Đơn giản biểu thức :
a) \(x+25+\left(-17\right)+63\)
b) \(\left(-75\right)-\left(p+20\right)+95\)
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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
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
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 #
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arigatou bẹn Hà nghen
mặc rầu chả bt là thèng nèo
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 \(abc\ne0\))
\(N=\dfrac{\left(a-b\right)\left(b+c\right)\left(a+c\right)+\left(b-c\right)\left(a+b\right)\left(c+a\right)+\left(c-a\right)\left(a+b\right)\left(b+c\right)+\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)\(=\dfrac{\left(a+c\right)\left(ab-b^2+ac-bc+ab-ac+b^2-cb\right)+\left(c-a\right)\left(ab+b^2+ac+bc+ab-b^2-ac+cb\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)\(=\dfrac{\left(a+c\right)\left(2ab-2bc\right)+\left(c-a\right)\left(2ab+2bc\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(=\dfrac{2b\left(a+c\right)\left(a-c\right)+2b\left(c-a\right)\left(a+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=\dfrac{2b\left(c+a\right)\left(a-c+c-a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)
a, \(x+25+\left(-17\right)+63\)
\(=x+25-17+63\)
\(=x+8+63\)
\(=x+71\)
b, \(\left(-75\right)-\left(p+20\right)+95\)
\(=-75-p-20+95\)
\(=-p-75-20+95\)
\(=-p-55+95\)
\(=-p-150\)
a) x + 25 + (- 17) + 63
x + 42 + 63
x + 105
b) (- 75) - (p + 20) + 95
= - 75 - p - 20 + 95
= p - 75 - 20 + 95
= p - 55 + 95
= p - 150