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

=>a=bk; c=dk

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

\(\dfrac{3a+2c}{3b+2d}=\dfrac{3bk+2dk}{3b+2d}=k\)

Do đó: a/b=3a+2c/3b+2d

Bài 1: Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)

\(\dfrac{a}{a+c}=\dfrac{ck}{ck+c}=\dfrac{ck}{c\left(k+1\right)}=\dfrac{k}{k+1}\)

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

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

Ta có:

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

b/ \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{-2a}{-2b}=\dfrac{7c}{7d}=\dfrac{-2a+7c}{-2b+7d}\)

PS: Xong

Y chang câu mới giải nhé

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

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

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

`#3107.101107`

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

Ta có:

\(\dfrac{3b}{a}=\dfrac{3d}{c}\Rightarrow3bc=3da\Rightarrow bc=da\)

Vậy, từ tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\) ta có thể suy ra tỉ lệ thức \(\dfrac{3b}{a}=\dfrac{3d}{c}\)

\(\Rightarrow B.\)

a) ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{b}=\dfrac{4c}{4d}=\dfrac{a+4c}{b+4d}\left(đpcm\right)\)

b;c;d tương tự hết

b: a/b=c/d

nên 3a/3b=2c/2d

=>a/b=c/d=(3a+2c)/(3b+2d)

c: a/c=b/d nên a/c=2b/2d=(a-2b)/(c-2d)

d: a/c=b/d

nên 5a/5c=2b/2d

=>a/c=b/d=(5a-2b)/(5c-2d)

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

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

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

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

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

Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:

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

\(=\dfrac{a^2}{ab+2ac+3ad}+\dfrac{b^2}{bc+2bd+3ab}+\dfrac{c^2}{cd+2ac+3bc}+\dfrac{d^2}{ad+2bd+3cd}\)

\(\ge\dfrac{\left(a+b+c+d\right)^2}{4\left(ab+ad+bc+bd+ca+cd\right)}\ge\dfrac{\left(a+b+c+d\right)^2}{\dfrac{3}{2}\left(a+b+c+d\right)^2}=\dfrac{2}{3}\)

*Chứng minh \(4\left(ab+ad+bc+bd+ca+cd\right)\le\dfrac{3}{2}\left(a+b+c+d\right)^2\)

\(\Leftrightarrow\left(a-b\right)^2+\left(a-d\right)^2+\left(b-c\right)^2+\left(b-d\right)^2+\left(a-c\right)^2+\left(c-d\right)^2\ge0\)

Bài 1: Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)

\(\dfrac{a}{a+c}=\dfrac{ck}{ck+c}=\dfrac{ck}{c\left(k+1\right)}=\dfrac{k}{k+1}\)

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

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

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) ⇒ a = b.k ; c = d.k

\(\dfrac{3a-2c}{5a+4c}=\dfrac{3.b.k-2.d.k}{5.b.k+4.d.k}=\dfrac{k\left(3.b-2.d\right)}{k\left(5b+4d\right)}=\dfrac{3b-2d}{5b+4d}\)

\(\dfrac{3b-2d}{5b+4d}=\dfrac{3b-2d}{5b+4d}\Rightarrow\dfrac{3a-2c}{5a+4c}=\dfrac{3b-2d}{5b+4d}\)