Question

chung minh rang:\(\frac{a}{b}=\frac{c}{d}\) thì\(\frac{a^2+b^2}{c^2+d^2}\)=\(\frac{ab}{c\text{d}}\)

Answer 1

Ta có\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\) (1)

Ta lại có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\frac{a}{c}.\frac{b}{d}=\frac{ab}{cd}\)(2)

Từ 1 và 2 \(\Rightarrow\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)

Answer 2

Ta có: \(\frac{a}{b}=\frac{c}{d}=>\frac{a}{c}=\frac{b}{d}\)

=>\(\frac{a}{c}.\frac{b}{d}=\frac{a}{c}.\frac{a}{c}=>\frac{ab}{cd}=\frac{a^2}{c^2}\)

\(\frac{a}{c}.\frac{b}{d}=\frac{b}{d}.\frac{b}{d}=>\frac{ab}{cd}=\frac{b^2}{d^2}\)

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

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

=>\(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}\)