Rút Gọn toán 8

M

muzik1999

T

thong7enghiaha

c)

Xét tử thức:

$x^3-y^3+z^3+3xyz$

$=(x-y)^3+z^3+3x^2y-3xy^2+3xyz$

$=(x-y+z)(x^2-2xy+y^2-zx+yz+z^2)+3xy(x-y+z)$

$=(x-y+z)(x^2+y^2+z^2+xy+yz-zx)$

$=\dfrac{1}{2}.(x-y+z)[(x+y)^2+(y+z)^2+(z-x)^2]$

Thay vào biểu thức ta có:

$\dfrac{\dfrac{1}{2}(x-y+z)[(x+y)^2+(y+z)^2+(z-x)^2]}{(x+y)^2+(y+z)^2+(z-x)^2}$

$=\dfrac{1}{2}(x-y+z)$
 
N

nguyenhanhnt2012

hì

mẫu $=(x^3+4^3)=(x+4)(x^2-4x+16)$
tử $=2x(x^2-4x+16)$
Kết quả $=\dfrac{2x}{x+4}$
 
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A

anh_anh_1321

a. [TEX]\frac{32x - 8x^2 + 2x^3}{x^3 + 64 }[/TEX]

[TEX]=\frac{2x(x^2-4x+16)}{x^3+4^3 }[/TEX][TEX]=\frac{2x(x^2-4x+16)}{(x+4)(x^2-4x+16) }[/TEX][TEX]=\frac{2x}{x+4 }[/TEX]

b. [TEX]\frac{2x^3 - 7x^2 - 12x + 45}{3x^3 - 19x^2 + 33x -9}[/TEX]

[TEX]=\frac{2x^3-6x^2-x^2+3x-15x+45}{3x^3-9x^2-10x^2+30x+3x-9}[/TEX][TEX]=\frac{2x^2(x-3)-x(x-3)-15(x-3)}{3x^2(x-3)-10x(x-3)-3(x-3)}[/TEX]
[TEX]=\frac{(x-3)(2x^2-x-15)}{(x-3)(3x^2-10x-3)}[/TEX][TEX]=\frac{2x^2-x-15}{3x^2-10x-3}[/TEX][TEX]=\frac{2x^2-6x+5x-15}{3x^2-x-9x-3}[/TEX][TEX]=\frac{(x-3)(2x+5)}{(x-3)(3x-1)}[/TEX][TEX]=\frac{2x+5}{3x-1}[/TEX]
 
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