Đặt a/b=c/d=k suy ra a=bk ; c=dk

Có : (c+2d).(a+b) = (dk+2d).(bk+b)

                            = (d(2+k)).(b(k+1))     

                            = d.b.(k+1).(k+2)     <1>

       (c+d).(a+2b) = (dk+d).(bk+2b)

                           = (d(k+1)).(b(k+2))

                           = d.b.(k+1).(k+2)     <2>

Từ <1> và <2> suy ra (c+2d).(a+b) = (c+d).(a+2b)

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

Suy ra: \(\left(a+b\right)\left(c-2d\right)=\left(c+d\right)\left(a-2b\right)\)

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

\(\Rightarrow ac-2ad+bc-2bd=ac-2bc+ad-2bd\)

\(\Rightarrow ac-2ad+bc=ac-2bc+ad\)

\(\Rightarrow2ad+bc=2bc+ad\)

\(\Rightarrow2ad-ad=2bc-bc\)

\(\Rightarrow ad=bc\)

\(\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}\left(đpcm\right)\)

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

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

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

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

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

a, ta có :

\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}=\dfrac{2b}{2d}\)

áp dụng tính chất dă y tỉ số bằng nhau ta có:

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

\(\Rightarrow\dfrac{a+2b}{c+2d}=\dfrac{2a-b}{2c-d}\Rightarrow\dfrac{a+2b}{2a-b}=\dfrac{c+2d}{2c-d}\) (ĐPCM)

b, ta có:

\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{3c}{3d}\)

áp dụng tính chất dă tỉ số bằng nhau ta có :

\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{3c}{3d}=\dfrac{a+3c}{b+3d}=\dfrac{a-c}{b-d}\)

\(\Rightarrow\dfrac{a+3c}{b+3d}=\dfrac{a-c}{b-d}\)

\(\Rightarrow\left(a+3c\right)\left(b-d\right)=\left(b+3d\right)\left(a-c\right)\) (ĐPCM)

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

= \(\frac{3a-2b}{3c-2d}=\frac{3a+2b}{3c+2d}\)=> \(\frac{3a-2b}{3a+2b}=\frac{3c-2d}{3c+2d}\)

tíc mình nhé! Thanks

Đặt a/b=c/d=k=>a=kb;c=kd

Khi đó ta có:3a-2b/3a+2b=3kb-2b/3kb+2b=b(3k-2)/b(3k+2)=3k-2/3k+2 (1)

                  3c-2d/3c+2d=3kd-2d/3kd+2d=d(3k-2)/d(3k+2)=3k-2/3k+2 (2)

Từ (1) và (2) =>....

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

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

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

\(\Rightarrow ad+ad+bc=bc+ad+bc\)

\(\Rightarrow2ad+bc=2bc+ad\)

\(\Rightarrow ab+2ad+bc+2cd=ab+2bc+ad+2cd\)

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

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

DỄ MÀ

(a+2c)(b+d)=ab+ad+2bc+2cd

(a+c)(b+2d)=ab+2ad+bc+2cd

Vì a/b=c/d nên ad=bc

suy ra đpcm

sao ko chứng minh ik

mn ơi giúp mk với