Ta có:

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

\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=k^2\)

\(\Rightarrow\dfrac{1}{a^2}+\dfrac{2}{ab}+\dfrac{1}{b^2}+\dfrac{2}{bc}+\dfrac{1}{c^2}+\dfrac{2}{ac}=k^2\)

\(\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{1\left(a+b+c\right)}{abc}=k^2\)

\(\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=k^2-k\)

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

k=6
a,b,c=2

\(\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}\ge3\sqrt[3]{\dfrac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)

\(\dfrac{a}{1+a}+\dfrac{b}{1+b}+\dfrac{c}{1+c}\ge3\sqrt[3]{\dfrac{abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)

Cộng vế và rút gọn:

\(\Rightarrow1\ge\dfrac{1+\sqrt[3]{abc}}{\sqrt[3]{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)

\(\Rightarrow\left(1+a\right)\left(1+b\right)\left(1+c\right)\ge\left(1+\sqrt[3]{abc}\right)^3\)

\(\Rightarrow\dfrac{\left(1+a\right)\left(1+b\right)\left(1+c\right)}{abc}\ge\dfrac{\left(1+\sqrt[3]{abc}\right)^3}{abc}=\left(\dfrac{1}{\sqrt[3]{abc}}+1\right)^3\ge\left(\dfrac{3}{a+b+c}+1\right)^3\ge\left(\dfrac{3}{k}+1\right)^3\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

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

Vậy \(k=3\)

a/c=b/d=k

=>a=ck; b=dk

=>\(\dfrac{c\cdot a^2+d\cdot b^2}{c^3+d^3}\)

\(=\dfrac{c\cdot c^2k^2+d\cdot d^2k^2}{c^3+d^3}=k^2\)

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

\(\Rightarrow a=c\times k\)

\(b=d\times k\)

\(\dfrac{c.\left(c.k\right)^2+d.\left(d.k\right)^2}{c^3+d^3}\)

=\(\dfrac{c^3.k^2+d^3.k^2}{c^3+d^3}\)

=\(\dfrac{k^2\left(c^3+d^3\right)}{1\left(c^3+d^3\right)}\)=k²

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