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

\(\Rightarrow\frac{5a}{5c}=\frac{2b}{2d}\)

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

\(\frac{5a}{5c}=\frac{2b}{2d}=\frac{5a+2b}{5c+2d}=\frac{5a-2b}{5c-2d}\)

                 \(\Rightarrow\frac{5a+2b}{5a-2b}=\frac{5c+2d}{5c-2d}\left(đpcm\right)\)

ta có:

\(\frac{5a+2b}{5a-2b}=\frac{5c+2d}{5c-2d}\Rightarrow\frac{5a+2b}{5c+2d}=\frac{5a-2b}{5c-2d}\)

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{5a}{5c}=\frac{2b}{2d}=\frac{5a-2b}{5c-2d}=\frac{5a+2b}{5c+2d}\)(đpcm)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow a=bk;c=dk\)

\(\Rightarrow\frac{5a+2b}{5a-2b}=\frac{5bk+2b}{5bk-2b}=\frac{b\left(5k+2\right)}{b\left(5k-2\right)}=\frac{5k+2}{5k-2}\left(1\right)\)

\(\Rightarrow\frac{5c+2d}{5c-2d}=\frac{5dk+2d}{5dk-2d}=\frac{d\left(5k+2\right)}{d\left(5k-2\right)}=\frac{5k+2}{5k-2}\left(2\right)\)

Từ (1) và (2) 

\(\Rightarrow\frac{5a+2b}{5a-2b}=\frac{5c+2d}{5c-2d}\left(\text{đpcm}\right)\)

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

\(\Rightarrow\frac{a^3}{c^3}=\frac{b^3}{d^3}=\frac{a^2b}{c^2d}=\frac{2b^3}{2d^3}=\frac{a^3+2b^3}{c^3+2d^3}\)

=>đpcm

ta có : ab=cd⇔ad=bc⇔4ad=4bc⇔2ad+2ad=2bc+2bcab=cd⇔ad=bc⇔4ad=4bc⇔2ad+2ad=2bc+2bc

⇔2ad−2bc=2bc−2ad⇔ac+2ad−2bc−4bd=ac+2bc−2ad−4bd⇔2ad−2bc=2bc−2ad⇔ac+2ad−2bc−4bd=ac+2bc−2ad−4bd

⇔(c+2d)(a−2b)=(a+2b)(c−2d)⇔a+2bc+2d=a−2bc−2d(đpcm) 

Bạn ơi! Phải chứng minh \(\frac{a}{b}=\frac{c}{d}\) chứ!

Ta có:

a/b =c/d

⟹a/c=b/d

Áp dụng tính chất dãy tỉ số bằng nhau ta có

a/c=b/d=5a+2b/5c+2b=5a-2b/5c-2d

Vì 5a=2b/5c=2d=5a-2b/5c-2d

⟹5a+2b/5a-2b=5c+2d/5c-2d

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

\(\Rightarrow\frac{5a}{5c}=\frac{2b}{2d}.\)

Áp dụng tính chất dãy tỉ số bằng nhau ta được:

\(\frac{5a}{5c}=\frac{2b}{2d}=\frac{5a+2b}{5c+2d}\) (1)

\(\frac{5a}{5c}=\frac{2b}{2d}=\frac{5a-2b}{5c-2d}\) (2)

Từ (1) và (2) \(\Rightarrow\frac{5a+2b}{5c+2d}=\frac{5a-2b}{5c-2d}\)

\(\Rightarrow\frac{5a+2b}{5a-2b}=\frac{5c+2d}{5c-2d}\left(đpcm\right).\)

Chúc bạn học tốt!

Bài này có 2 cách nè:

Cách 1:

\(\frac{a+b}{c+d}=\frac{a-2b}{c-2d}=\frac{\left(a+b\right)-\left(a-2b\right)}{\left(c+d\right)-\left(c-2d\right)}=\frac{a+b-a+2b}{c+d-c+2d}=\frac{3b}{3d}=\frac{b}{d}\left(1\right)\)

\(\frac{a+b}{c+a}=\frac{a-2b}{c-d}=\frac{2a+2b}{2c+2d}=\frac{2-ab+2a+2b}{c-2d+2c+2d}=\frac{3a}{3c}=\frac{a}{c}\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Rightarrow\frac{b}{d}=\frac{a}{c}\) Suy ra: \(\frac{a}{b}=\frac{c}{d}\)

Cách 2:

\(\frac{a+b}{c+d}=\frac{a-2b}{c-2d}\)

\(\Rightarrow\left(a+b\right).\left(c-2d\right)=\left(a-2b\right).\left(c+d\right)\)

\(\Rightarrow a.c-2a.d+b.c-2b.d=a.c+a.d-2b.c-2b.d\)

\(\Rightarrow a.c-a.c-2a.d-a.d+b.c-2b.c-2b.d+2b.d=0\)

\(\Rightarrow-3a.d+3b.d=0\)

\(\Rightarrow3b.c=3a.d\)

\(\Rightarrow b.c=a.d\)

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