a) Vì \(\dfrac{a}{b} = \dfrac{c}{d}\) nên \(ad = bc\)

Ta có \(\dfrac{{a + b}}{b} = \dfrac{{c + d}}{d}\)\( \Rightarrow d(a + b) = b(c + d)\)\( \Rightarrow ad + bd = bc + bd\)

\( \Rightarrow ad = bc\) (luôn đúng)

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

b) Vì \(\dfrac{a}{b} = \dfrac{c}{d}\) nên \(ad = bc\)

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

\(\begin{array}{l} \Rightarrow d(a - b) = b(c - d)\\ \Leftrightarrow ad - bd = bc - bd\\ \Leftrightarrow ad = bc\end{array}\) ( luôn đúng)

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

c)  Vì \(\dfrac{a}{b} = \dfrac{c}{d}\) nên \(ad = bc\)

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

\(\begin{array}{l} \Rightarrow a(c + d) = c(a + b)\\ \Leftrightarrow ac + ad = ac + bc\\ \Leftrightarrow ad = bc\end{array}\) (luôn đúng)

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

vi ab = cd

=>a/b=c/d

=>a+c/b+d =a/b = c/d

=>a-c/b-d =a/b = c/d

(sgk s8 )

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

\(\Rightarrow\dfrac{a+b}{b}=\dfrac{bk+b}{b}=\dfrac{b\left(k+1\right)}{b}=k+1\) và \(\dfrac{c+d}{d}=\dfrac{dk+d}{d}=\dfrac{d\left(k+1\right)}{d}=k+1\)

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

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

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a-b}{b}=\dfrac{b\left(k-1\right)}{b}=k-1\\\dfrac{c-d}{d}=\dfrac{d\left(k-1\right)}{d}=k-1\end{matrix}\right.\)\(\Rightarrow\dfrac{a-b}{b}=\dfrac{c-d}{d}\)

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

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

e: Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)

nên \(\dfrac{a}{c}=\dfrac{b}{d}\)

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

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

hay \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)

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

=>\(\dfrac{a}{b}-1=\dfrac{c}{d}-1\)

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

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

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

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

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

phần b đề kiểu gì vậy??//

hết rồi còn đâu mà lập

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

Ta có:\(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=bc\)

\(\Rightarrow bd-ad=bd-bc\)

\(\Rightarrow d\left(b-a\right)=b\left(d-c\right)\)

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

a, Áp dụng t/c dtsbn:

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

b, Áp dụng t/c dtsbn:

\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}=\dfrac{5b}{5d}=\dfrac{3a}{4c}=\dfrac{4b}{4d}=\dfrac{2a+5b}{2c+5d}=\dfrac{3a-4b}{3c-4d}\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)

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

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

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

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

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

Ta có \(\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=k^2\)

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

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