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

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

=>\(\frac{a}{c}.\frac{a}{c}=\frac{b}{d}.\frac{s}{c}=>\frac{a^2}{c^2}=\frac{ab}{cd}\)

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

\(=>\frac{ab}{cd}\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\)

\(=>\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)\(\)

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

\(\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}\)

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

\(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\)

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

=> \(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{ab}{cd}\)

Áp dụng dãy tỉ số bằng nhau ta có;

\(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)

=> đpcm

Chúc bạn làm bài tốt

b) Vì \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2\)

Áp dụng tính chất dãy tỉ số bằng nhau :

\(\frac{a}{c}=\frac{b}{d}=\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\) (đpcm)

(a² + b²) / (c² + d²) = ab/cd 
<=> (a² + b²)cd = ab(c² + d²) 
<=> a²cd + b²cd = abc² + abd² 
<=> a²cd - abc² - abd² + b²cd = 0 
<=> ac(ad - bc) - bd(ad - bc) = 0 
<=> (ac - bd)(ad - bc) = 0 
<=> ac - bd = 0 hoặc ad - bc = 0 
<=> ac = bd hoặc ad = bc 
<=> a/b = d/c hoặc a/b = c/d (đpcm)

hello ib ko

Đặt \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=k\Rightarrow\hept{\begin{cases}a=ck\\b=dk\end{cases}}\)

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

và \(\frac{ab}{cd}=\frac{cdk^2}{cd}=k^2\)(2)

Từ (1) và (2) suy ra \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

Khi đó \(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\frac{b^2.k^2+b^2}{d^2.k^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}\)(1)

Lại có \(\frac{ab}{cd}=\frac{bkb}{dkd}=\frac{b^2}{d^2}\)(2)

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

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=kb\\c=kd\end{cases}}\)

Ta có : \(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(kb\right)^2+b^2}{\left(kd\right)^2+d^2}=\frac{k^2b^2+b^2}{k^2d^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}\)(1)

            \(\frac{ab}{cd}=\frac{kb\cdot b}{kd\cdot d}=\frac{kb^2}{kd^2}=\frac{b^2}{d^2}\)(2)

Từ (1) và (2) => đpcm