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

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

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

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

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

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

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

\(\rightarrow\left(a+b\right)\left(c-d\right)=\left(a-b\right)\left(c+d\right)\)

\(\rightarrow ac-ad+bc-bd=ac+ad-bc-bd\)

\(\rightarrow-ad+bc=ad-bc\)

\(\rightarrow bc+bc=ad+ad\)

\(\rightarrow2bc=2ad\)

\(\rightarrow bc=ad\)

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

Chúc bạn học tốt!

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

Áp dụng tính chất dãy tỉ số bằng nhau có:

\(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}=\dfrac{a+b+a-b}{c+d+c-d}=\dfrac{a+a+b-b}{c+c+d-d}=\dfrac{2a}{2c}=\dfrac{a}{c}_{\left(1\right)}.\)

\(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}=\dfrac{a+b-a+b}{c+d-c+d}=\dfrac{a-a+b+b}{c-c+d+d}=\dfrac{2b}{2d}=\dfrac{b}{d}_{\left(2\right)}.\)

Từ \(_{\left(1\right)+\left(2\right)}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\Leftrightarrow\dfrac{a}{b}=\dfrac{c}{d}\) (t/c tỉ lệ thức).

\(\Rightarrowđpcm.\)

a=b*k

c=d*k

thì b*k+b/b*k-b=b*(k+1)/b*(k-1)=k+1/k-1

thì d*k+d/d*k-d=d*(k+1)/d*(k-1)=k+1/k-1

nen suy ra a+b/a-b=c+d/c-d

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

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

\(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}=\dfrac{a+b+a-b}{c+d+c-d}=\dfrac{2a}{2c}=\dfrac{a}{c}\)(1)

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

Từ (1) và (2) ta có: \(\dfrac{a}{c}=\dfrac{b}{d}\Leftrightarrow\dfrac{a}{b}=\dfrac{c}{d}\left(đpcm\right)\)

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

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

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

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

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

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 \(\dfrac{a-b}{b}=\dfrac{c-d}{d}\)

chọn C

C

7 c

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

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

Theo tính chất dãy tỉ số bằng nhau ta có

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

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

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Rightarrow a=bk\) và \(c=dk\)

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

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

\(\Rightarrow\dfrac{a+b}{c-d}=\dfrac{c+d}{c-d}\) (với \(a-b\ne0,c-d\ne0\))

Vậy \(\dfrac{a}{b}=\dfrac{c}{d}thì\)\(\dfrac{a+b}{c-d}=\dfrac{c+d}{c-d}\) ( \(a-b\ne0,c-d\ne0\))

Theo tính chất của dãy tỉ số bằng nhau ta có