Sửa đề: Chứng minh \(\dfrac{ab}{cd}=\left(\dfrac{2a+3b}{2c+3d}\right)^2\)

Đặt a/b=c/d=k

=>a=bk; c=dk

\(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2}{d^2}\)

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

Do đó: \(\dfrac{ab}{cd}=\left(\dfrac{2a+3b}{2c+3d}\right)^2\)

Sửa đề: Chứng minh \(\frac{2a^2+c^2}{b^2+2c^2}=\frac{a}{b}\)

Đặt \(\frac{a}{c}=\frac{c}{b}=k\)

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

\(\frac{2a^2+c^2}{b^2+2c^2}=\frac{2\cdot\left(bk^2\right)^2+\left(bk\right)^2}{b^2+2\cdot\left(bk\right)^2}=\frac{2b^2k^4+b^2k^2}{b^2+2b^2k^2}=\frac{b^2k^2\left(2k^2+1\right)}{b^2\left(1+2k^2\right)}=k^2\)

\(\frac{a}{b}=\frac{bk^2}{b}=k^2\)

Do đó; \(\frac{2a^2+c^2}{b^2+2c^2}=\frac{a}{b}\)

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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: 

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

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

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

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:

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