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: \(\dfrac{a}{a+c}=\dfrac{b}{b+d}\)

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: \(\dfrac{a}{a+c}=\dfrac{b}{b+d}\)

1) Ta có:
\(\dfrac{a}{a+b}\)=\(\dfrac{c}{c+d}\)
=>a.(c+d) = c.(a+b)
a.c+a.d = a.c+b.d
Do đó a.d=b.d
=>\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)( đpcm)

Câu 2: 

Đặt a/b=c/d=k

=>a=bk; c=dk

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

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

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

b: \(\dfrac{a^2}{b^2}=\dfrac{b^2k^2}{b^2}=k^2\)

\(\dfrac{2c^2-ac}{2d^2-bd}=\dfrac{c\left(2c-a\right)}{d\left(2d-b\right)}=\dfrac{dk}{d}\cdot\dfrac{2dk-bk}{2d-b}=k^2\)

=>\(\dfrac{a^2}{b^2}=\dfrac{2c^2-ac}{2d^2-bd}\)

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Sửa đề: Cho \(\frac{a}{a'}=\frac{b}{b'}=\frac{c}{c'}\)

Giải:

Dặt \(\frac{a}{a'}=\frac{b}{b'}=\frac{c}{c'}=k\Rightarrow\hept{\begin{cases}a=ka'\\b=kb'\\c=kc'\end{cases}}\)

Ta có:

\(\frac{a-3b+2c}{a'-3b'-2c'}=\frac{ka'-3kb'+2kc'}{a'-3b'+2c'}=\frac{k\left(a'-3b'+2c'\right)}{a'-3b'+2c'}=k=\frac{a}{a'}=\frac{b}{b'}=\frac{c}{c'}\)

\(từ:\frac{a}{a'}=\frac{b}{b'}=\frac{c}{c'}\Rightarrow\frac{a}{a'}=\frac{3b}{3b'}=\frac{2c}{2c'}=2018\)

Áp dụng tính chất của dãy tỉ số bằng nhau:

\(\frac{a}{a'}=\frac{3b}{3b'}=\frac{2c}{2c'}=\frac{a-3b+2c}{a-3b'+2c}=2018\)