a: Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)

nên \(\dfrac{a}{c}=\dfrac{b}{d}\)

d: Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)

nên \(\dfrac{a}{c}=\dfrac{b}{d}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

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

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

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

hay \(\dfrac{a}{b}=\dfrac{a+c}{b+d}\)

Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)

nên \(\dfrac{a}{c}=\dfrac{b}{d}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

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

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

Ta có: \(\left(a+b+c+d\right)\left(a-b-c+d\right)=\left(a-b+c-d\right)\left(a+b-c-d\right)\)

\(\Leftrightarrow\left(a+d\right)^2-\left(b+c\right)^2=\left(a-d\right)^2-\left(b-c\right)^2\)

\(\Leftrightarrow\left(a+d-a+d\right)\left(a+d+a-d\right)=\left(b+c-b+c\right)\left(b+c+b-c\right)\)

\(\Leftrightarrow2d\cdot2a=2c\cdot2b\)

\(\Leftrightarrow ad=bc\)

hay \(\dfrac{a}{c}=\dfrac{b}{d}\)

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

   \(\Leftrightarrow1+\dfrac{b}{a}=1+\dfrac{d}{c}\)

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

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

Áp dụng t/c dtsbn:

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

Từ \(\dfrac{a}{d}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}\Rightarrow\left(\dfrac{a}{c}\right)^{2014}=\left(\dfrac{b}{d}\right)^{2014}=\left(\dfrac{a-b}{c-d}\right)^{2014}\left(1\right)\)

Từ \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\Rightarrow\left(\dfrac{a}{c}\right)^{2014}=\left(\dfrac{b}{d}\right)^{2014}=\dfrac{a^{2014}+b^{2014}}{c^{2014}+d^{2014}}\left(2\right)\)

Từ (1) và (2) suy ra \(\dfrac{a^{2014}+b^{2014}}{c^{2014}+d^{2014}}=\left(\dfrac{a-b}{c-d}\right)^{2014}\)

Đề bài sai

Ví dụ: với \(a=1;b=2;c=3,d=4\) thì \(x=\dfrac{1}{2}\) ; \(y=\dfrac{3}{4}\) ; \(z=\dfrac{2}{3}\)

Khi đó  \(x< y\) nhưng \(z< y\)

\(\text{Vì }\dfrac{a}{b}< \dfrac{c}{d}\text{ nên }ad< bc\left(1\right)\)

\(\text{Xét tích}:a\left(b+d\right)=ab+ad\left(2\right)\)

                \(b\left(a+c\right)=ba+bc\left(3\right)\)

\(\text{Từ(1);(2);(3)}\Rightarrow a\left(b+d\right)< b\left(a+c\right)\text{ do đó }\dfrac{a}{b}< \dfrac{a+c}{b+d}\left(4\right)\)

\(\text{Tương tự ta có:}\dfrac{a+c}{b+d}< \dfrac{c}{d}\left(5\right)\)

\(\text{Từ (4);(5) ta được }\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{c}{d}\)

\(\Rightarrow x< y< z\)

a) \(\dfrac{a}{c}=\dfrac{a+b}{c+d}\)

=> a(c + d) = c(a + b)

=> ac + ad = ac + bc

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

b) \(\dfrac{b}{d}=\dfrac{a-b}{c-d}\)

=> b(c - d) = d(a - b)

=> bc - bd = ad - bd

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

Theo đề bài ta có:

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

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

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

\(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}=\dfrac{a+b-a+b}{c+d-c+d}=\dfrac{b}{d}\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\) ta có:

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

- Theo đề bài:

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

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

\(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)\(=\dfrac{a+b+a-b}{c+d+c-d}\)\(=\dfrac{2a}{2c}=\dfrac{a}{c}\) (1)

\(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)\(=\dfrac{a+b-\left(a-b\right)}{c+d-\left(c-d\right)}=\dfrac{a+b-a+b}{c+d-c+d}=\dfrac{2b}{2d}=\dfrac{b}{d}\) (2)

- Từ (1) và (2)

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

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

\(\left\{{}\begin{matrix}\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}\\\dfrac{a}{c}=\dfrac{b}{d}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\left(\dfrac{a}{c}\right)^2=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\\\left(\dfrac{a}{c}\right)^2=\dfrac{ab}{cd}\end{matrix}\right.\)

\(\Rightarrow\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)