b, Ta có \(m=a+b+c\)

          \(\Rightarrow am+bc=a\left(a+b+c\right)+bc=a\left(a+b\right)+ac+bc=\left(a+c\right)\left(a+b\right)\)

CMTT \(bm+ac=\left(b+c\right)\left(b+a\right)\);\(cm+ab=\left(c+a\right)\left(c+b\right)\)

Suy ra \(\left(am+bc\right)\left(bm+ac\right)\left(cm+ab\right)=\left(a+b\right)^2\left(a+c\right)^2\left(b+c\right)^2\)

Ta có: \(\left(a+b+c+d\right)\left(a-b-c+d\right)=\left(a-b+c-d\right)\left(a+b-c-d\right)\)

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

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

\(\Leftrightarrow2d\cdot2a=2c\cdot2b\)

\(\Leftrightarrow ad=bc\)

hay \(\dfrac{a}{c}=\dfrac{b}{d}\)

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

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

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

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

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

\(\cdot\frac{a+b}{a-b}=\frac{bk+b}{bk-b}=\frac{b\left(k+1\right)}{b\left(k-1\right)}=\frac{k+1}{k-1}\left(1\right)\)

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

Từ (1) và (2) \(\Rightarrow\)\(\frac{a+b}{a-b}=\frac{c+d}{c-d}\)

Chúc bạn học tốt!!! k cho mk nha !!

Thấy đúng thì chọn cho mk nha

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

Áp dụng t/c dtsbn:

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

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

a: \(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\)

=>(a+5)(b-6)=(a-5)(b+6)

=>ab-6a+5b-30=ab+6a-5b-30

=>-6a+5b=6a-5b

=>-12a=-10b

=>6a=5b

=>\(\dfrac{a}{b}=\dfrac{5}{6}\)

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

=>\(a=bk;c=dk\)

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

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

Do đó: \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\)

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

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

<=> \(\frac{a+b+c+d}{c+d}=\frac{a+b+c+d}{d+a}\)

<=> \(\frac{a+b+c+d}{c+d}-\frac{a+b+c+d}{d+a}=0\)

<=> \(\left(a+b+c+d\right)\left(\frac{1}{c+d}-\frac{1}{d+a}\right)=0\)

<=> \(\orbr{\begin{cases}a+b+c+d=0\\\frac{1}{c+d}-\frac{1}{d+a}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}a+b+c+d=0\\c=a\end{cases}\left(đpcm\right)}}\)

Đặt = t => a = bt ; c = dt thay vào từng vế  

Đặt a/b=c/d= t suy ra a=bt; c=dt

(a+b)/(a-b)= bt+b/bt-b = b(t+1)/b(t-1)=t+1/t-1 (1)

(c+d)/(c-d)= dt+d/dt-d = d(t+1)/d(t-1)=t+1/t-1 (2)

Từ (1) và (2) suy ra (a+b)/(a-b)= (c+d)/(c-d)