Giải

a) $\frac{a-2b}{b}=\frac{c-2d}{d}$

Đặt $\frac{a}{b}=\frac{c}{d}=k\Rightarrow \left\{ \begin{align}& a=b.k \\ & c=d.k \\ \end{align} \right.$

Khi đó $\frac{a-2b}{b}=\frac{b.k-2b}{b}=\frac{b\left( k-2 \right)}{b}=k-2$
$\frac{c-2d}{d}=\frac{d.k-2d}{d}=\frac{d\left( k-2 \right)}{d}=k-2$

Vậy $\frac{a-2b}{b}=\frac{c-2d}{d}$

b) $\frac{a-2c}{3a+c}=\frac{b-2d}{3b+d}$

Ta có : $\frac{a}{b}=\frac{c}{d}\Leftrightarrow \frac{a}{c}=\frac{b}{d}=k\Rightarrow \left\{ \begin{align}& a=c.k \\ & b=d.k \\ \end{align} \right.$

Khi đó : $\frac{a-2c}{3a+c}=\frac{c.k-2c}{3.c.k+c}=\frac{c(k-2)}{c(3k+1)}=\frac{k-2}{3k+1}$

$\frac{b-2d}{3b+d}=\frac{d.k-2d}{3.d.k+d}=\frac{d\left( k-2 \right)}{d\left( 3k+1 \right)}=\frac{k-2}{3k+1}$

Vậy $\frac{a-2c}{3a+c}=\frac{b-2d}{3b+d}$

c) $\frac{{{a}^{2}}-2{{b}^{2}}}{{{(a+4b)}^{2}}}=\frac{{{c}^{2}}-2{{d}^{2}}}{{{(c+4d)}^{2}}}$

Đặt $\frac{a}{b}=\frac{c}{d}=k\Rightarrow \left\{ \begin{align}& a=b.k \\ & c=d.k \\
\end{align} \right.$

Khi đó $\frac{{{a}^{2}}-2{{b}^{2}}}{{{(a+4b)}^{2}}}=\frac{{{(b.k)}^{2}}-2{{b}^{2}}}{{{(b.k+4b)}^{2}}}=\frac{{{b}^{2}}.\left( {{k}^{2}}-2 \right)}{{{b}^{2}}.{{\left( k+4 \right)}^{2}}}=\frac{{{k}^{2}}-2}{{{\left( k+4 \right)}^{2}}}$ (1)

$\frac{{{c}^{2}}-2{{d}^{2}}}{{{(c+4d)}^{2}}}=\frac{{{(d.k)}^{2}}-2{{d}^{2}}}{{{(d.k+4d)}^{2}}}=\frac{{{d}^{2}}.\left( {{k}^{2}}-2 \right)}{{{d}^{2}}.{{\left( k+4 \right)}^{2}}}=\frac{{{k}^{2}}-2}{{{(k+4)}^{2}}}$ (2)

Từ (1) và (2) suy ra $\frac{{{a}^{2}}-2{{b}^{2}}}{{{(a+4b)}^{2}}}=\frac{{{c}^{2}}-2{{d}^{2}}}{{{(c+4d)}^{2}}}$


d) $(a+4c)(2b-3d)=(b+4d)(2a-3c)$

Ta có : $\frac{a}{b}=\frac{c}{d}\Leftrightarrow \frac{a}{c}=\frac{b}{d}=k\Rightarrow \left\{ \begin{align}& a=c.k \\ & b=d.k \\ \end{align} \right.$

Khi đó : $\frac{a+4c}{2a-3c}=\frac{c.k+4c}{2.c.k-3c}=\frac{c(k+4)}{c(2k-3)}=\frac{k+4}{2k-3}$

$\frac{b+4d}{2b-3d}=\frac{d.k+4d}{2.d.k-3d}=\frac{d\left( k+4 \right)}{d\left( 2k-3 \right)}=\frac{k+4}{2k-3}$

Do đó : $\frac{a+4c}{2a-3c}=\frac{b+4d}{2b-3d}\Leftrightarrow $$(a+4c)(2b-3d)=(b+4d)(2a-3c)$

Vậy $(a+4c)(2b-3d)=(b+4d)(2a-3c)$

e) $\frac{ac}{bd}=\frac{{{a}^{2}}-{{c}^{2}}}{{{b}^{2}}-{{d}^{2}}}$

Ta có: $\frac{a}{b}=\frac{c}{d}\Rightarrow \frac{a}{b}.\frac{c}{d}=\frac{a}{b}.\frac{a}{b}=\frac{c}{d}.\frac{c}{d}\Rightarrow \frac{ac}{bd}=\frac{{{a}^{2}}}{{{b}^{2}}}=\frac{{{c}^{2}}}{{{d}^{2}}}$ (1)

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

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

Từ (1) và (2) suy ra $\frac{ac}{bd}=\frac{{{a}^{2}}-{{c}^{2}}}{{{b}^{2}}-{{d}^{2}}}$

10 . Cho $\frac{a}{b}=\frac{c}{d}$ ( giả thiết các tỉ số đều có nghĩa ) . Chứng minh
a) $\frac{a-2b}{b}=\frac{c-2d}{d}$
b) $\frac{a-2c}{3a+c}=\frac{b-2d}{3b+d}$
c) $\frac{{{a}^{2}}-2{{b}^{2}}}{{{(a+4b)}^{2}}}=\frac{{{c}^{2}}-2{{d}^{2}}}{{{(c+4d)}^{2}}}$
d) $(a+4c)(2b-3d)=(b+4d)(2a-3c)$
e) $\frac{ac}{bd}=\frac{{{a}^{2}}-{{c}^{2}}}{{{b}^{2}}-{{d}^{2}}}$

Giải

a) $\frac{a-2b}{b}=\frac{c-2d}{d}$
Cách 1 :
Đặt $\frac{a}{b}=\frac{c}{d}=k\Rightarrow \left\{ \begin{align}& a=b.k \\ & c=d.k \\ \end{align} \right.$
Khi đó $\frac{a-2b}{b}=\frac{b.k-2b}{b}=\frac{b\left( k-2 \right)}{b}=k-2$
$\frac{c-2d}{d}=\frac{d.k-2d}{d}=\frac{d\left( k-2 \right)}{d}=k-2$
Vậy $\frac{a-2b}{b}=\frac{c-2d}{d}$

Cách 2 :
Ta có : $\frac{a}{b}=\frac{c}{d}\Rightarrow \frac{a}{b}-2=\frac{c}{d}-2\Leftrightarrow \frac{a-2b}{b}=\frac{c-2d}{d}$
Vậy $\frac{a-2b}{b}=\frac{c-2d}{d}$


b) $\frac{a-2c}{3a+c}=\frac{b-2d}{3b+d}$
Cách 1 :
Ta có : $\frac{a}{b}=\frac{c}{d}\Leftrightarrow \frac{a}{c}=\frac{b}{d}=k\Rightarrow \left\{ \begin{align}& a=c.k \\ & b=d.k \\ \end{align} \right.$
Khi đó : $\frac{a-2c}{3a+c}=\frac{c.k-2c}{3.c.k+c}=\frac{c(k-2)}{c(3k+1)}=\frac{k-2}{3k+1}$
$\frac{b-2d}{3b+d}=\frac{d.k-2d}{3.d.k+d}=\frac{d\left( k-2 \right)}{d\left( 3k+1 \right)}=\frac{k-2}{3k+1}$
Vậy $\frac{a-2c}{3a+c}=\frac{b-2d}{3b+d}$

Cách 2 :
Ta có : $\frac{a}{b}=\frac{c}{d}\Leftrightarrow \frac{a}{b}=\frac{c}{d}=\frac{3a}{3b}=\frac{3a+c}{3b+c}$ (1)
$\frac{a}{b}=\frac{c}{d}\Leftrightarrow \frac{a}{b}=\frac{c}{d}=\frac{2c}{2d}=\frac{a-2c}{b-2d}$ (2)
Từ (1) và (2) suy ra $\frac{a-2c}{b-2d}=\frac{3a+c}{3b+d}\Leftrightarrow \frac{a-2c}{3a+c}=\frac{b-2d}{3b+d}$
Vậy $\frac{a-2c}{3a+c}=\frac{b-2d}{3b+d}$


c)
$\frac{{{a}^{2}}-2{{b}^{2}}}{{{(a+4b)}^{2}}}=\frac{{{c}^{2}}-2{{d}^{2}}}{{{(c+4d)}^{2}}}$

Cách 1 :

Đặt $\frac{a}{b}=\frac{c}{d}=k\Rightarrow \left\{ \begin{align}& a=b.k \\ & c=d.k \\
\end{align} \right.$
Khi đó $\frac{{{a}^{2}}-2{{b}^{2}}}{{{(a+4b)}^{2}}}=\frac{{{(b.k)}^{2}}-2{{b}^{2}}}{{{(b.k+4b)}^{2}}}=\frac{{{b}^{2}}.\left( {{k}^{2}}-2 \right)}{{{b}^{2}}.{{\left( k+4 \right)}^{2}}}=\frac{{{k}^{2}}-2}{{{\left( k+4 \right)}^{2}}}$ (1)
$\frac{{{c}^{2}}-2{{d}^{2}}}{{{(c+4d)}^{2}}}=\frac{{{(d.k)}^{2}}-2{{d}^{2}}}{{{(d.k+4d)}^{2}}}=\frac{{{d}^{2}}.\left( {{k}^{2}}-2 \right)}{{{d}^{2}}.{{\left( k+4 \right)}^{2}}}=\frac{{{k}^{2}}-2}{{{(k+4)}^{2}}}$ (2)
Từ (1) và (2) suy ra $\frac{{{a}^{2}}-2{{b}^{2}}}{{{(a+4b)}^{2}}}=\frac{{{c}^{2}}-2{{d}^{2}}}{{{(c+4d)}^{2}}}$

Cách 2 :
Ta có : $\frac{a}{b}=\frac{c}{d}\Leftrightarrow \frac{a}{c}=\frac{b}{d}$ . Áp dụng tính chất dãy tỉ số bằng nhau , ta có :
$\frac{a}{c}=\frac{b}{d}=\frac{4b}{4d}=\frac{a+4b}{c+4d}\Leftrightarrow \frac{{{a}^{2}}}{{{c}^{2}}}=\frac{{{(a+4b)}^{2}}}{{{(c+4d)}^{2}}}$(1)
$\frac{a}{c}=\frac{b}{d}\Leftrightarrow \frac{{{a}^{2}}}{{{c}^{2}}}=\frac{{{b}^{2}}}{{{d}^{2}}}=\frac{2{{b}^{2}}}{2{{d}^{2}}}=\frac{{{a}^{2}}-2{{b}^{2}}}{{{c}^{2}}-2{{d}^{2}}}$ (2).

Từ (1) và (2) suy ra $\frac{{{a}^{2}}-2{{b}^{2}}}{{{c}^{2}}-2{{d}^{2}}}=\frac{{{(a+4b)}^{2}}}{{{(c+4d)}^{2}}}\Leftrightarrow \frac{{{a}^{2}}-2{{b}^{2}}}{{{(a+4b)}^{2}}}=\frac{{{c}^{2}}-2{{d}^{2}}}{{{(c+4d)}^{2}}}$



d) $(a+4c)(2b-3d)=(b+4d)(2a-3c)$
Cách 1 :


Ta có : $\frac{a}{b}=\frac{c}{d}\Leftrightarrow \frac{a}{c}=\frac{b}{d}=k\Rightarrow \left\{ \begin{align}& a=c.k \\ & b=d.k \\ \end{align} \right.$
Khi đó : $\frac{a+4c}{2a-3c}=\frac{c.k+4c}{2.c.k-3c}=\frac{c(k+4)}{c(2k-3)}=\frac{k+4}{2k-3}$

$\frac{b+4d}{2b-3d}=\frac{d.k+4d}{2.d.k-3d}=\frac{d\left( k+4 \right)}{d\left( 2k-3 \right)}=\frac{k+4}{2k-3}$
Do đó : $\frac{a+4c}{2a-3c}=\frac{b+4d}{2b-3d}\Leftrightarrow $$(a+4c)(2b-3d)=(b+4d)(2a-3c)$

Vậy $(a+4c)(2b-3d)=(b+4d)(2a-3c)$
Cách 2 : Tương tự Cách 2 ý b)


e) $\frac{ac}{bd}=\frac{{{a}^{2}}-{{c}^{2}}}{{{b}^{2}}-{{d}^{2}}}$

Ta có: $\frac{a}{b}=\frac{c}{d}\Rightarrow \frac{a}{b}.\frac{c}{d}=\frac{a}{b}.\frac{a}{b}=\frac{c}{d}.\frac{c}{d}\Rightarrow \frac{ac}{bd}=\frac{{{a}^{2}}}{{{b}^{2}}}=\frac{{{c}^{2}}}{{{d}^{2}}}$ (1)
Áp dụng tính chất dãy tỉ số bằng nhau ta có : $\frac{{{a}^{2}}}{{{b}^{2}}}=\frac{{{c}^{2}}}{{{d}^{2}}}=\frac{{{a}^{2}}-{{c}^{2}}}{{{b}^{2}}-{{d}^{2}}}$ (2)
Từ (1) và (2) suy ra $\frac{ac}{bd}=\frac{{{a}^{2}}-{{c}^{2}}}{{{b}^{2}}-{{d}^{2}}}$