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

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

Do a + b + c + d khác 0 nên: b+c+d = a+c+d = a+b+d = a+b+c  => a = b = c = d

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

\(\Rightarrow A=1+1+1+1=4\)

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

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

c) \(a\left(b-c-d\right)-a\left(b+c-d\right)\\ =ab-ac-ad-ab-ac+ad\\ =\left(ab-ab\right)-\left(ac+ac\right)+\left(ad-ad\right)\\ =-2ac\)

d) \(\left(a+b\right)\left(c+d\right)-\left(a+d\right)\left(b+c\right)\\ =a\left(c+d\right)+b\left(c+d\right)-\left[a\left(b+c\right)+d\left(b+c\right)\right]\\ =ac+ad+bc+bd-\left(ab+ac+bd+dc\right)\\ =ac+ad+bc+bd-ab-ac-bd-dc\\ =\left(ac-ac\right)+ad+bc+\left(bd-bd\right)-dc\\ =ad+bc-dc\)

e) \(\left(a+b\right)\left(c-d\right)-\left(a-b\right)\left(c+d\right)\\ =a\left(c-d\right)+b\left(c-d\right)-\left[a\left(c+d\right)-b\left(c+d\right)\right]\\ =ac-ad+bc-bd-\left(ac+ad-bc-bd\right)\\ =ac-ad+bc-bd-ac-ad+bc+bd\\ =\left(ac-ac\right)-\left(ad+ad\right)+\left(bc+bc\right)+\left(bd-bd\right)\\ =-2ad+2bc\)

f) \(\left(a+b\right)^2-\left(a-b\right)^2\\ =\left(a+b\right)\left(a+b\right)-\left(a-b\right)\left(a-b\right)\\ =a\left(a+b\right)+b\left(a+b\right)-\left[a\left(a-b\right)-b\left(a-b\right)\right]\\ =a^2+ab+ab+b^2-\left(a^2-ab-ab+b^2\right)\\ =a^2+2ab+b^2-\left(a^2-2ab+b^2\right)\\ =a^2+2ab+b^2-a^2+2ab-b^2\\ =\left(a^2-a^2\right)+\left(2ab+2ab\right)+\left(b^2-b^2\right)\\ =4ab\)

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

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

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

\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=1\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=d\end{cases}}\)

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

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

A=a-b+c-b+c+d+c-d+a

A=2a-2b-3c

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

B=a + b - c + b + c - a - a + c

B=2b + c - a

C = - ( 4a + 5b + c) - ( 5b + 3c )

C = -4a - 5b - c - 5b -3c

C= -4a - 10b - 4c

D= ( a - 3b + c) - ( 2a -b +c)

D= a - 3b +c - 2a + b -c

D= a - 2b

\(=\frac{1}{a^2}+\frac{2}{ab}+\frac{1}{b^2}+\frac{2}{bc}+\frac{1}{c^2}+\frac{2}{ac}\)

\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2a+2b+2c}{abc}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)

\(A=\frac{1}{a^2}+\frac{2}{ab}+\frac{1}{b^2}+\frac{2}{bc}+\frac{1}{c^2}+\frac{2}{ac}=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)

Linh không biết a + b + c = 0 để làm gì?

Theo tính chất của tỉ lệ thức, ta có thể viết: \(\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}\)

Cộng 1 vào 2 vế được: \(1+\dfrac{a^2}{b^2}=1+\dfrac{c^2}{d^2}\) hay \(\dfrac{a^2+b^2}{b^2}=\dfrac{c^2+d^2}{d^2}\)

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

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

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

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

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