ĐƠN GIẢN CÁC BIỂU THỨC
\(a,-b-\left(b-a+c\right)\) \(b,-\left(a-b+c\right)-\left(c-a\right)\)
\(c,b-\left(b+a-c\right)\) \(d,a-\left(-b+a-c\right)\)
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BA
17 tháng 11 2019
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 \(abc\ne0\))
21 tháng 12 2017
\(N=\dfrac{\left(a-b\right)\left(b+c\right)\left(a+c\right)+\left(b-c\right)\left(a+b\right)\left(c+a\right)+\left(c-a\right)\left(a+b\right)\left(b+c\right)+\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)\(=\dfrac{\left(a+c\right)\left(ab-b^2+ac-bc+ab-ac+b^2-cb\right)+\left(c-a\right)\left(ab+b^2+ac+bc+ab-b^2-ac+cb\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)\(=\dfrac{\left(a+c\right)\left(2ab-2bc\right)+\left(c-a\right)\left(2ab+2bc\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(=\dfrac{2b\left(a+c\right)\left(a-c\right)+2b\left(c-a\right)\left(a+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=\dfrac{2b\left(c+a\right)\left(a-c+c-a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)
TT
8 tháng 1 2019
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
TT
8 tháng 1 2019
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
18 tháng 6 2018
a. \(\left(a+b+c\right)^2+\left(a-b-c\right)^2+\left(b-c-a\right)^2+\left(c-a-b\right)^2=a^2+b^2+c^2+2ab+2bc+2ac+a^2+b^2+c^2-2ab+2bc-2ac+c^2+a^2+b^2-2bc+2ac-2ab+a^2+b^2+c^2+2ab-2ac-2bc=4\left(a^2+b^2+c^2\right)\)b. Bạn làm tương tự câu a , đáp số ra : \(4\left(a^2+b^2+c^2+d^2\right)\)
TA
31 tháng 12 2017
\(M=\dfrac{\left(a+b\right)\left(b+c\right)}{\left(a-b\right)\left(b-c\right)}.\dfrac{\left(b+c\right)\left(c+a\right)}{\left(b-c\right)\left(c-a\right)}.\dfrac{\left(c+a\right)\left(a+b\right)}{\left(c-a\right)\left(a-b\right)}\) (loạn mắt quá !!! 
)
\(M=\dfrac{\left(a+b\right)\left(b+c\right)\left(c-a\right)\left(b+c\right)\left(c+a\right)\left(a-b\right)\left(c+a\right)\left(a+b\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Thôi để bữa sau a làm (đau mắt quá :)))
a/ - b - (b - a + c)
= - b - b + a -c
b/ - (a - b + c) - (c - a)
= - a + b - c - c + a
c/ b - (b + a - c)
= b - b - a + c
d/ a - (- b + a - c)
= a + b - a + c
a) -b - ( b - a + c )
= -b - b + a - c
=-2b + a - c
b) - ( a- b + c ) - ( c - a )
= - a + b -c -c + a
= b-2c
c) b - ( b + a - c )
= b - b - a + c
= -a + c
d) a - ( -b + a - c )
= a + b - a + c
= b + c