Câu 7 trang 6 Sách bài tập (SBT) Toán 8 tập 1

Thực hiện phép tính:

a. \(\left( {\dfrac{1}{2}x - 1} \right)\left( {2x - 3} \right)\)

b. \(\left( {x - 7} \right)\left( {x - 5} \right)\)

c. \(\left( {x - \dfrac{1}{2}} \right)\left( {x + \dfrac{1}{2}} \right)\left( {4x - 1} \right)\)

Giải:

a. 

\(\eqalign{
& \left( {{1 \over 2}x - 1} \right)\left( {2x - 3} \right) \cr
& = {1 \over 2}x.2x + {1 \over 2}x.\left( { - 3} \right) + \left( { - 1} \right).2x + \left( { - 1} \right).\left( { - 3} \right) \cr
& = {x^2} - {3 \over 2}x - 2x + 3 \cr
& = {x^2} - {7 \over 2}x + 3 \cr} \)

b. 

\(\eqalign{
& \left( {x - 7} \right)\left( {x - 5} \right) \cr
& = x.x + x.\left( { - 5} \right) + \left( { - 7} \right).x + \left( { - 7} \right).\left( { - 5} \right) \cr
& = {x^2} - 5x - 7x + 35 \cr
& = {x^2} - 12x + 35 \cr} \)

c. 

\(\eqalign{
& \left( {x - {1 \over 2}} \right)\left( {x + {1 \over 2}} \right)\left( {4x - 1} \right) \cr
& = \left[ {x.x + x.{1 \over 2} + \left( { - {1 \over 2}} \right).x + \left( { - {1 \over 2}} \right).{1 \over 2}} \right]\left( {4x - 1} \right) \cr
& = \left( {{x^2} + {1 \over 2}x - {1 \over 2}x - {1 \over 4}} \right)\left( {4x - 1} \right) \cr
& = \left( {{x^2} - {1 \over 4}} \right)\left( {4x - 1} \right) \cr
& = {x^2}.4x + {x^2}.\left( { - 1} \right) + \left( { - {1 \over 4}} \right).4x + \left( { - {1 \over 4}} \right).\left( { - 1} \right) \cr
& = 4{x^3} - {x^2} - x + {1 \over 4} \cr} \)

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