Question

Cho \(\frac{a}{b}=\frac{c}{d}\) chứng minh rằng :

\(\frac{ab}{cd}=\left(\frac{a-b}{c-d}\right)^2\)

Answer 1

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

Suy ra \(\begin{cases}a=bk\\c=dk\end{cases}\)\(\Rightarrow\frac{ab}{cd}=\left(\frac{a-b}{c-d}\right)^2\Leftrightarrow\frac{bkb}{dkd}=\left(\frac{bk-b}{dk-d}\right)^2\)

Xét VT \(\frac{bkb}{dkd}=\frac{b^2}{d^2}\left(1\right)\)

Xét VP \(\left(\frac{bk-b}{dk-d}\right)^2=\left(\frac{b\left(k-1\right)}{d\left(k-1\right)}\right)^2=\left(\frac{b}{d}\right)^2=\frac{b^2}{d^2}\left(2\right)\)

Từ (1) và (2) -->Đpcm

Answer 2

Giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

Ta có:
\(a=b.k\)

\(c=d.k\)

Theo bài ra ta có:
\(\frac{ab}{cd}=\frac{b.k.b}{d.k.d}=\frac{b^2.k}{d^2.k}=\frac{b^2}{d^2}=\left(\frac{b}{d}\right)^2\)   (1)

\(\left(\frac{a-b}{c-d}\right)^2=\left(\frac{b.k-b}{d.k-d}\right)^2=\left[\frac{b.\left(k-1\right)}{d.\left(k-1\right)}\right]^2=\left(\frac{b}{d}\right)^2\)   (2)

Từ (1) và (2) suy ra \(\frac{ab}{cd}=\left(\frac{a-b}{c-d}\right)^2\)

\(\Rightarrowđpcm\)

Answer 3

Answer 4

Ta có

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

Áp dụng tc của dãy tỉ số bằng nhau ta có

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

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

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

 

Answer 5

thank you very much !