Tháng Tư 20, 2024

Thực hiện phép tính: \(a)\left( {x + 3} \right)\left( {x – 1} \right) – \left( {x – 3} \right)\left( {x + 1} \right)\) \(b)({x^2} – 2xy + {y^2})\left( {x – 3y} \right)\) \(c)\left( {2{x^2} + 3xy + 5} \right)\left( {8x + 2} \right)\) \(d)(x + y)\left( {x – 12} \right) – (x – 1)\left( {x + 8{y^2}} \right)\)

Thực hiện phép tính:

\(a)\left( {x + 3} \right)\left( {x – 1} \right) – \left( {x – 3} \right)\left( {x + 1} \right)\)

\(b)({x^2} – 2xy + {y^2})\left( {x – 3y} \right)\)

\(c)\left( {2{x^2} + 3xy + 5} \right)\left( {8x + 2} \right)\)

\(d)(x + y)\left( {x – 12} \right) – (x – 1)\left( {x + 8{y^2}} \right)\)

A. \(a) 4x\)

\(b) {x^3} – 5{x^2}y + 7x{y^2} – 3{y^3}\)

\(c) 16{x^3} + 4{x^2} + 24{x^2}y + 6xy + 40x + 10 \)

\(d) 8{y^2} – 8x{y^2} + xy – 11x – 12y \)

B. \(a) 4x\)

\(b) {x^3} + 5{x^2}y + 7x{y^2} – 3{y^3}\)

\(c) 4{x^2} + 24{x^2}y + 6xy + 40x + 10 \)

\(d) 8{y^2} – 8x{y^2} + xy – 11x – 12y \)

C. \(a) 4x\)

\(b) 5{x^2}y + 7x{y^2} – 3{y^3}\)

\(c) 16{x^3} + 4{x^2} + 24{x^2}y + 6xy + 40x + 10 \)

\(d) 8{y^2} – 8x{y^2} + xy – 11x – 12y \)

D. \(a) 4x\)

\(b) {x^3} – 5{x^2}y + 7x{y^2} – 3{y^3}\)

\(c) 16{x^3} + 4{x^2} + 24{x^2}y + 6xy + 40x + 10 \)

\(d) 8{y^2} + xy – 11x – 12y \)

Hướng dẫn Chọn đáp án là: A

Lời giải chi tiết:

Hướng dẫn giải chi tiết

\(\begin{array}{l}a)\left( {x + 3} \right)\left( {x – 1} \right) – \left( {x – 3} \right)\left( {x + 1} \right)\\ = x.x – 1.x + 3.x – 3.1 – x.x – 1.x + 3.x + 3.1\\ = {x^2} – x + 3x – 3 – {x^2} – x + 3x + 3\\ = 4x\end{array}\)

\(\begin{array}{l}b)\left( {{x^2} – 2xy + {y^2}} \right)\left( {x – 3y} \right)\\ = {x^2}.x – {x^2}.3y – 2xy.x + 2xy.3y + {y^2}.x – 3y.{y^2}\\ = {x^3} – 3{x^2}y – 2{x^2}y + 6x{y^2} + x{y^2} – 3{y^3}\\ = {x^3} – 5{x^2}y + 7x{y^2} – 3{y^3}\end{array}\)

\(\begin{array}{l}c)\left( {2{x^2} + 3xy + 5} \right)\left( {8x + 2} \right)\\ = 2{x^2}.8x + 2{x^2}.2 + 3xy.8x + 3xy.2 + 5.8x + 5.2\\ = 16{x^3} + 4{x^2} + 24{x^2}y + 6xy + 40x + 10\end{array}\)

\(\begin{array}{l}d)\left( {x + y} \right)\left( {x – 12} \right) – \left( {x – 1} \right)\left( {x + 8{y^2}} \right)\\ = {x^2} – 12x + xy – 12y – {x^2} – 8x{y^2} + x + 8{y^2}\\ = 8{y^2} – 8x{y^2} + xy – 11x – 12y\end{array}\)

Chọn A.