Question

1. Cho tỉ lệ thức: \(\frac{a}{b}=\frac{c}{d}\). CMR: \(\frac{2a^2-3ab+4b^2}{5b^2+6ab}=\frac{2c^2-3cd+4d^2}{5d^2+6cd}\)

2. Cho \(\frac{a}{b}=\frac{c}{d}\). CMR:

a. \(\frac{a^2+b^2}{b^2+c^2}=\frac{a}{c}\)

b. \(\frac{c^2-a^2}{a^2+b^2}=\frac{c-a}{a}\)

Answer 1

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

\(\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}\)

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

\(\frac{a}{c}=\frac{b}{d}=\frac{2a^2}{2c^2}=\frac{3ab}{3cd}=\frac{4b^2}{4d^2}=\frac{2a^2-3ab+4b^2}{2c^2-3cd+4d^2}=\frac{5b^2}{5d^2}=\frac{6ab}{6cd}=\frac{5b^2+6ab}{5d^2+6cd}\)

Suy ra : \(\frac{2a^2-3ab+4b^2}{2c^2-3cd+4d^2}=\frac{5b^2+6ab}{5d^2+6cd}\)

\(\Rightarrow\frac{2a^2-3ab+4b^2}{5b^2+6ab}=\frac{2c^2-3cd+4d^2}{5d^2+6cd}\) \(\left(dpcm\right)\)