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a)
b) (x – 1)(x + 1)(x2 + 1)
= [x .(x + 1) – 1 .(x + 1)] . (x2 + 1)
= {x.x + x.1 + (-1).x + (-1).1}. (x2 + 1)
= (x2 + x – x – 1) . (x2 + 1)
= (x2 – 1) . (x2 + 1)
= x2 . (x2 +1) – 1.(x2 + 1)
= x2 . x2 + x2 . 1 – (1.x2 + 1.1)
= x4 + x2 – (x2 + 1)
= x4 + x2 – x2 – 1
= x4 – 1
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=\dfrac{2x^4+x^3-5x^2-3x-3}{x^2-3}\)
\(=\dfrac{2x^4-6x^2+x^3-3x+x^2-3}{x^2-3}\)
\(=2x^2+x+1\)
b: \(=\dfrac{x^5+x^2+x^3+1}{x^3+1}=x^2+1\)
c: \(=\dfrac{2x^3-x^2-x+6x^2-3x-3+2x+6}{2x^2-x-1}\)
\(=x+3+\dfrac{2x+6}{2x^2-x-1}\)
d: \(=\dfrac{3x^4-8x^3-10x^2+8x-5}{3x^2-2x+1}\)
\(=\dfrac{3x^4-2x^3+x^2-6x^3+4x^2-2x-15x^2+10x-5}{3x^2-2x+1}\)
\(=x^2-2x-5\)
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x(x2 – y) – x2 (x + y) + y (x2– x) = x3 – xy – x3 – x2y + yx2 – yx= (2x-2y) – (x2 -2xy +y2) =2(x-y) – (x-y)2
Với x =1/2, y = -100 biểu thức có giá trị là -2 . 1/2. (-100) = 100.
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(\left(5x-2y\right)\left(x^2-xy+1\right)\)
\(=5x^3-5x^2y+5x-2x^2y+2xy^2-2y\)
\(=5x^3-7x^2y+2xy^2+5x-2y\)
b) Ta có: \(\left(x-1\right)\left(x+1\right)\left(x+2\right)\)
\(=\left(x^2-1\right)\left(x+2\right)\)
\(=x^3+2x^2-x-2\)
c) Ta có: \(\dfrac{1}{2}x^2y^2\cdot\left(2x+y\right)\left(2x-y\right)\)
\(=\dfrac{1}{2}x^2y^2\left(4x^2-y^2\right)\)
\(=2x^4y^2-\dfrac{1}{2}x^2y^4\)
`@` `\text {Ans}`
`\downarrow`
`(x-3)^2 - x^2 + 10x - 7`
`= x^2 - 6x + 9 - x^2 + 10x - 7`
`= (x^2 - x^2) + (-6x + 10x) + (9-7)`
`= 4x + 2`
\(\left(x-3\right)^2-x^2+10x-7\)
\(=\left(x^2-6x+9\right)-x^2+10x-7\)
\(=\left(x^2-x^2\right)-\left(6x-10x\right)+\left(9-7\right)\)
\(=4x+2\)
\(=2\left(2x+1\right)\)