Đặt a/b=c/d=k

=>a=bk; c=dk

a: \(\dfrac{2a+b}{2a-b}=\dfrac{2bk+b}{2bk-b}=\dfrac{2k+1}{2k-1}\)

\(\dfrac{2c+d}{2c-d}=\dfrac{2dk+d}{2dk-d}=\dfrac{2k+1}{2k-1}\)

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

b: \(\dfrac{2a+b}{a-2b}=\dfrac{2bk+b}{bk-2b}=\dfrac{2k+1}{k-2}\)
\(\dfrac{2c+d}{c-2d}=\dfrac{2dk+d}{dk-2d}=\dfrac{2k+1}{k-2}\)

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

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)

Mình hướng dẫn thôi nhé:

Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=kb\\c=kd\end{matrix}\right.\) . Sau đó thế vào biểu thức tính rồi suy ra đpcm

Ví dụ bài đầu tiên: Thế a = kb; c=kd vào biểu thức,ta có:

\(\dfrac{a}{a+b}=\dfrac{kb}{kb+b}=\dfrac{kb}{b\left(k+1\right)}=\dfrac{k}{k+1}\) (1)

\(\dfrac{c}{c+d}=\dfrac{kd}{kd+d}=\dfrac{kd}{d\left(k+1\right)}=\dfrac{k}{k+1}\) (2)

Từ (1) và (2) ,ta có đpcm: \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)

Các bài sau làm tương tự:Thế a=kb ; c=kd vào biểu thức rồi tính từng vế . Sau đó so sánh hai vế. Thấy hai vế = nhau => đpcm

Helpp 

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

\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có: \(\dfrac{5a+2b}{5a-2b}=\dfrac{5bk+2b}{5bk-2b}=\dfrac{5k+2}{5k-2}\)

\(\dfrac{5c+2d}{5c-2d}=\dfrac{5dk+2d}{5dk-2d}=\dfrac{5k+2}{5k-2}\)

Do đó: \(\dfrac{5a+2b}{5a-2b}=\dfrac{5c+2d}{5c-2d}\)

giải nốt câu b

Đặt:

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

\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

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

\(\Rightarrow\dfrac{a-2b}{c-2d}=\dfrac{bk-2b}{dk-2d}=\dfrac{b\left(k-2\right)}{d\left(k-2\right)}=\dfrac{b}{d}\)

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

ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow ad=bc\Leftrightarrow4ad=4bc\Leftrightarrow2ad+2ad=2bc+2bc\)

\(\Leftrightarrow2ad-2bc=2bc-2ad\Leftrightarrow ac+2ad-2bc-4bd=ac+2bc-2ad-4bd\)

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

Giải:

Theo đề ra, ta có:

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

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

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

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

\(\Rightarrowđpcm\).

Chúc bạn học tốt!!!

ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow ad=bc\Leftrightarrow2ab+ad=2ab+bc\)

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

Theo bài ra ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{2a}{2b}=\dfrac{c}{d}\)

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

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

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

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

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

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:

\(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

\(\dfrac{a+2b}{b}=\dfrac{bk+2b}{b}=\dfrac{b\left(k+2\right)}{b}=k+2\)

\(\dfrac{c+2d}{d}=\dfrac{dk+2d}{d}=\dfrac{d\left(k+2\right)}{d}=k+2\)

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

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

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

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

a) ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow ad=bc\Leftrightarrow ad+2bd=bc+2bd\)

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

b) ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow ad=bc\Leftrightarrow2ad=2bc\Leftrightarrow ad+ad=bc+bc\)

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

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

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

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

\(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow\left\{{}\begin{matrix}a=b.k\\c=dk\end{matrix}\right.\)

\(\dfrac{a+2b}{b}=\dfrac{bk+2b}{b}=\dfrac{b\left(2+k\right)}{b}=k+2\left(1\right)\)

\(\dfrac{c+2d}{d}=\dfrac{dk+2d}{d}=\dfrac{d\left(k+2\right)}{d}=k+2\left(2\right)\)

Từ (1) và (2)\(\Rightarrow\dfrac{a+2b}{b}=\dfrac{c+2d}{d}\)

Điều kiện đã cho có thể được viết lại thành \(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+d}+\dfrac{d}{d+a}=2\)

hay \(1-\dfrac{a}{a+b}-\dfrac{b}{b+c}+1-\dfrac{c}{c+d}-\dfrac{d}{d+a}=0\)

\(\Leftrightarrow\dfrac{b}{a+b}-\dfrac{b}{b+c}+\dfrac{d}{c+d}-\dfrac{d}{d+a}=0\)

\(\Leftrightarrow\dfrac{b^2+bc-ab-b^2}{\left(a+b\right)\left(b+c\right)}+\dfrac{d^2+da-cd-d^2}{\left(c+d\right)\left(d+a\right)}=0\)

\(\Leftrightarrow\dfrac{b\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}+\dfrac{d\left(a-c\right)}{\left(c+d\right)\left(d+a\right)}=0\)

\(\Leftrightarrow\left(c-a\right)\left[\dfrac{b}{\left(a+b\right)\left(b+c\right)}-\dfrac{d}{\left(c+d\right)\left(d+a\right)}\right]=0\)

\(\Leftrightarrow\dfrac{b}{\left(a+b\right)\left(b+c\right)}=\dfrac{d}{\left(c+d\right)\left(d+a\right)}\) (do \(c\ne a\))

\(\Leftrightarrow b\left(cd+ca+d^2+da\right)=d\left(ab+ac+b^2+bc\right)\)

\(\Leftrightarrow bcd+abc+bd^2+abd=abd+acd+b^2d+bcd\)

\(\Leftrightarrow abc+bd^2-acd-b^2d=0\)

\(\Leftrightarrow ac\left(b-d\right)-bd\left(b-d\right)=0\)

\(\Leftrightarrow\left(b-d\right)\left(ac-bd\right)=0\)

\(\Leftrightarrow ac=bd\) (do \(b\ne d\))

 Do đó \(A=abcd=ac.ac=\left(ac\right)^2\), mà \(a,c\inℕ^∗\) nên A là SCP (đpcm)