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a) \(\left(x^2+2x+1\right)\left(x+1\right)\)
\(=x^3+x^2+2x^2+2x+x+1\)
\(=x^3+3x^2+3x+1\)
b) Ta có: \(\left(x^3-x^2+2x-1\right)\left(5-x\right)\)
\(=5x^3-x^4-5x^2+x^3+10x-2x^2-5+5x\)
\(=-x^4+6x^3-7x^2+15x-5\)
Ta có: \(\left(x-5\right)\left(x^3-x^2+2x-1\right)\)
\(=-\left(5-x\right)\left(x^3-x^2+2x-1\right)\)
\(=x^4-6x^3+7x^2-15x+5\)
(x3 – 2x2 + x – 1)(5 – x)
= (x3 – 2x2 + x – 1).5 + (x3 – 2x2 + x – 1).(–x)
= x3.5 + (–2x2).5 + x.5 + (–1).5 + x3.(–x) + (–2x2).(–x) + x.(–x) + (–1).(–x)
= 5x3 – 10x2 + 5x – 5 – x4 + 2x3 – x2 + x
= –x4 + (5x3 + 2x3) – (10x2 + x2) + (5x + x) – 5
= –x4 + 7x3 – 11x2 + 6x – 5
Ta có:
(x3 – 2x2 + x – 1).(x – 5)
= (x3 – 2x2 + x – 1).[–(5 – x)]
= – (x3 – 2x2 + x – 1).(5 – x)
= – (–x4 + 7x3 – 11x2 + 6x – 5)
= x4 – 7x3 + 11x2 – 6x + 5.
a) x 3 – 3 x 2 + 3x – 1;
b) – x 4 + 7 x 3 – 11 x 2 + 6x – 5;
c) c 3 + 2 c 2 – 5c – 6.
a) \(\left(x^2-2x+1\right)\left(x-1\right)=\left(x-1\right)^2\left(x-1\right)=\left(x-1\right)^3=x^3-3x^2+27x-1\)
\(a,\)\(\left(x^2-2x+1\right)\left(x-1\right)\)
\(=\left(x-1\right)^2\left(x-1\right)\)
\(=\left(x-1\right)^3\)
\(=x^3-3x^2+3x-1\)
\(b,\)\(\left(x^3-2x^2+x-1\right)\left(5-x\right)\)
\(=5x^3-x^4-10x^2+2x^3+5x-x^2-5+x\)
\(-x^4+7x^3-11x^2+6x-5\)
\(a.\left(x^2-3x+2\right)\left(x-2\right)\)
\(=x^3-2x^2-3x^2+6x+2x-4\)
\(=x^3-5x^2+8x-4\)
b. Từ kết quả ở câu a, ta có:
\(\left(x^2-3x+2\right)\left(x-2\right)\)
\(=\left(-x^2-3\left(-x\right)+2\right)\left(2-\left(-x\right)\right)\) (ta thay x bằng -x nha)
\(=\left(-x^2+3x+2\right)\left(2+x\right)\)
\(=-2x^2-x^3+6x+3x^2+2+2x\)
\(=-x^3+x^2+8x+2\)
( x3 - 2x2 + x - 1 ) ( 5-x )
= 5x3 - x4 - 10x2 + 2x3 + 5x - x2 - 5 + x
= - x4 + 7x3 -11x2 + 6x - 5
=> ( x3 - 2x2 + x -1 ) ( x - 5 )
= x4 - 7x3 + 11x2 - 6x + 5