\(\dfrac{a}{b}\) = \(\dfrac{c}{d}\)

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

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

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

Làm tương tự với 3 phân số còn lại và cộng vế với vế

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

Làm tương tự với 3 phân số còn lại và cộng vế với vế

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

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

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

= \(\dfrac{ab-ad+bc-cd-ab+bc-da+cd}{\left(b+d\right)\left(b-d\right)}\) 

= 0

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

lol

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

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

\(VT\ge\dfrac{\left(2a+2b+2c+2d\right)^2}{\left(a+b\right)\left(a+c\right)+\left(b+c\right)\left(b+d\right)+\left(a+c\right)\left(c+d\right)+\left(a+d\right)\left(b+d\right)}=\dfrac{4\left(a+b+c+d\right)^2}{\left(a+b+c+d\right)^2}=4\)

Dấu "=" xảy ra khi \(a=b=c=d\)

Đặt a/b=c/d=k

=>a=bk; c=dk

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

\(\dfrac{c+d}{c}=\dfrac{dk+d}{d}=\dfrac{k+1}{k}\)

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

Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)

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

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

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

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

Bài 1: Nhân chéo

Bài 2:

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

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

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

\(\Rightarrowđpcm\)

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

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

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

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

\(=\dfrac{2b}{2b}=1\)

\(\Rightarrow a+b+c=a+b-c\)

\(\Rightarrow c=-c\)

\(\Rightarrow c+c=0\)

\(\Rightarrow2c=0\Rightarrow c=0\)

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

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

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

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

\(=\left(\dfrac{a+b+c}{b+c+d}\right)^3\left(2\right)\)

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

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

a: Đặt a/b=c/d=k

=>a=bk; c=dk

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

\(\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{k}{k-1}\)

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

b: Đặt a/b=c/d=k

=>a=bk; c=dk

\(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bk+b}{dk+d}\right)^2=\dfrac{b^2}{d^2}\)

\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2}{d^2}\)

DO đó: \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)