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

Bài 1: Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)

\(\dfrac{a}{a+c}=\dfrac{ck}{ck+c}=\dfrac{ck}{c\left(k+1\right)}=\dfrac{k}{k+1}\)

\(\dfrac{b}{b+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)

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

Bài 1: Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)

\(\dfrac{a}{a+c}=\dfrac{ck}{ck+c}=\dfrac{ck}{c\left(k+1\right)}=\dfrac{k}{k+1}\)

\(\dfrac{b}{b+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)

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

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

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

\(\dfrac{a}{b}=\dfrac{c}{d}=>ad=cb=>ad+ac=cb+ac\)

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

a) \(\dfrac{a}{b}=\dfrac{c}{d}\)

nên \(\dfrac{a}{b}+1=\dfrac{c}{d}+1\)

hay \(\dfrac{a+b}{b}=\dfrac{c+d}{d}\)

b) \(\dfrac{a}{b}=\dfrac{c}{d}\)

nên \(\dfrac{a}{b}-1=\dfrac{c}{d}-1\)

hay \(\dfrac{a-b}{b}=\dfrac{c-d}{d}\)

a/ Đặt :

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

Ta có :

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

\(VP=\dfrac{c-d}{c+d}=\dfrac{dk-d}{dk+d}=\dfrac{d\left(k-1\right)}{d\left(k+1\right)}=\dfrac{k-1}{k+1}\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)

b/ Đặt :

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

\(VT=\dfrac{2a+5b}{3a-4b}=\dfrac{2bk+5b}{3bk-4b}=\dfrac{b\left(2k+5\right)}{b\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\left(1\right)\)

\(VP=\dfrac{2c+5d}{3c-4d}=\dfrac{2dk+5d}{3dk-4d}=\dfrac{d\left(2k+5\right)}{d\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)

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

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

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

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

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

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

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

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

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

Cày bài vt

Cày bài vt

Ta đặt:

\(\dfrac{a}{b}=\dfrac{c}{d}=k\) => \(a=b\times k\) ; \(c=d\times k\) 

a) Ta có:  \(\dfrac{a}{b}=\dfrac{b\times k}{d\times k}=\dfrac{b}{d}\)  (1)

=> \(\dfrac{a+b}{c+d}=\dfrac{b\times k+b}{d\times k+d}=\dfrac{b\times\left(k+1\right)}{d\times\left(k+1\right)}=\dfrac{b}{d}\) (2)

Từ (1),(2) => đpcm

b)

\(\dfrac{a+b}{a}=\dfrac{b\times k+b}{b\times k}=\dfrac{b\times\left(k+1\right)}{b\times k}=\dfrac{k+1}{k}\) (1)

\(\dfrac{c+d}{c}=\dfrac{d\times k+d}{d\times k}=\dfrac{d\times\left(k+1\right)}{d\times k}=\dfrac{k+1}{k}\) (2)

Từ (1),(2) => đpcm

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

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

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