Question

Cho \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\). Chứng minh rằng:

\(\dfrac{2020a^2+2021b^2}{2020a^2-2021b^2}\)=\(\dfrac{2020c^2+2021d^2}{2020c^2-2021d^2}\)

Answer 1

Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)

\(\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}\)

\(\Leftrightarrow\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}\)

\(\Leftrightarrow\dfrac{2020a^2}{2020c^2}=\dfrac{2021b^2}{2021d^2}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{2020a^2}{2020c^2}=\dfrac{2021b^2}{2021d^2}=\dfrac{2020a^2+2021b^2}{2020c^2+2021d^2}=\dfrac{2020a^2-2021b^2}{2020c^2-2021d^2}\)

Ta có: \(\dfrac{2020a^2+2021b^2}{2020c^2+2021d^2}=\dfrac{2020a^2-2021b^2}{2020c^2-2021d^2}\)(cmt)

nên \(\dfrac{2020a^2+2021b^2}{2020a^2-2021b^2}=\dfrac{2020c^2+2021d^2}{2020c^2-2021d^2}\)(đpcm)