Bài 1:

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=>a=bk; c=dk

\(\frac{5a+4b}{5a-4b}=\frac{5\cdot bk+4b}{5\cdot bk-4b}=\frac{b\left(5k+4\right)}{b\left(5k-4\right)}=\frac{5k+4}{5k-4}\)

\(\frac{5c+4d}{5c-4d}=\frac{5\cdot dk+4d}{5\cdot dk-4d}=\frac{d\left(5k+4\right)}{d\left(5k-4\right)}=\frac{5k+4}{5k-4}\)

Do đó: \(\frac{5a+4b}{5a-4b}=\frac{5c+4d}{5c-4d}\)

Bài 2:

a: Đặt \(\frac{a}{b}=\frac{b}{c}=k\)

=>b=ck; a=bk=ck^2

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

\(\frac{a}{c}=\frac{ck^2}{c}=k^2\)

Do đó: \(\frac{a}{c}=\frac{a^2+b^2}{b^2+c^2}\)

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

=>\(\begin{cases}c=dk\\ b=ck=dk\cdot k=dk^2\\ a=bk=dk^2\cdot k=dk^3\end{cases}\)

\(\left(\frac{a+b-c}{b+c-d}\right)^3=\left(\frac{dk^3+dk^2-dk}{dk^2+dk-d}\right)^3\)

\(=\left\lbrack\frac{dk\left(k^2+k-1\right)}{d\left(k^2+k-1\right)}\right\rbrack^3=k^3\)

\(\frac{a}{d}=\frac{dk^3}{d}=k^3\)

Do đó: \(\left(\frac{a+b-c}{b+c-d}\right)^3=\frac{a}{d}\)

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

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

\(\frac{ab}{cd}=\frac{kbb}{kdd}=\frac{k.b^2}{k.d^2}=\frac{b^2}{d^2}\) (1)

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

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

Ta có: \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=k^3\)

Mà: \(k^3=\frac{a}{d}\) => \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d}\)

a)Ta có:\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)

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

\(\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\left(đpcm\right)\)

Đặt \(\frac{a}{b}=\frac{b}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\b=dk\end{cases}}\Leftrightarrow a=bk^2\)

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

và \(\frac{a}{d}=\frac{dk^2}{d}=k^2\)(2)

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

a=bk^2 là sai rồi

a. Áp dụng tính chất dãy tỉ số = nhau ta có:
x/4 = y/3 = z/9 = (x - 3y + 4z)/(4 - 9 + 36) = 62/31 = 2
=> x/4 = 2 -> x = 8
     y/3 = 2 -> y = 6
     z/9 = 2 -> z = 18