\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{2a}{2c}=\frac{3b}{3d}\)

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

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

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

\(\dfrac{2a+3b}{2a-3b}=\dfrac{2bk+3b}{2bk-3b}=\dfrac{b\left(2k+3\right)}{b\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\left(1\right)\)

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

Từ (1) và (2) suy ra \(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3b}\left(=\dfrac{2k+3}{2k-3}\right)\)

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

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

a, Có: \(\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{a}{c}=\frac{b}{d}\)

Áp dụng tính chất dãy tỉ số bằng nhau

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

Có: \(\frac{2a-3b}{2c-3d}=\frac{2a+3b}{2c+3d}\Leftrightarrow\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)

b, Co: \(\frac{a}{c}=\frac{b}{d}\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2\Rightarrow\frac{ab}{cd}\)

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

Áp dụng tính chất dãy tỉ số bằng nhau

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

Tu (1)&(2),có: \(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

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

Ta có: \(\dfrac{a-b}{c-d}=\dfrac{bk-b}{dk-d}=\dfrac{b}{d}\)

\(\dfrac{2a-3b}{2c-3d}=\dfrac{2bk-3b}{2dk-3d}=\dfrac{b}{d}\)

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

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a) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)

\(\Rightarrow\frac{2a}{2c}=\frac{3b}{3d}\)

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

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

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

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)

a)\(\frac{2a+3b}{2a-3b}=\frac{2bk+3b}{2bk-3b}=\frac{b\left(2k+3\right)}{b\left(2k-3\right)}=\frac{2k+3}{2k-3}\)(1)

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

Từ (1) và (2) \(\Rightarrow\)\(\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)

b)\(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\)(1)

\(\frac{a^2-b^2}{c^2-d^2}=\frac{b^2.k^2-b^2}{d^2.k^2-d^2}=\frac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\frac{b^2}{d^2}\)(2)

Từ (1) và(2)\(\Rightarrow\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)

c)\(\left(\frac{a+b}{c+d}\right)^2=\frac{\left(bk+b\right)^2}{\left(ck+d\right)^2}=\frac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\frac{b^2}{d^2}\)(1)

\(\frac{a^2+b^2}{c^2+d^2}=\frac{b^2.k^2+b^2}{d^2.k^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}\)(2)

Từ (1) và(2)\(\Rightarrow\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)

k cho mình nhé