Bài 7 trang 6 SBT toán 8 tập 1

Đề bài

Thực hiện phép tính:

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

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

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

Lời giải

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

\(=\dfrac{1}{2}x.2x+\dfrac{1}{2}x.(-3)+(-1).2x\)\(+(-1).(-3)\)

\(=x^2-\dfrac{3}{2}x-2x+3\)

\(=x^2-\dfrac{7}{2}x+3\)

\(b)\)

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

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

\(= \left[ x.x + x.\dfrac{1}{2} - \dfrac{1}{2} .x  - \dfrac{1}{2}.\dfrac{1}{2}\right]\)\(.\left( {4x - 1} \right) \)
\( = \left( {{x^2} + \dfrac{1}{2}x - \dfrac{1}{2}x - \dfrac{1}{4}} \right)\left( {4x - 1} \right) \)
\(= \left( {{x^2} - \dfrac{1}{4}} \right)\left( {4x - 1} \right) \)
\(= {x^2}.4x + {x^2}.\left( { - 1} \right) + \left( { - \dfrac{1}{4}} \right).4x \)\(+ \left( { - \dfrac{1}{4}} \right).\left( { - 1} \right) \)
\(= 4{x^3} - {x^2} - x +\dfrac{1}{4} \)