3x3-14x2+4x+3
a7+a2+1
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c) Ta có: \(\dfrac{5x^4+9x^3-2x^2-4x-8}{x-1}\)
\(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)
\(=\dfrac{5x^3\left(x-1\right)+14x^2\left(x-1\right)+12x\left(x-1\right)+8\left(x-1\right)}{x-1}\)
\(=5x^3+14x^2+12x+8\)
d) Ta có: \(\dfrac{5x^3+14x^2+12x+8}{x+2}\)
\(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}\)
\(=\dfrac{5x^2\left(x+2\right)+4x\left(x+2\right)+4\left(x+2\right)}{x+2}\)
\(=5x^2+4x+4\)
c) Ta có: \(\dfrac{5x^4+9x^3-2x^2-4x-8}{x-1}\)
\(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)
\(=\dfrac{5x^3\left(x-1\right)+14x^2\left(x-1\right)+12x\left(x-1\right)+8\left(x-1\right)}{x-1}\)
\(=5x^3+14x^2+12x+8\)
1: Sửa đề: 3x-5
\(=\dfrac{-x^2\left(3x-5\right)-3\left(3x-5\right)}{3x-5}=-x^2-3\)
2: \(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)
=5x^2+14x^2+12x+8
3: \(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}=5x^2+4x+4\)
4: \(=\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}=x^2+1-2x\)
5: \(=\dfrac{x^2\left(5-3x\right)+3\left(5-3x\right)}{5-3x}=x^2+3\)
Bài 3:
Ta có: \(2n^2+n-7⋮n-2\)
\(\Leftrightarrow2n^2-4n+5n-10+3⋮n-2\)
\(\Leftrightarrow n-2\in\left\{1;-1;3;-3\right\}\)
hay \(n\in\left\{3;1;5;-1\right\}\)
a) ( x 2 – 4x + 1)( x 2 – 2x + 3). b) (3x – y – 1)(x – 7y – 1).
b: \(\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}\)
\(=x^2-2x+1\)
\(=\left(x-1\right)^2\)
c: \(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)
\(=5x^3+14x^2+12x+8\)
\(\dfrac{3x^3+4x-1}{x+1}\)
\(=\dfrac{3x^3+3x^2-3x^2-3x+7x+7-8}{x+1}\)
\(=3x^2-3x+7-\dfrac{8}{x+1}\)
lim x → − 1 + x 2 + 4 x + 3 x 3 + x 2 = lim x → − 1 + x + 1 x + 3 x 2 x + 1 = lim x → − 1 + x + 1 x + 3 x 2 = 0 1 = 0.
Chọn đáp án D
\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}...\frac{2018}{2019}\)
\(=\frac{1\cdot2\cdot3\cdot4\cdot\cdot\cdot2018}{2\cdot3\cdot4\cdot5\cdot\cdot\cdot2019}\)
\(=\frac{1\cdot\left(2\cdot3\cdot4\cdot\cdot\cdot2018\right)}{\left(2\cdot4\cdot5\cdot\cdot\cdot2018\right)\cdot2019}\)
\(=\frac{1}{2019}\)
Vậy .......................................
Ta có ; \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.......\frac{2015}{2016}\)
\(=\frac{1.2.3......2015}{2.3.4.....2016}\)
\(=\frac{1}{2016}\)