a, A = ( -a - b + c) - ( -a - b - c)

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

= 2c

b, c = -2

=> A = 2.-2 = -4

Bạn có thể làm trình bày cho mình luôn được hông

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

   = -a - b + c + a + b + c

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

   = 2c

b) Thay a = 1; b = -1; c = -2 vào A, ta có :

A = [ -1 - ( -1 ) + ( -2 ) ] - [ -1 - ( -1 ) - ( -2 ) ] 

A = -2 + 2

A = 0

Vậy A = 0 khi a = 1; b = -1; c = -2

Giải:

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow a=bk,c=dk\)

Ta có:

\(\left(\frac{a+b}{c+d}\right)^2=\left(\frac{bk+b}{dk+d}\right)^2=\left[\frac{b.\left(k+1\right)}{d.\left(k+1\right)}\right]^2=\left(\frac{b}{d}\right)^2\) (1)

\(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\frac{b^2.k^2+b^2}{d^2.k^2+d^2}=\frac{b^2.\left(k^2+1\right)}{d^2.\left(k^2+1\right)}=\frac{b^2}{d^2}=\left(\frac{b}{d}\right)^2\) (2)

Từ (1) và (2) suy ra \(\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)

Vậy \(\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)

theo đề bài ta có
\(ab\left(c^2+d^2\right)=ab.c^2+ab.d^2=\left(a.c\right).\left(b.c\right)+\left(a.d\right).\left(b.d\right)\\ cd\left(a^2+b^2\right)=cd.a^2+cd.b^2=\left(c.a\right).\left(d.a\right)+\left(c.b\right).\left(d.b\right)\)
\(\left(a.c\right)\left(b.c\right)+\left(a.d\right)\left(b.d\right)=\left(c.a\right)\left(d.a\right)+\left(c.b\right)\left(d.b\right)\) vì mỗi vế đều bằng nhau
- Cnứng minh \(\frac{\left(a^2+b^2\right)}{c^2+d^2}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
ta có vì \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{\left(a+b\right)}{\left(c+d\right)}=\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{a^2}{c^2}=\frac{b^2}{d^2}\Rightarrow\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{\left(a^2+b^2\right)}{\left(c^2+d^2\right)}\)

Gọi a/b=c/d=k(k khác 0)

Ta có:

a=bk

c=dk

VT:(\(\frac{a+b}{c+d}\))² =(\(\frac{bk+b}{dk+d}\))² =(\(\frac{b\left(k+1\right)}{d\left(k+1\right)}\))² =(\(\frac{b}{d}\))^{2 (1)}

VP:\(\frac{a^2+b^2}{c^2+d^2}\)=\(\frac{b^2k^2+b^2}{d^2k^2+d^2}\)=\(\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}\)=\(\frac{b^2}{d^2}\)=(\(\frac{b}{d}\))² (2)

Từ (1) và (2) suy ra bằng nhau

a.(b-c)+c.(a-b)

= ab - ac + ac - bc

= ab - bc

= b(a - c)

a.(b-c)-b.(a+c)

= ab - ac - ba - bc

= -ac - bc

= -c(a + b)

a.(b+c)-b.(a-c)

= ab + ac - ba + bc

= ac + bc

= c(a + b)

không cần k đâu bạn à

2. a(b - c) + c(a - b) = ab - ac + ac - bc = ab - bc = b(a - c)

3. a(b - c) - b(a + c) = ab - ac - ab - bc = -ac - bc = -c(a + b)

4. a(b + c) - b(a - c) = ab + ac - ab + bc = ac + bc = c(a + b)

~~ Học tốt ~~ 

thank các bạn nhiều lắm

Mình tính thử a ,b ,c bằng nhau đó

Mình nghĩ là 0,037037037037037037

ko biet lam

bạn khá thông minh 

nhưg sorry mình k thể k cho bb đc nha

a) (a+b)-(a-b)+(a-c)-(a+c)

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

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

= 0 + 2b + (-2a )+ 0

= 2b + (-2a)

b) (a+b-c)+(a-b+c)-(b+c-a)-(a-b-c)

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

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

= 0 + 0 + 0 + 2a + 0 + 0

= 2a

Nhớ k cho mk nha

a,(a+b)-(a-b)+(a-c)-(a+c)

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

=2b

b,(a+b-c)+(a-b+c)-(b+c-a)-(a-b-c)

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

=2a

\(a,\left(a+b\right)-\left(a-b\right)+\left(a-c\right)-\left(a+c\right)\)

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

\(=2b-2c\)

\(b,\left(a+b-c\right)+\left(a-b+c\right)-\left(b+c-a\right)-\left(a-b-c\right)\)

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

\(=2a\)