Ta có: 2bd = c(b + d)

=> (a + c).d = bc + cd

=> ad + cd = bc + cd

=> ad = bc

=> \(\frac{a}{b}=\frac{c}{d}\left(đpcm\right)\)

Ta có : 2bd = c (b + d )

=) ( a + c ). d = bc + cd

=) ad + cd = bc + cd

=) ad = bc

=) a/b = c/ d ( đpcm)

Ta có : 2bd = c (b + d )

 => ( a + c ). d = bc + cd 

=>ad + cd = bc + cd 

=>ad = bc 

=> a/b = c/ d ( đpcm)

Ta có 2bd=c(b+d) \(=>\frac{2b}{c}=\frac{b+d}{d}\)

Mà a+c=2b nên \(\frac{a+c}{c}=\frac{b+d}{d}=>\frac{a+c}{b+d}=\frac{c}{d}\)

Áp dụng t/c dãy tỉ số bằng nhau ta có:

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

Vậy \(\frac{a}{b}=\frac{c}{d}\)

Ta có:

\(2bd=c\left(b+d\right)\)

\(\Rightarrow\left(a+c\right).d=bc+cd\)

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

\(\Rightarrow ad=bc\)

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

Giúp mik nha

😁😁😁😁😁

Ta có: 2bd = c(b + d)
=> (a + c).d = bc + cd
=> ad + cd = bc + cd
=> ad = bc
=> a/b = c/d (đpcm)

\(a+c=2b\) (*)

\(2bd=c\left(b+d\right)\)(**)

Thế (*) vào (**)

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

Theo tính chất phân phối ta có:

\(ad+cd=cb+cd\)

\(\Leftrightarrow ad=cb\)

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

\(2b.d=c\left(b+d\right)\Leftrightarrow\left(a+c\right)d=c\left(b+d\right)\Leftrightarrow\frac{a+c}{c}=\frac{b+d}{d}\Leftrightarrow\frac{a}{c}+1=\frac{b}{d}+1\Leftrightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}\)

\(\frac{24}{85}\)

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