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

=>\(a=b\cdot k;c=d\cdot k\)

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

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

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

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

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

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

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

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

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

4: \(\dfrac{5a+3b}{5c+3d}=\dfrac{5\cdot bk+3b}{5dk+3d}=\dfrac{b\left(5k+3\right)}{d\left(5k+3\right)}=\dfrac{b}{d}\)

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

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

Bài 1:

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Ta có: \(b^2=a\cdot c\)

=>\(\frac{a}{b}=\frac{b}{c}\) (1)

Ta có: \(c^2=bd\)

=>\(\frac{b}{c}=\frac{c}{d}\) (2)

Từ (1),(2) suy ra \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)

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

=>\(\begin{cases}c=dk\\ b=ck=dk\cdot k=dk^2\\ a=bk=dk^2\cdot k=dk^3\end{cases}\)

a: \(\frac{a^3+b^3-c^3}{b^3+c^3-d^3}=\frac{\left(dk^3\right)^3+\left(dk^2\right)^3-\left(dk\right)^3}{\left(dk^2\right)^3+\left(dk\right)^3-d^3}=\frac{d^3k^3\left(k^6+k^3-1\right)}{d^3\left(k^6+k^3-1\right)}=k^3\)

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

\(=\left\lbrack\frac{dk\left(k^2+k-1\right)}{d\left(k^2+k-1\right)}\right\rbrack^3=k^3\)

Do đó: \(\frac{a^3+b^3-c^3}{b^3+c^3-d^3}=\left(\frac{a+b-c}{b+c-d}\right)^3\)

b: \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\)

\(=\frac{\left(dk^3\right)^3+\left(dk^2\right)^3+\left(dk\right)^3}{\left(dk^2\right)^3+\left(dk\right)^3+d^3}=\frac{d^3k^3\left(k^6+k^3+1\right)}{d^3\left(k^6+k^3+1\right)}=k^3\)

\(\frac{a}{d}=\frac{dk^3}{d}=k^3\)

Do đó: \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d}\)

Bài 1: 
Đặt a/b=c/d=k

=>a=bk; c=dk

a: \(\dfrac{a}{3a+b}=\dfrac{bk}{3bk+b}=\dfrac{k}{3k+1}\)

\(\dfrac{c}{3c+d}=\dfrac{dk}{3dk+d}=\dfrac{k}{3k+1}\)

Do đó: \(\dfrac{a}{3a+b}=\dfrac{c}{3c+d}\)

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

\(\dfrac{2c+3d}{2c-3d}=\dfrac{2dk+3d}{2dk-3d}=\dfrac{2k+3}{2k-3}\)

Do đó: \(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\)