LD
H9
11 tháng 8 2024 lúc 8:59

\(1,x^2+2x+1-y^2\\ =\left(x^2+2x+1\right)-y^2\\ =\left(x+1\right)^2-y^2\\ =\left(x-y+1\right)\left(x+y+1\right)\\ 2,x^2+2x-3\\ =\left(x^2+3x\right)+\left(-x-3\right)\\ =x\left(x+3\right)-\left(x+3\right)\\ =\left(x-1\right)\left(x+3\right)\\ 3,4x^2-4xy+y^2-7\\ =\left(4x^2-4xy+y^2\right)-\left(\sqrt{7}\right)^2\\ =\left(2x-y\right)^2-\left(\sqrt{7}\right)^2\\ =\left(2x-y-\sqrt{7}\right)\left(2x-y+\sqrt{7}\right)\\ 4,8x^3-\dfrac{1}{8}\\ =\left(2x\right)^3-\left(\dfrac{1}{2}\right)^3\\ =\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\\ 5,\dfrac{1}{25}x^2-64y^2\\ =\left(\dfrac{1}{5}x\right)^2-\left(8y\right)^2\\ =\left(\dfrac{1}{5}x-8y\right)\left(\dfrac{1}{5}x+8y\right)\\ 6,x^3+\dfrac{1}{27}\\ =x^3+\left(\dfrac{1}{3}\right)^3\\ =\left(x+\dfrac{1}{3}\right)\left(x^2+\dfrac{1}{3}x+\dfrac{1}{9}\right)\)

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H9
11 tháng 8 2024 lúc 9:00

\(7,x^3+y^3+z^3-3xyz\\ =\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\\ =\left[\left(x+y\right)^3+z^3\right]+\left[-3xy\left(x+y\right)-3xyz\right]\\ =\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\\ =\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2-3xy\right]\\ =\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\\ =\left(x^2-xy+y^2-xz-yz+z^2\right)\)

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NT
11 tháng 8 2024 lúc 9:01

1: \(x^2+2x+1-y^2\)

\(=\left(x+1\right)^2-y^2\)

=(x+1+y)(x+1-y)

2: \(x^2+2x-3\)

\(=x^2+3x-x-3\)

=x(x+3)-(x+3)

=(x+3)(x-1)

3: \(4x^2-4xy+y^2-7\)

\(=\left(2x-y\right)^2-7\)

\(=\left(2x-y-\sqrt{7}\right)\left(2x-y+\sqrt{7}\right)\)

4: \(8x^3-\dfrac{1}{8}=\left(2x\right)^3-\left(\dfrac{1}{2}\right)^3=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)

5: \(\dfrac{1}{25}x^2-64y^2=\left(\dfrac{1}{5}x\right)^2-\left(8y\right)^2\)

\(=\left(\dfrac{1}{5}x-8y\right)\left(\dfrac{1}{5}x+8y\right)\)

6: \(x^3+\dfrac{1}{27}=x^3+\left(\dfrac{1}{3}\right)^3\)

\(=\left(x+\dfrac{1}{3}\right)\left(x^2-\dfrac{1}{3}x+\dfrac{1}{9}\right)\)

7: \(x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz\)

\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)

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TH
11 tháng 8 2024 lúc 9:01

`1, x^2 + 2x + 1 - y^2`

`= x^2 + 2 . x . 1 +  1 - y^2`

`= (x + 1)^2 - y^2`

`= (x+1-y)(x+1+y)`

`2, x^2 + 2x - 3`

`= x^2 + 3x - x - 3`

`= x(x + 3)-(x+3)`

`= (x-1)(x+3)`

`3, 4x^2-4xy+y^2-7`

`= (2x)^2 - 2 . 2x . y + y^2 - sqrt(7)^2`

`= (2x - y)^2 - sqrt(7)^2`

`= (2x - y - sqrt(7))(2x - y + sqrt(7))`

 

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