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BÀI 1:

\(\dfrac{a}{k}=\dfrac{x}{a}\Rightarrow a^2=kx\)

\(\dfrac{b}{k}=\dfrac{y}{b}\Rightarrow b^2\)=ky

Vay \(\dfrac{a^2}{b^2}=\dfrac{kx}{ky}=\dfrac{x}{y}\)

Bài 2:

Vì a=b+c nên ad=(b+c)d=bd+cd (1)

Vi c=\(\dfrac{bd}{b-d}\)nen \(bd=\)c.(b-d)=bc-cd hay bc=bd+cd (2)

Từ (1),(2) =>ad=bc=>\(\dfrac{a}{b}=\dfrac{c}{d}\)

Bài 1: Nhân chéo

Bài 2:

Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)

\(\Rightarrow\left(\dfrac{a}{b}\right)^3=\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}\)

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

\(\Rightarrowđpcm\)

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

\(\dfrac{a+b+c}{a+b-c}=\dfrac{a-b+c}{a-b-c}\)

\(=\dfrac{a+b+c-a+b-c}{a+b-c-a+b+c}\)

\(=\dfrac{\left(a-a\right)+\left(b+b\right)+\left(c-c\right)}{\left(a-a\right)+\left(b+b\right)+\left(c-c\right)}\)

\(=\dfrac{2b}{2b}=1\)

\(\Rightarrow a+b+c=a+b-c\)

\(\Rightarrow c=-c\)

\(\Rightarrow c+c=0\)

\(\Rightarrow2c=0\Rightarrow c=0\)

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)

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

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a.b.c}{b.c.d}=\dfrac{a}{d}\left(1\right)\)

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\Rightarrow\left(\dfrac{a}{b}\right)^3=\left(\dfrac{b}{c}\right)^3=\left(\dfrac{c}{d}\right)^3\)

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

Từ \(\left(1\right)\) và \(\left(2\right)\) ta có:

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

Theo t/c dãy tỉ số bằng nhau ta có :

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

Vậy \(\dfrac{a}{b}=\dfrac{c+c}{b+d}\left(đpcm\right)\)

It's show time :)

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

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

\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}\Rightarrow\dfrac{a}{c}.\dfrac{b}{d}=\dfrac{a-b}{c-d}.\dfrac{a-b}{c-d}\)

hay \(\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\) (đpcm)

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

Từ \(\dfrac{a}{b}\) = \(\dfrac{c}{d}\) = k ( k \(\in\) Q, k \(\ne\) 0 )

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

VP = \(\dfrac{2a+3c}{2b+3d}\) = \(\dfrac{2.b.k+3.d.k}{2b+3d}\) = \(\dfrac{k.\left(2b+3d\right)}{2b+3d}\) = k (1)

VT = \(\dfrac{2a-3c}{2b-3d}\) = \(\dfrac{2.b.k-3.d.k}{2b-3d}\) = \(\dfrac{k.\left(2b-3d\right)}{2b-3d}\) = k (2)

Từ (1) và (2) ta có: \(\dfrac{2a+3c}{2b+3d}\) = \(\dfrac{2a-3c}{2b-3d}\)

hay: (2a+3c).(3b-3d) = (2a-3c).(2b+3d)

b: Đặt a/b=c/d=k

=>a=bk; c=dk

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

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

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

c: \(\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{b^3k^3+b^3}{d^3k^3+d^3}=\dfrac{b^3}{d^3}\)

\(\dfrac{a^3-b^3}{c^3-d^3}=\dfrac{b^3k^3-b^3}{d^3k^3-d^3}=\dfrac{b^3}{d^3}\)

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

d: \(\dfrac{a^{2018}-b^{2018}}{a^{2018}+b^{2018}}=\dfrac{b^{2018}k^{2018}-b^{2018}}{b^{2018}k^{2018}+b^{2018}}=\dfrac{k^{2018}-1}{k^{2018}+1}\)

\(\dfrac{c^{2018}-d^{2018}}{c^{2018}+d^{2018}}=\dfrac{k^{2018}-1}{k^{2018}+1}\)

Do đó: \(\dfrac{a^{2018}-b^{2018}}{a^{2018}+b^{2018}}=\dfrac{c^{2018}-d^{2018}}{c^{2018}+d^{2018}}\)

Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Rightarrow a=bk,c=dk\)

Ta có VT:

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

\(=\dfrac{b^2\left(k-1\right)^2}{d^2\left(k-1\right)^2}=\dfrac{b^2}{d^2}\) (1)

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

Từ (1) và (2) 

\(\Rightarrow\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{ab}{cd}\left(đpcm\right)\)

Có: \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow ab=cd\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}\)

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

\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}\)\(\Leftrightarrow\left(\dfrac{a}{c}\right)^2=\left(\dfrac{b}{d}\right)^2=\dfrac{ab}{cd}=\left(\dfrac{a-b}{c-d}\right)^2\)

Vậy...

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

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

\(VT=\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=\dfrac{bd.k^2}{bd}=k^2\)

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

\(\Rightarrow\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\left(đpcm\right)\)

Vậy \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)