đại số 8 , bài khó

H

huuthuyenrop2

Bài 2:
$\dfrac{x^3+y^3+z^3-3xyz}{(x-y)^2+(y-z)^2+(z-x)^2}$
\Leftrightarrow $\dfrac{(x+y+z)(x^2+y^2+z^2-xy-xz-yz)}{(x-y)^2+(y-z)^2+(z-x)^2}$
\Leftrightarrow $\dfrac{2(x+y+z)(x^2+y^2+z^2-xy-xz-yz)}{2.[(x-y)^2+(y-z)^2+(z-x)^2]}$
\Leftrightarrow $\dfrac{(x+y+z)(2x^2+2y^2+2z^2-2xy-2xz-2yz}{2.[(x-y)^2+(y-z)^2+(z-x)^2]}$
\Leftrightarrow $\dfrac{(x+y+z).[(x-y)^2+(y-z)^2+(z-x)^2]}{2.[(x-y)^2+(y-z)^2+(z-x)^2]}$
\Leftrightarrow $\dfrac{x+y+z}{2}$
Mà x+y+z= 2
\Rightarrow $\dfrac{x^3+y^3+z^3-3xyz}{(x-y)^2+(y-z)^2+(z-x)^2} = 1$
 
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S

sam_chuoi

Umbala

1. $3a^2+3b^2=10ab <=> 3a^2-10ab+3b^2=0 <=> (a-3b)(3a-b)=0$ suy ra a=3b hoặc b=3a. Thay từng TH vào rồi tính. 2. Có $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)$ , $(x-y)^2+(x-z)^2+(y-z)^2=2(x^2+y^2+z^2-xz-yz-xy$. Suy ra kq là 1.
 
H

huy14112

1.

$\dfrac{a^2}{ab+b^2}+\dfrac{b^2}{ab-a^2}+\dfrac{a^2+b^2}{ab}$

$=\dfrac{a^2}{b(b+a)}+\dfrac{b^2}{a(b-a)}+\dfrac{a^2+b^2}{ab}$

$=\dfrac{a^2.a(b-a)+b.b^2(a+b)+(a^2+b^2)(b^2-a^2)}{ab(a+b)(b-a)}$

$=\dfrac{a^3(b-a)+b^3(a+b)+(a^2+b^2)(b^2-a^2)}{ab(b^2-a^2)}$

$=\dfrac{a^3b-a^4+b^3a+b^4+(a^2+b^2)(b^2-a^2)}{ab(b^2-a^2)}$

$=\dfrac{ab(a^2+b^2)+(b^4-a^4)+(a^2+b^2)(b^2-a^2)}{ab(b^2-a^2)}$

$=\dfrac{ab(a^2+b^2)}{ab(a^2-b^2)}+\dfrac{2(a^2+b^2)(b^2-a^2)}{ab(b^2-a^2)}$

$=\dfrac{a^2+b^2}{b^2-a^2}+\dfrac{2(a^2+b^2)}{ab}=\dfrac{a^2+b^2}{b^2-a^2}+\dfrac{3a^2+3b^2}{1,5ab}=\dfrac{a^2+b^2}{b^2-a^2}+\dfrac{20}{3}$

Lại có : $3a^2+3b^2=10ab$

$\leftrightarrow 3a^2+3b^2-10ab=0$

$3a(a-3b)-b(a-3b)=0$

$(3a-b)(a-3b)=0$

Vậy xảy ra 2TH :

$3a=b$ hoặc $a=3b$

Thay vào $\dfrac{a^2+b^2}{b^2-a^2}+\dfrac{20}{3}$

có :

_$\dfrac{a^2+9a^2}{9a^2-a^2}+\dfrac{20}{3}=\dfrac{10a^2}{8a^2}+\dfrac{20}{3}=\dfrac{10}{8}+\dfrac{20}{3}=\dfrac{112}{15}$

hoặc :

_$\dfrac{9b^2+b^2}{b^2-9b^2}+\dfrac{20}{3}=\dfrac{10b^2}{-8b^2}+\dfrac{20}{3}=\dfrac{-10}{8}+\dfrac{20}{3}=\dfrac{65}{12}$

 
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