l}b)\;{\left( {a - b + c} \right)^2} - {\left( {b - c} \right)^2} + 2ab - 2ac\\ = {a^2} + {b^2} + {c^2} - 2ab - 2bc + 2ac - \left( {{b^2} - 2.b.c + {c^2}} \right) + 2ab - 2ac\\ = {a^2} + {b^2} + {c^2} - 2ab - 2bc + 2ac - {b^2} + 2bc - {c^2} + 2ab - 2ac\\ = {a^2}\end{array}\)

\(\begin{array}{l}c){\left( {3x + 1} \right)^2} - 2\left( {3x + 1} \right)\left( {3x + 5} \right) + {\left( {3x + 5} \right)^2}\\ = {\left( {\left( {3x + 1} \right) - \left( {3x + 5} \right)} \right)^2}\\ = {\left( {3x + 1 - 3x - 5} \right)^2}\\ = {( - 4)^2} = 16\end{array}\)

\(\begin{array}{l}d){\left( {a + b + c} \right)^2} + {\left( {a - b - c} \right)^2} + {\left( {b - c - a} \right)^2} + {\left( {c - a - b} \right)^2}\\ = {a^2} + {b^2} + {c^2} + 2ab + 2ac + 2bc + {a^2} + {b^2} + {c^2} - 2ab - 2ac + 2bc\\ + {a^2} + {b^2} + {c^2} - 2ab - 2bc + 2ac + {a^2} + {b^2} + {c^2} + 2ab - 2ac - 2bc\\ = 4{a^2} + 4{b^2} + 4{c^2} \end{array}\)