Question

[Nguyễn Ngọc Thảo Linh]

cho tỉ lệ thức a/b=c/d .chứng minh ac/bd=a²+c²/b²+d²

Answer 1

[Minh Triều]

đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)

suy ra:\(\frac{ac}{bd}=\frac{bk.dk}{bd}=k.k=k^2\)

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

vậy \(\frac{ab}{bd}=\frac{a^2+c^2}{b^2+d^2}\)

Answer 2

[Lê Chí Cường]

Ta có:\(\frac{a}{b}=\frac{c}{d}=>\frac{a}{b}.\frac{c}{d}=\frac{c}{d}.\frac{c}{d}=>\frac{ac}{bd}=\frac{c^2}{d^2}\)

          \(\frac{c}{d}=\frac{a}{b}=>\frac{a}{b}.\frac{c}{d}=\frac{a}{b}.\frac{a}{b}=>\frac{ac}{bd}=\frac{a^2}{b^2}\)

=>\(\frac{ac}{bd}=\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}\)

=>\(\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\)