theo bài ra ta có:

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

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

vì a + b + c + d khác 0 => a = b = c = d

ta có:

\(\Rightarrow\frac{4a}{a}=\frac{4b}{b}=\frac{4c}{c}=\frac{4d}{d}=k+1\)

=> 4 = 4 = 4 = 4 = k + 1

=> k + 1 = 4

=> k = 3

vật k = 3

theo đầu bài

=>\(\dfrac{b+c+d}{a}\)=\(\dfrac{c+d+a}{b}\)=\(\dfrac{d+a+b}{c}\)=\(\dfrac{a+b+c}{d}\)=\(\dfrac{b+c+d+c+d+a+d+a+b+a+b+c}{a+b+c+d}\)=\(\dfrac{3\left[a+b+c+d\right]}{a+b+c+d}\)=>=3

=>k=3

Theo tính chất dãy tỉ số bằng nhau ,ta có :

\(\dfrac{b+c+d}{a}=\dfrac{c+d+a}{b}=\dfrac{d+a+b}{c}=\dfrac{a+b+c}{d}=\dfrac{b+c+d+c+d+a+d+a+b+a+b+c}{a+b+c+d}\)

\(=\dfrac{3\left(a+b+c+d\right)}{a+b+c+d}=3\)

=> k = 3

sửa: \(\dfrac{b+c+d}{a}=\dfrac{c+d+a}{b}=\dfrac{d+a+b}{c}=\dfrac{a+b+c}{d}=k\)

giải:

\(\dfrac{b+c+d}{a}=\dfrac{c+d+a}{b}=\dfrac{d+a+b}{c}=\dfrac{a+b+c}{d}\\ =\dfrac{b+c+d+c+d+a+d+a+b+a+b+c}{a+b+c+d}\\ =\dfrac{3\left(a+b+c+d\right)}{a+b+c+d}=3=k\)

vậy k=3

Giải :

Cộng thêm 1 vào mỗi tỉ số đã cho ta được:

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

Vì a+b+c+d \(\ne\)0 nên a=b=c=d

=>k=\(\dfrac{3a}{a}=3\)

Cách 1:

Ta xét tích a(c-d) và c(a-b)

Ta có: a(c-d)=ac-ad (1)

           c(a-b)=ac-bc(2)

Ta lại có \(\dfrac{a}{c}=\dfrac{c}{d}\)=>ad=bc (3)

Từ (1), (2), (3) ta có a(c-d)=c(a-d). Do đó \(\dfrac{a}{a-b}=\dfrac{c}{c-d}\)

Cách 2:

 Đặt \(\dfrac{a}{b}=\dfrac{c}{d}\)=k thì a=bk, c=dk. 

Xét \(\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\left(1\right)\)

Xét \(\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\left(2\right)\)

Từ (1) và (2)=> \(\dfrac{a}{a-b}=\dfrac{c}{c-d}\)

Cách 3: Ta có

\(\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{a}{c}=\dfrac{b}{d}\)

Aps dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{a}{c}=\dfrac{b}{d}=>\dfrac{a-b}{c-d}\)

=>\(\dfrac{a}{c}=\dfrac{a-b}{c-d}=>\dfrac{a}{a-b}=\dfrac{c}{c-d}\)

Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)

\(\Leftrightarrow\dfrac{b}{a}=\dfrac{d}{c}\)

\(\Leftrightarrow\dfrac{b}{a}-1=\dfrac{d}{c}-1\)

\(\Leftrightarrow\dfrac{b-a}{a}=\dfrac{d-c}{c}\)

\(\Leftrightarrow\dfrac{a-b}{a}=\dfrac{c-d}{c}\)

hay \(\dfrac{a}{a-b}=\dfrac{c}{c-d}\)(đpcm)

Chọn đẳng thức \(\dfrac{d}{b}=\dfrac{c}{a}\) nhé bạn

\(\dfrac{a}{b}=\dfrac{c}{d}\)

\(\Rightarrow\dfrac{d}{b}=\dfrac{c}{a}\)

sorry đăng nhầm,cái này mk hỏi có bn trả lời rồi

\(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}=\dfrac{1}{3}\\ \Rightarrow\left\{{}\begin{matrix}b+c+d=3a\\a+c+d=3b\\a+b+d=3c\\a+b+c=3d\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b+c+d=2a\\a+b+c+d=2b\\a+b+c+d=2c\\a+b+c+d=2d\end{matrix}\right.\\ \Rightarrow2a=2b=2c=2d\\ \Rightarrow a=b=c=d\\ \Rightarrow A=\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}=1+1+1+1=4\)

ab+c+d=ba+c+d=ca+b+d=da+b+c=a+b+c+d3(a+b+c+d)=13⇒⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩b+c+d=3aa+c+d=3ba+b+d=3ca+b+c=3d⇒⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩a+b+c+d=2aa+b+c+d=2ba+b+c+d=2ca+b+c+d=2d⇒2a=2b=2c=2d⇒a=b=c=d⇒A=a+aa+a+a+aa+a+a+aa+a+a+aa+a=1+1+1+1=4

C. \(\dfrac{b}{d}=\dfrac{c}{a}\)

Chúc bạn học tốt!!