Question

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

Chứng minh rằng: \(a,\frac{a+2c}{a-c}=\frac{b+2d}{b-d}\)

                            \(b,\frac{a.c}{b.d}=\frac{\left(a+c\right)^2}{\left(b-d\right)^2}\)

Answer 1

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

Đặt \(\frac{a}{c}=\frac{b}{d}=k\Rightarrow\hept{\begin{cases}a=ck\\b=dk\end{cases}}\)(*)

Khi đó \(\frac{a+2c}{a-c}=\frac{ck+2c}{ck-c}=\frac{c\left(k+2\right)}{c\left(k-1\right)}=\frac{k+2}{k-1}\)(1) ; 

Lại có \(\frac{b+2d}{b-d}=\frac{dk+2d}{dk-d}=\frac{d\left(k+2\right)}{d\left(k-1\right)}=\frac{k+2}{k-1}\)(2)

Từ (1)(2) = > \(\frac{a+2c}{a-c}=\frac{b+2d}{b-d}\left(\text{đpcm}\right)\)

Answer 2

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=kb\\c=kd\end{cases}}\)

\(\Rightarrow VT=\frac{a+2c}{a-c}=\frac{kb+2kd}{kb-kd}=\frac{k\left(b+2d\right)}{k\left(b-d\right)}=\frac{b+2d}{b-d}=VP\)

=> đpcm