\(a+c=2b\\ \Leftrightarrow d\left(a+c\right)=2bd\\\Leftrightarrow d\left(a+c\right)=c\left(b+d\right) \\ \Leftrightarrow ad+cd=cb+cd\\ \Leftrightarrow ad=cb\\ \Leftrightarrow\dfrac{a}{b}=\dfrac{c}{d}\)

Ta có:

\(a+c=2b_{\left(1\right)}.\)

\(2bd=c\left(b+d\right)_{\left(2\right)}.\)

Từ \(_{\left(1\right)}\) và \(_{\left(2\right)}\Rightarrow\left(a+c\right)d=c\left(b+d\right).\)

\(\Rightarrow ad+cd=cb+cd\) (t/c phân phối).

\(\Rightarrow ad=bc\) (rút gọn cả 2 vế cho \(cd\)).

\(\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}\) (t/c cơ bản của tỉ lệ thức).

\(\Rightarrowđpcm.\)

\(2bd=c\left(b+d\right)\Rightarrow2b=\frac{c\left(b+d\right)}{d}\)

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

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

Ta có: 

\(a+c=2b_{\left(1\right)}\)

\(2bd=c\left(b+d\right)_2\)

Từ \(\left(1\right)\)và \(\left(2\right)\)\(\Rightarrow\)\(\left(a+c\right).d=c.\left(b+d\right)\)

\(\Rightarrow\)\(ad+cd=cb+cd\)( tính chất phân phối )

\(\Rightarrow\)\(ad=bc\)( rút gọn cả 2 vế cho \(cd\))

\(\Rightarrow\)\(\frac{a}{b}=\frac{c}{d}\)( tính chất cơ bản của tỉ lệ thức )

\(\Rightarrow\)\(\left(đpcm\right)\)

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)

Ta có: \(\left(a+b+c+d\right)\left(a-b-c+d\right)=\left(a-b+c-d\right)\left(a+b-c-d\right)\)

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

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

\(\Leftrightarrow2d\cdot2a=2c\cdot2b\)

\(\Leftrightarrow ad=bc\)

hay \(\dfrac{a}{c}=\dfrac{b}{d}\)