Giải thích các bước giải:
a, Ta có: $\frac{2a+13b}{3a-7b}$ = $\frac{2c+13d}{3c-7d}$
⇔ (2a+13b)(3c-7d) = (2c+13d)(3a-7b)
⇔ 6ac - 14ad + 39bc - 91bd = 6ac - 14bc + 39ad - 91bd
⇔ 53bc = 53ad
⇔ bc = ad
⇔ $\frac{a}{b}$ = $\frac{c}{d}$ (đpcm)
b, Ta có: $\frac{cy-bz}{x}$ = $\frac{az-cx}{y}$ = $\frac{bx-ay}{z}$
⇔ $\frac{cxy-bxz}{x^{2}}$ = $\frac{ayz-cxy}{y^{2}}$ = $\frac{bxz-ayz}{z^{2}}$
⇔ $\frac{cxy-bxz}{x^{2}}$ = $\frac{ayz-cxy}{y^{2}}$ = $\frac{bxz-ayz}{z^{2}}$ = $\frac{cxy-bxz+ayz-cxy+bxz-ayz}{x^{2}+y^{2}+z^{2}}$ = 0
⇒ $\frac{cxy-bxz}{x^{2}}$ = 0 ⇔ cy = bz ⇔ $\frac{b}{y}$ = $\frac{c}{z}$
$\frac{ayz-cxy}{y^{2}}$ = 0 ⇔ az = cx ⇔ $\frac{a}{x}$ = $\frac{c}{z}$
Vậy $\frac{a}{x}$ = $\frac{b}{y}$ = $\frac{c}{z}$ (đpcm)
