Question

\(\dfrac{a}{b}=\dfrac{c}{d}\left(b+d\ne0\right)\)

Chứng tỏ: \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\)

Answer 1

Vì \(\dfrac{a}{b}=\dfrac{c}{d}\) (theo đề bài)

Áp dụng tính chất dãy tỉ số bằng nhau

\(\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\)

\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}=\dfrac{a^2+c^2}{b^2+d^2}\)

Vậy \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}.\)

Answer 2

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

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

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

\(\Rightarrow\left(\dfrac{a}{b}\right)^2=\left(\dfrac{c}{d}\right)^2=\left(\dfrac{a+c}{b+d}\right)^2=\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}=\dfrac{a^2+c^2}{b^2+d^2}\)

\(\Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\)

Answer 3

Theo bài ra, ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)

ADTC của dãy tỉ số bằng nhau, ta có:

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

\(\Rightarrow\left(\dfrac{a}{b}\right)^2=\left(\dfrac{c}{d}\right)^2=\left(\dfrac{a+c}{b+d}\right)^2\)

\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}=\left(\dfrac{a+c}{b+d}\right)^2\)

ADTC của dãy tỉ số bằng nhau, ta có:

\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}=\left(\dfrac{a+c}{b+d}\right)^2=\dfrac{a^2+c^2}{b^2+d^2}\)

Vậy ....