a) x + 25 +(-17) + 63 = x + 8+63 = x + 71

b) (-75) - (p +20) + 95 =-75 -p -20 + 95 =-p -75 +75 = -p

a) x+25+(-17)+63=x+88-17=x+71
b) (-75)-(p+20)+95=-75-p-20+95=20+(-p)-20=-p 

a. x+25+(-17)+63

= x+(25-17+63)

= x+71

b. (-75)+(p+20)+95

= -75+p+20+95

= p+(20+95-75)

= p+40

câu b kia mới đúng

câu b hình như sai:Phải là 

b)(-75)-(p+20)+95

=(-75)+95-(p+20)

=20-(p+20)

=20-p-20

=20-20-p

=0-7

=-p

a) x + 25 + (-17) + 63 = x + 71

b) (-75) - (p + 20) + 95 = - 75 - p - 20 + 95 = - p

a) = x + 8+63 = x +71 

b) (-75 )  - p - 20 +95 

=(-75) -20 + 95 - p = 0-p = -p

các bạn cho mình vài li-ke cho tròn 510 với  

vào toán mà hỏi Pham thi linh chi

a,x+25+(-17)+63

= x + (25 - 17 + 63)

= x + 71

b,(-75)-(p+20)+95

= -75 - p - 20 + 95

= p + (95 - 75 - 20)

= p + 0

# HOK TỐT #

dù hêm hỉu lém nhưng vưỡn thấy hay hay 

arigatou bẹn Hà nghen

mặc rầu chả bt là thèng nèo

a)x+25+(-17)+63

= x + (25 - 17 + 63)

= x + (8 + 63) 

= x + 71

b)( -75 ) - ( p + 20 ) + 95

=   -75 - p - 20 + 95

=    p + (95 - 75 - 20)

=    p + (20 - 20)

=    p + 0

=    p

a) \(\left(a+h\right)-\left(-c+a+b\right)=a+h+c-a-b\)

\(=\left(a-a\right)+\left(h-h\right)+c=0+0+c=c\)

b) \(-\left(x+y\right)+\left(-z+x+y\right)=-x-y-z+x+y\)

= \(\left(-x+x\right)+\left(-y+y\right)-z=0+0-z=-z\)

Tích cho cái

MTC: \(abc\left(a-b\right)\left(b-c\right)\left(a-c\right)\)nên

\(A=\frac{bc\left(b-c\right)\left(a-2\right)\left(a-1014\right)}{abc\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{ac\left(a-c\right)\left(b-2\right)\left(b-1004\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{ab\left(a-b\right)\left(c-2\right)\left(c-1004\right)}{abc\left(a-c\right)\left(a-b\right)\left(b-c\right)}\)

\(=\frac{2008b^2c+2008a^2c+2008a^2b-2008bc^2-2008a^2c-2008ab^2}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\frac{2008\left[\left(c^2a-c^2b\right)+\left(a^2b-a^2c\right)+\left(b^2a-b^2c\right)\right]}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\frac{2008\left(a-b\right)\left(b-c\right)\left(a-c\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\frac{2008}{abc}\) ( với \(abc\ne0\))

\(N=\dfrac{\left(a-b\right)\left(b+c\right)\left(a+c\right)+\left(b-c\right)\left(a+b\right)\left(c+a\right)+\left(c-a\right)\left(a+b\right)\left(b+c\right)+\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)\(=\dfrac{\left(a+c\right)\left(ab-b^2+ac-bc+ab-ac+b^2-cb\right)+\left(c-a\right)\left(ab+b^2+ac+bc+ab-b^2-ac+cb\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)\(=\dfrac{\left(a+c\right)\left(2ab-2bc\right)+\left(c-a\right)\left(2ab+2bc\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

\(=\dfrac{2b\left(a+c\right)\left(a-c\right)+2b\left(c-a\right)\left(a+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=\dfrac{2b\left(c+a\right)\left(a-c+c-a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)

Mình chỉ biết mỗi cách quy đồng...... Rồi kết hợp ....