## cho a/(b+c)+b/(c+a)+c/(a+b)=1.CMR a^2/(b+c)+b^2/(c+a)+c^2/(a+b)=0

Question

cho a/(b+c)+b/(c+a)+c/(a+b)=1.CMR a^2/(b+c)+b^2/(c+a)+c^2/(a+b)=0

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2 tháng 2021-10-20T02:24:38+00:00 2 Answers 6 views 0

## Answers ( )

1. Giải thích các bước giải :

a/(b+c)+b/(c+a)+c/(a+b)=1

<=>(a+b+c)(a/(b+c)+b/(c+a)+c/(a+b))=a+b+c

<=>(a.(a+b+c))/(b+c)+(b.(a+b+c))/(c+a)+(c.(a+b+c))/(a+b)=a+b+c

<=>(a^2+a.(b+c))/(b+c)+(b^2+b.(c+a))/(c+a)+(c^2+c.(a+b))/(a+b)=a+b+c

<=>a^2/(a+b)+a+b^2/(c+a)+b+c^2/(a+b)+c=a+b+c

<=>a^2/(a+b)+b^2/(c+a)+c^2/(a+b)=a+b+c-a-b-c

<=>a^2/(a+b)+b^2/(c+a)+c^2/(a+b)=0

Vậy : a/(b+c)+b/(c+a)+c/(a+b)=1 thì a^2/(a+b)+b^2/(c+a)+c^2/(a+b)=0

2. Giải thích các bước giải:

Ta có:

$\dfrac{a^2}{b+c}$

$=a\cdot \dfrac{a}{b+c}$

$=a\cdot (\dfrac{a}{b+c}+1-1)$

$=a\cdot (\dfrac{a+b+c}{b+c}-1)$

$=\dfrac{a}{b+c}\cdot (a+b+c)-a$

$\to \dfrac{a^2}{b+c}=\dfrac{a}{b+c}\cdot (a+b+c)-a(1)$

Tương tự

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

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

Cộng vế với vế của $(1), (2), (3)$

$\to \dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=(a+b+c)\cdot (\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b})-(a+b+c)$

$\to \dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=(a+b+c)\cdot 1-(a+b+c)$

$\to \dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=0$