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

\(=-\left(b+c\right)^2+\left(a+d\right)^2\)   ( 1 )

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

Từ ( 1 ) và ( 2 ) suy ra 

\(b^2+2bc+c^2-a^2-2ad-d^2=a^2-2ad+d^2-b^2+2bc-c^2\)

\(4ad=4ac\Rightarrow ad=bc\)

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

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

=>(a+b)(c-d)=(a-b)(c+d)

=>ac-ad+bc-bd=ac+ad-bc-bd

=>-ad+bc=ad-bc

=>-2ad=-2bc

=>ad=bc

=>a/b=c/d

câu a sai đề nha!!

bài này bn chỉ cần bỏ ngoặc là ra hết thôi mà

a, đề của bạn sai

b, ta có : (a - b ) - (c - d ) = a - b -c -(- d )

                                     = a - b - c + d

                                     = (a + d ) + (-b - c )

                                     =(a + d) - (b + c)

=> (a - b)-(c - d ) = (a + d) - (b + c)

câu a) (a-b) + (c-d) = (a+c) - (b-d)

 (a + b)(c + d) - (a + d)(b + c)

= ac + ad + bc + bd - ab - ac- bd - dc

= ad - ab + bc - dc

= a(d - b) + c(b- d)

= a(d - b) - c(d - b)

= (a - c)(d - b) (=vế phải)

Vậy đpcm

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

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

\(\Rightarrow\dfrac{a^3}{b^3}=\dfrac{\left(a+b+c\right)^3}{\left(b+c+d\right)^3}=\dfrac{a}{d}\).