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theo cách tính tổng (bn có thể xem lại ở toán 7 hay 6 j đấy) thì bt trên bằng 1/x - 1/(x+5)
từ đó tính tiếp nha bn
1) (x2 - 2x - 1)(x - 3)
= x2(x - 3) - 2x(x - 3) - 1(x - 3)
= x3 - 3x2 - 2x2 + 6x - x + 3
= x3 - 5x2 + 5x + 3
2. (-x + 4)(-x2 + 4x - 1)
= -x(-x2 + 4x - 1) + 4(-x2 + 4x - 1)
= x3 - 4x2 + x - 4x2 + 16x - 4
= x3 - 8x2 + 17x - 4
3 ) (2x - 1)(x2 - 5x + 3)
= 2x(x2 - 5x + 3) - 1(x2 - 5x + 3)
= 2x3 - 10x2 + 6x - x2 + 5x - 3
= 2x3 - 11x2 + 11x - 3
Bài làm :
1) (x2 - 2x - 1)(x - 3)
= x2(x - 3) - 2x(x - 3) - 1(x - 3)
= x3 - 3x2 - 2x2 + 6x - x + 3
= x3 - 5x2 + 5x + 3
2) (-x + 4)(-x2 + 4x - 1)
= -x(-x2 + 4x - 1) + 4(-x2 + 4x - 1)
= x3 - 4x2 + x - 4x2 + 16x - 4
= x3 - 8x2 + 17x - 4
3 ) (2x - 1)(x2 - 5x + 3)
= 2x(x2 - 5x + 3) - 1(x2 - 5x + 3)
= 2x3 - 10x2 + 6x - x2 + 5x - 3
= 2x3 - 11x2 + 11x - 3
Bài 1:
\(3a.\left(2a^2-ab\right)=6a^3-3a^2b\)
\(\left(4-7b^2\right).\left(2a+5b\right)=8a+20b-14ab^2-35b^3\)
Bài 2:
\(2x^2-6x+xy-3y=2x.\left(x-3\right)+y.\left(x-3\right)=\left(x-3\right).\left(2x+y\right)\)
Bài 3: Tại x = 3/2, y =1/3 thì Q = 67/9
Bài 4:
\(\left(\frac{1}{x+1}+\frac{2x}{1-x^2}\right).\left(\frac{1}{x-1}\right)\) \(\frac{1}{\left(x+1\right).\left(x-1\right)}+\frac{2x}{\left(1-x^2\right).\left(x-1\right)}=\frac{x-1}{\left(x+1\right).\left(x-1\right)^2}+\frac{-2x}{\left(x-1\right)^2.\left(x+1\right)}\)
= \(\frac{x-1-2x}{\left(x+1\right).\left(x-1\right)^2}=\frac{-\left(x+1\right)}{\left(x+1\right).\left(x-1\right)^2}=\frac{-1}{\left(x-1\right)^2}\)
1) \(\left(x^3-8\right):\left(x-2\right)=\left[\left(x-2\right)\left(x^2+2x+4\right)\right]:\left(x-2\right)=x^2+2x+4\)
2) \(\left(x^3-1\right):\left(x^2+x+1\right)=\left[\left(x-1\right)\left(x^2+x+1\right)\right]:\left(x^2+x+1\right)=x-1\)
3) \(\left(x^3+3x^2+3x+1\right):\left(x^2+2x+1\right)=\left(x+1\right)^3:\left(x+1\right)^2=x+1\)
4) \(\left(25x^2-4y^2\right):\left(5x-2y\right)=\left[\left(5x-2y\right)\left(5x+2y\right)\right]:\left(5x-2y\right)=5x+2y\)
\(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right).....\left(1-\dfrac{1}{2008^2}\right)\)
\(=\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}....\dfrac{2008^2-1}{2008^2}\)
\(=\dfrac{1.3}{4}.\dfrac{2.4}{9}.\dfrac{3.5}{16}....\dfrac{2007.2009}{2008^2}\)
\(=\left(\dfrac{1.2.3...2007}{2.3.4....2008}\right).\dfrac{3.4.5...2009}{2.3.4...2008}\)
\(=\dfrac{1}{2008}.\dfrac{2009}{2}=\dfrac{2009}{4016}\)