a: Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=>a=bk; c=dk

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

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

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

b: \(\frac{7a-4b}{3a+5b}=\frac{7\cdot bk-4b}{3\cdot bk+5b}=\frac{b\left(7k-4\right)}{b\left(3k+5\right)}=\frac{7k-4}{3k+5}\)

\(\frac{7c-4d}{3c+5d}=\frac{7\cdot dk-4d}{3\cdot dk+5d}=\frac{d\left(7k-4\right)}{d\left(3k+5\right)}=\frac{7k-4}{3k+5}\)

Do đó: \(\frac{7a-4b}{3a+5b}=\frac{7c-4d}{3c+5d}\)

c: \(\frac{ac}{bd}=\frac{bk\cdot dk}{bd}=k^2\)

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

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

Do đó; \(\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}=\frac{\left(c-a\right)^2}{\left(d-b\right)^2}\)

d: \(\frac{a^3+b^3}{c^3+d^3}=\frac{\left(bk\right)^3+b^3}{\left(dk\right)^3+d^3}=\frac{b^3\left(k^3+1\right)}{d^3\left(k^3+1\right)}=\frac{b^3}{d^3}\)

\(\frac{\left(a+b\right)^3}{\left(c+d\right)^3}=\frac{\left(bk+b\right)^3}{\left(dk+d\right)^3}=\frac{b^3\left(k+1\right)^3}{d^3\left(k+1\right)^3}=\frac{b^3}{d^3}\)

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

Do đó:

\(b,a^2+b^2=c^2+d^2\)

\(\Rightarrow a^2+b^2+c^2+d^2=2c^2+2d^2⋮2\)

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

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

Ta có \(a^2-a=\left(a-1\right)a⋮2\)(vì tích của 2 số nguyên liên tiếp)

Tương tự ta có \(\left(b^2-b\right)⋮2;\left(c^2-c\right)⋮2;\left(d^2-d\right)⋮2\)

\(\Rightarrow\left(a^2-a\right)+\left(b^2-b\right)+\left(c^2-c\right)+\left(d^2-d\right)⋮2\)

\(\Rightarrow\left(a^2+b^2+c^2+d^2\right)-\left(a+b+c+d\right)⋮2\)

mà \(a^2+b^2+c^2+d^2⋮2\)nên \(a+b+c+d⋮2\)

Câu a để nghĩ tiếp 

bn làm câu b được không

k phải toán violympic

mik chỉ nhấn đại thôi chớ nó ko phải là violympic

Ta có:

b² = a.c \(\Rightarrow\frac{b}{c}=\frac{a}{b}\)

c² = b.d \(\Rightarrow\frac{c}{d}=\frac{b}{c}\)

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

\(\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}=\frac{a}{d}\left(1\right)\)

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

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

Từ (1) và (2) \(\Rightarrow\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d}\left(đpcm\right)\)

làm ơn làm phước tick cho mk lên 180