Bài 42 trang 26 SBT toán 7 tập 2

Đề bài

Tính \(f (x) + g (x) - h (x)\) biết:

\(f(x) = {x^5} - 4{{\rm{x}}^3} + {x^2} - 2{\rm{x}} + 1\)

\(g(x) = {x^5} - 2{{\rm{x}}^4} + {x^2} - 5{\rm{x}} + 3\)

\(h(x) = {x^4} - 3{{\rm{x}}^2} + 2{\rm{x}} - 5\)

Lời giải

Ta có:

\(f (x) + g (x) - h (x)\)

\( = \left( {{x^5} - 4{{\rm{x}}^3} + {x^2} - 2{\rm{x}} + 1} \right)\)\( + \left( {{x^5} - 2{{\rm{x}}^4} + {x^2} - 5{\rm{x}} + 3} \right) \)\(- \left( {{x^4} - 3{{\rm{x}}^2} + 2{\rm{x}} - 5} \right)\)

\( = {x^5} - 4{{\rm{x}}^3} + {x^2} - 2{\rm{x}} + 1 \)\(+ {x^5} - 2{{\rm{x}}^4} + {x^2} - 5{\rm{x}} + 3\)\( - {x^4} + 3{{\rm{x}}^2} - 2{\rm{x}} + 5 \)

\(= (x^5+x^5) +(-2x^4 - x^4) - 4{{\rm{x}}^3} \)\(+ (x^2 + x^2 + 3x^2) + (-2x - 5x - 2x) \)\(+ (1 + 3 + 5)\)

\( = 2{{\rm{x}}^5} - 3{{\rm{x}}^4} - 4{{\rm{x}}^3} + 5{{\rm{x}}^2} - 9{\rm{x}} + 9  \)