a+b-c/a+b-c + 2c/a+b-c = a-b-c/a-b-c + 2c/a-b-c

suy ra 2c/a+b-c = 2c/a-b-c

Dấu = xảy ra khi c=0

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

\(\Leftrightarrow\left(a+b+c\right)\left(a-b-c\right)=\left(a-b+c\right)\left(a+b-c\right)\) 

\(\Leftrightarrow a^2-\left(b+c\right)^2=a^2-\left(b-c\right)^2\)

\(\Leftrightarrow\left(b+c\right)^2-\left(b-c\right)^2=0\)

\(\Leftrightarrow\left(b+c-b+c\right)\left(b+c+b-c\right)=0\)

\(\Leftrightarrow4bc=0\)

Do b\(\ne\) 0\(\Rightarrow c=0\)

Vậy c=0 thì thỏa tỉ lệ thức (đcpcm)

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

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

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

\(\Rightarrow a+b+c-a-b+c=0\)

\(\Rightarrow2c=0\)

\(\Rightarrow c=0\) (đpcm)

Cách 1:

Ta xét tích a(c-d) và c(a-b)

Ta có: a(c-d)=ac-ad (1)

           c(a-b)=ac-bc(2)

Ta lại có \(\dfrac{a}{c}=\dfrac{c}{d}\)=>ad=bc (3)

Từ (1), (2), (3) ta có a(c-d)=c(a-d). Do đó \(\dfrac{a}{a-b}=\dfrac{c}{c-d}\)

Cách 2:

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

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

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

Từ (1) và (2)=> \(\dfrac{a}{a-b}=\dfrac{c}{c-d}\)

Cách 3: Ta có

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

Aps dụng tính chất dãy tỉ số bằng nhau:

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

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

Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)

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

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

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

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

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

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

Áp dụng tc dtsbn:

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

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

Vậy ta lập đc tỉ lệ thức \(\dfrac{a}{c}=\dfrac{b}{a}\)

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

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

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

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

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

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

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

\(\Rightarrow c=-c\)

\(\Rightarrow c-\left(-c\right)=0\)

\(\Rightarrow c+c=0\)

\(\Rightarrow2c=0\)

\(\Rightarrow c=0\)

\(\Rightarrow\) Đpcm.

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

=> Ta có: \(\dfrac{a+b}{b}=\dfrac{bk+b}{b}=\dfrac{b\left(k+1\right)}{b}=k+1\) (1)

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

Từ (1) và (2) => \(\dfrac{a+b}{b}=\dfrac{c+d}{d}\) ( đpcm)

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Rightarrow\left[{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\) (1)

Thay (1) vào đề bài:

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

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

Khi đó: \(VT=VP\)

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

Vậy \(\dfrac{a+b}{b}=\dfrac{c+d}{d}\) khi \(\left[{}\begin{matrix}a,b,c,d\ne0\\a\ne b;c\ne d\end{matrix}\right.\).

Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{b}+1=\dfrac{c}{d}+1\)

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

Vậy...

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

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

Ta có:

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

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

Từ (1) và (2) suy ra:

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

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

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

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

Từ (1) và (2) suy ra:

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

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

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

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

từ (1);(2) nên \(\dfrac{a}{a-b}=\dfrac{c}{c-d}\)

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

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

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