mn trả lời chi tiết cho e vs ạ

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

\(\Rightarrow\left(b+d\right)c=\left(a+c\right)d\)

\(\Rightarrow dpcm\)

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

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

\(\Rightarrow dpcm\)

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

\(\Rightarrow\left(b-3d\right)\left(b-3d\right)=\left(3b+5d\right)\left(a-3c\right)\)

\(\Rightarrow dpcm\)

Đính chính câu c

\(\Rightarrow\left(3a+5c\right)\left(b-3d\right)=\left(3b+5d\right)\left(a-3c\right)\)

tu gia thiet \(\frac{a}{b}=\frac{c}{d}\Rightarrow a.d=b.c\)  (1)

\(\frac{a-b}{a}=\frac{c-d}{c}\) (*)

<=> \(\left(a-b\right)c=\left(c-d\right)a\)

<=> \(ac-bc=ac-ad\)

<=>  \(bc=ad\) (2)

do (1) nen (2) dung => (*) duoc chung minh

Vay neu \(\frac{a}{b}=\frac{c}{d}\)  thi \(\frac{a-b}{a}=\frac{c-d}{c}\)

Chuc em hoc tot !!!

Ta có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a.b}{c.d}\left(1\right)\)

Áp dụng tính chất dãy tỉ số bằng nhau tao có

\(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\left(2\right)\)

Từ (1) và (2) ta có ĐPCM

 A = a + b  -5

 B = - b - c  + 1

D  = b - a

A + B + D = (a + b  -5) + (-b - c + 1) + (b - a)

A + B + D = a + b - 5 - b - c + 1 + b - a

A + B + D = (a - a) + (b - b) + b - c  - (5 - 1)

A + B + D = 0 +  0 + b - c - 4

A + B + D = b - c - 4  = C 

Vậy A + B + D = C (đpcm) 

\(\frac{a}{b}=\frac{c}{d}\)(\(b,d\ne0\))

\(\Leftrightarrow ad=bc\)

\(\Leftrightarrow2ad=2bc\)

\(\Leftrightarrow ad-bc=bc-ad\)

\(\Leftrightarrow ad-bc+ac-bd=bc-ad+ac-bd\)

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

\(\Leftrightarrow\frac{a+b}{a-b}=\frac{c+d}{c-d}\)(\(a-b,c-d\ne0\))