Question

[Thiên Minh Hoàng]

cho a+b=c=2024 và 1/a+b + 1/b+c + 1/c+a = 1/2024. tính M=a/b+c + b/c+a + c/a+b

Answer 1

[Lâm Gia Linh]

Điều kiện xác định $:$

$a\ne -b;b\ne -c;c\ne -a$

Theo giả thuyết, ta có :
$+)a+b+c=2024$

$+)\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{2}$

$\Rightarrow (a+b+c)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})=2024.\frac{1}{2}($ Nhân vế $-$ vế $)$

$\iff \frac{a}{a+b}+\frac{a}{b+c}+\frac{a}{c+a}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{c}{b+c}+\frac{c}{c+a}=1012$

$\iff (\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})+\frac{a}{a+b}+\frac{a}{c+a}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{c}{b+c}+\frac{c}{c+a}=1012$

$\iff N+(\frac{a}{a+b}+\frac{b}{a+b})+(\frac{b}{b+c}+\frac{c}{b+c})+(\frac{c}{c+a}+\frac{a}{c+a})=1012$

$\iff N=1012-\frac{a+b}{a+b}-\frac{b+c}{b+c}-\frac{c+a}{c+a}$

$\iff N=1012-1-1-1$

$\iff N=1009$

Vậy $N=1009$

Answer 2

[lê anh]

Ta có M=c/a+b +a/b+c + b/c+a
M+1+1+1=c/a+b +1+ a/b+c +1 +b/c+a+1
M+3=c/a+b + a+b/a+b + a/b+c + b+c/b+c + b/c+a + c+a/c+a
M+3=c+a+b/a+b + a+b+c/b+c + b+a+c/c+a
mà c+a+b=a+b+c=b+a+c=2024
=>M+3=2024/a+b + 2024/b+c + 2024/a+c
M+3=2024x(1/a+b + 1/b+c + 1/a+c)
mà 1/a+b + 1/b+c + 1/a+c=2024
=>M+3=2024x1/2024
M+3=1
M=1-3
M=-2