Toán 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 20/10/2021 By Maya 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
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` Trả lời
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 $(1), (2), (3)$ $\to \dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=(a+b+c)\cdot (\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b})-(a+b+c)$ $\to \dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=(a+b+c)\cdot 1-(a+b+c)$ $\to \dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=0$ Trả lời
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`
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 $(1), (2), (3)$
$\to \dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=(a+b+c)\cdot (\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b})-(a+b+c)$
$\to \dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=(a+b+c)\cdot 1-(a+b+c)$
$\to \dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=0$