Bài 28 trang 9 SBT toán 8 tập 1

Đề bài

Phân tích thành nhân tử

a) \({\left( {x + y} \right)^2} - {\left( {x - y} \right)^2}\)

b) \({\left( {3x + 1} \right)^2} - {\left( {x + 1} \right)^2}\)

c) \({x^3} + {y^3} + {z^3} - 3xyz\)

Lời giải

\(a)\) \({\left( {x + y} \right)^2} - {\left( {x - y} \right)^2}\)

\( = \left[ {\left( {x + y} \right) + \left( {x - y} \right)} \right]\)\(\left[ {\left( {x + y} \right) - \left( {x - y} \right)} \right]\)

\( = \left( {x + y + x - y} \right)\left( {x + y - x + y} \right) \)

\(= 2x.2y = 4xy\)

\(b)\) \({\left( {3x + 1} \right)^2} - {\left( {x + 1} \right)^2}\)

\( = \left[ {\left( {3x + 1} \right) + \left( {x + 1} \right)} \right][ {\left( {3x + 1} \right) - \left( {x + 1} \right)} ]\)

\( = \left( {3x + 1 + x + 1} \right)\left( {3x + 1 - x - 1} \right) \)

\(= \left( {4x + 2} \right).2x \)

\(=2.(2x+1).2x\)

\(= 4x\left( {2x + 1} \right)\)

\(c)\) \({x^3} + {y^3} + {z^3} - 3xyz\)

\( = {\left( {x + y} \right)^3} - 3xy\left( {x + y} \right) \)\(+ {z^3} - 3xyz\)

\( = \left[ {{{\left( {x + y} \right)}^3} + {z^3}} \right]\)\( - 3xy\left( {x + y + z} \right) \)

\( = \left( {x + y + z} \right)\left[ {{{\left( {x + y} \right)}^2} - \left( {x + y} \right)z + {z^2}} \right]\)\( - 3xy\left( {x + y + z} \right)  \)

\(  = \left( {x + y + z} \right)( {x^2} + 2xy + {y^2} - xz - yz \)\(+ {z^2} - 3xy ) \)

\( = \left( {x + y + z} \right)( {x^2} + {y^2} + {z^2} - xy - xz\)\( - yz)  \)