Question

Cho \(\dfrac{a}{c}=\dfrac{c}{b}\), chứng minh rằng:

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

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

Answer 1

a) Ta có: \(\dfrac{a}{c}=\dfrac{c}{b}\Rightarrow ab=c^2\)

Khi đó ta có: \(\dfrac{a^2+c^2}{b^2+c^2}=\dfrac{a^2+ab}{b^2+ab}=\dfrac{a\left(a+b\right)}{b\left(a+b\right)}=\dfrac{a}{b}\left(đpcm\right)\)

câu b: https://hoc24.vn/hoi-dap/question/559910.html

Answer 2

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

\(\Rightarrow\dfrac{a}{b}=1\) (1)

\(\Rightarrow a=b\) (*)

\(\Rightarrow\dfrac{a^2+c^2}{b^2+c^2}=\dfrac{a^2+c^2}{a^2+c^2}=1\) (2)

Từ (1) và (2) suy ra \(\dfrac{a^2+c^2}{b^2+c^2}=\dfrac{a}{b}\)(đpcm)

Answer 3

Ta có:

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

\(\Rightarrow ab=c^2\left(1\right)\)

Thay (1) vào \(\dfrac{a^2+c^2}{b^2+c^2}\) ta được

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

\(\RightarrowĐpcm\)

b) Ta có: ab = c² ( Theo a ) (1)

Thay (1) vào biểu thức \(\dfrac{b^2-a^2}{a^2+c^2}\) ta được:

\(\dfrac{b^2-a^2}{a^2+c^2}=\dfrac{b^2-ab+ab-a^2}{a^2+ab}=\dfrac{b\left(b-a\right)+a\left(b-a\right)}{a\left(a+b\right)}=\dfrac{\left(a+b\right)\left(b-a\right)}{a\left(a+b\right)}=\dfrac{b-a}{a}\)

\(\RightarrowĐpcm\)