1/ (a – b + c) – (a + c) = -b

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

-b=-b

2/ (a + b) – (b – a) + c = 2a + c

a+b-b+a+c=2a+c

2a+c=2a+c

3/ - (a + b – c) + (a – b – c) = -2b

-a-b+c+a-b-c=-2b

-(b.2)=-2b

-2b=-2b

4/ a(b + c) – a(b + d) = a(c – d)

ab+ac-ab+ad=a(c-d)

ac-ad=a(c-d)

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

5/ a(b – c) + a(d + c) = a(b + d)

ab-ac+ad+ac=a(b+d)

ab+ad=a(b+d)

a(b+d)=a(b+d)

6/ a.(b – c) – a.(b + d) = -a.( c + d)

ab-ac-ab=ad=-a(c+d)

-ac+ad=-a(c+d)

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

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

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

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=(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

Bỏ ngoặc, Ta có: a+b-c+a-b-a+b+c=a+a-a+b-b+b+c-c=a+b

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

thế thôi

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

bài này tớ giải được nhung a,b,c,d\(\in\)N*

đăng lại làm gì

Theo đề bài ta có \(\dfrac{a}{b}=\dfrac{c}{d}\)

=> \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\) ( tính chất dãy tỉ số = nhau )

=> \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a-c}{b-d}\) ( tính chất dãy tỉ số = nhau )

Giải:

Vì \(\dfrac{a}{b}\) sau khi rút gọn sẽ bằng \(\dfrac{c}{d}.\)

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

Theo tính chất của dãy tỉ số bằng nhau, ta có:

\(1,\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}.\)

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

\(\Rightarrowđpcm.\)