toán nâng cao!- phân tích đa thức thành nhân tử

T

thaolovely1412

2) Ta có:
a)[TEX] x+1+\frac{2x+3}{3}+ \frac{3x+5}{5} +...+ \frac{20x+39}{39} = 22+\frac{4}{3} +\frac{6}{5}+...+\frac{40}{39}[/TEX]
[TEX]\Leftrightarrow ( x+ \frac{2}{3x}+\frac{3}{5x}+...+ \frac{20}{39x}+20 =22+\frac{4}{3} +\frac{6}{5}+...+\frac{40}{39}[/TEX]
[TEX]\Leftrightarrow (1+\frac{2}{3} +\frac{3}{5} +..+\frac{20}{39})x = 2(\frac{2}{3} +\frac{3}{5} +..+\frac{20}{39}) [/TEX]
[TEX]\Leftrightarrow x=2 [/TEX]
 
L

lehiepteo

Phân tích thành nhân tử

A= (x-y)³+(y-x)³ +(z-x)³
B=(ab+bc+ca)(a+b+c)-abc
C=a(b²-c²)+b(c²-a²)+c(a²-b²)
 
T

thaolovely1412

[TEX]A= (x-y)^3+(y-z)^3 +(z-x)^3[/TEX]
Đặt [TEX]x-y=a[/TEX], [TEX]y-z=b[/TEX], [TEX]z-x=c[/TEX] thì [TEX]a+b+c=0[/TEX]
\Rightarrow [TEX]a^3+b^3+c^3=3abc[/TEX]
Vậy [TEX]A= 3(x-y)(y-z)(z-x)[/TEX]
C= a(b²-c²)-b(c²-a²)+c(a²-b²)
=a(b²-c²)+b(a²-c²)+c(a²-b²)
=a(b²-c²)+b[(a²-b²)+(b²-c²)]+c(a²-b²)
=a(b²-c²)+b(b²-c²)+b(a²-b²)+c(a²-b²)
=(b²-c²)(a+b)+(a²-b²)(b+c)
=(b-c)(b+c)(a+b)+(a+b)(a-b)(b+c)
=(a+b)(b+c)(b-c+a-b)
=(a+b)(b+c)(a-c)
 
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