\(\dfrac{a}{b}\) = \(\dfrac{c}{d}\)

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

   \(\dfrac{a}{c}\)  =  \(\dfrac{5a}{5c}\) = \(\dfrac{3b}{3d}\) Áp dụng tính chất dãy tỉ số bằng nhau ta có:

      \(\dfrac{a}{c}\) =   \(\dfrac{5a+3b}{5c+3d}\) (1) 

       \(\dfrac{a}{c}\) = \(\dfrac{5a-3b}{5c-3d}\)  (2)

Kết hợp (1) và (2) ta có:

       \(\dfrac{5a+3b}{5c+3d}\) =  \(\dfrac{5a-3b}{5c-3d}\) 

⇒   \(\dfrac{5a+3b}{5a-3b}\) =  \(\dfrac{5c+3d}{5c-3d}\) (đpcm)

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

      \(\dfrac{a}{b}\) =  \(\dfrac{3a}{3b}\) = \(\dfrac{2c}{2d}\)

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

     \(\dfrac{a}{b}\) = \(\dfrac{3a+2c}{3b+2d}\) (đpcm)

Ta có:

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

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

Xét:

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

\(=1+\dfrac{c}{a-b}\left[\dfrac{b\left(b-c\right)+a\left(c-a\right)}{ab}\right]=1+\dfrac{c}{a-b}\left(\dfrac{b^2-bc+ac-a^2}{ab}\right)\)

\(=1+\dfrac{c}{a-b}\left[\dfrac{\left(b-a\right)\left(b+a\right)-c\left(b-a\right)}{ab}\right]=1+\dfrac{c}{a-b}.\dfrac{\left(b-a\right)\left(a+b-c\right)}{ab}\)

\(=1-\dfrac{c\left(a+b-c\right)}{ab}=1-\dfrac{c.\left(-2c\right)}{ab}=1+\dfrac{2c^2}{ab}\) (do \(a+b+c=0\Rightarrow a+b=-c\))

Tương tự:

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

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

\(\Rightarrow P=3+2\left(\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab}\right)=3+\dfrac{2\left(a^3+b^3+c^3\right)}{abc}\)

Mặt khác ta có đằng thức quen thuộc:

Khi \(a+b+c=0\) thì \(a^3+b^3+c^3=3abc\)

\(\Rightarrow P=3+\dfrac{2.3abc}{abc}=9\)

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 \(\dfrac{a}{c}=\dfrac{a-b}{c-d}\) 

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

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

\(\dfrac{a}{b}=\dfrac{c}{d}=>ad=cb=>ad+ac=cb+ac\)

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

a) \(\dfrac{a}{b}=\dfrac{c}{d}\)

nên \(\dfrac{a}{b}+1=\dfrac{c}{d}+1\)

hay \(\dfrac{a+b}{b}=\dfrac{c+d}{d}\)

b) \(\dfrac{a}{b}=\dfrac{c}{d}\)

nên \(\dfrac{a}{b}-1=\dfrac{c}{d}-1\)

hay \(\dfrac{a-b}{b}=\dfrac{c-d}{d}\)

Áp dụng BĐT BSC:

\(\dfrac{ab}{a+b}+\dfrac{bc}{b+c}+\dfrac{ca}{c+a}\)

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

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

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

Đẳng thức xảy ra khi \(a=b=c\)

4ab ≤ (a + b)² ⇒ \(\dfrac{4ab}{a+b}\le a+b\)

Tương tự \(\dfrac{4ac}{a+c}\le a+c\) ; \(\dfrac{4bc}{b+c}\le b+c\)

⇒ Cộng lại vế với vế :

4VT ≤ 2 (a+b+c) ⇒ VT ≤ \(\dfrac{a+b+c}{2}\)

Có \(\dfrac{a}{b}=\dfrac{c}{d}=>ad=bc\) => a² = ad => a=d

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

<=> (a+b)(c-a) = (a-b)(c+a)

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

<=> ac - ad + bc - bd = ac + ad -bc -bd

<=> 2bc = 2ad (luôn đúng) => đpcm

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

Áp dụng t/c dtsbn:

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

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