Question

[❖ミ★𝓣𝓻𝓪𝓷𝓰 (𝓯𝓪𝓷 𝓛𝓝𝓓)★彡『ʈєɑɱ❖๖ۣۜƝƘ☆...]

 Giúp tui với tui sẽ tick nha

Answer 1

[Xyz OLM]

Ta có \(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)

=> \(\frac{a+b+c+d}{a}+1=\frac{a+b+c+d}{b}+1=\frac{a+b+c+d}{c}+1=\frac{a+b+c+d}{d}+1\)

=> \(\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)

Nếu a + b + c + d = 0 

=> a + b = -(c + d) ; 

b + c = -(d + a)

Khi đó

M = \(\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=\frac{-\left(c+d\right)}{c+d}+\frac{-\left(d+a\right)}{d+a}+\frac{c+d}{-\left(c+d\right)}+\frac{d+a}{-\left(d+a\right)}\)

= (-1) + (-1) + (-1) + (-1) = -4

Nếu a + b + c + d \(\ne\)0

=> \(\frac{1}{a}=\frac{1}{b}=\frac{1}{c}=\frac{1}{d}\Rightarrow a=b=c=d\)

Khi đó M = \(\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=\frac{2a}{2a}+\frac{2b}{2b}+\frac{2c}{2c}+\frac{2d}{2d}=1+1+1+1=4\)

Vậy khi a + b + c + d = 0 thì M  =-4

khi a + b + c+ d \(\ne\)0 thì M = 4

Answer 2

[tran vinh]

 ta có: 2a+b+c+d/a=a+2b+c+d/b=a+b+2c+d/c=a+b+c+2d/d=2a+b+c+d+a+2b+c+d+a+b+2c+d+a+b+c+2d/a+b+c+d=3

suy ra 2a+b+c+d=3a suy ra a=b+c+d

tương tự: b=a+c+d,c=a+b+d,d=a+b+c

cộng lại ta có: a+b+c+d=b+c+d+a+c+d+a+b+d+a+b+c

suy ra 0=2(a+b+c+d) suy ra a+b+c+d=0 suy ra a+b=-(c+d),b+c=-(d+a),(c+d)=-(a+b),(d+a)=-(b+c) (1)

vì thế thay (1) vào M ta tính đc M=-4