Tháng Ba 29, 2024

Rút gọn các biểu thức sau: \(a)\;2x{\left( {2x – 1} \right)^2} – 3x\left( {x + 3} \right)\left( {x – 3} \right) – 4x{\left( {x + 1} \right)^2}\) \(b)\;{\left( {a – b + c} \right)^2} – {\left( {b – c} \right)^2} + 2ab – 2ac\) \(c)\;{(3x + 1)^2} – 2\left( {3x + 1} \right)\left( {3x + 5} \right) + {\left( {3x + 5} \right)^2}\) \(d)\;{\left( {a + b + c} \right)^2} + {\left( {a – b – c} \right)^2} + {(b – c – a)^2} + {\left( {c – a – b} \right)^2}\)

Rút gọn các biểu thức sau:

\(a)\;2x{\left( {2x – 1} \right)^2} – 3x\left( {x + 3} \right)\left( {x – 3} \right) – 4x{\left( {x + 1} \right)^2}\)

\(b)\;{\left( {a – b + c} \right)^2} – {\left( {b – c} \right)^2} + 2ab – 2ac\)

\(c)\;{(3x + 1)^2} – 2\left( {3x + 1} \right)\left( {3x + 5} \right) + {\left( {3x + 5} \right)^2}\)

\(d)\;{\left( {a + b + c} \right)^2} + {\left( {a – b – c} \right)^2} + {(b – c – a)^2} + {\left( {c – a – b} \right)^2}\)

A. \( a) {x^3} – 16{x^2} + 25x \)

\( b) {a^2} \)

\( c) 16 \)

\( d) 4{a^2} + 4{b^2} + 4{c^2} \)

B. \( a) {x^3} – 8{x^2} + 25x \)

\( b) {a^2}+4 \)

\( c) 16 \)

\( d) 4{a^2} + 4{b^2} + 4{c^2} \)

C. \( a) {x^3} – 16{x^2} \)

\( b) {a^2} \)

\( c) 16 \)

\( d) {a^2} + {b^2} + {c^2} \)

D. \( a) {x^3} – 16{x^2} + 25x \)

\( b) {a^2} \)

\( c) 16 \)

\( d) 2{a^2} + 2{b^2} + 2{c^2} \)

Hướng dẫn Chọn đáp án là: A

Lời giải chi tiết:

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

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

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

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