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a. A=(a-b)+(a+b-c)-(a-b-c)
=a-b+a+b-c-a+b+c
=(a+a-a)+(b+b-b)+(c-c)
=a+b
b. B=(a-b)-(b-c)+(c-a)-(a-b-c)
=a-b-b+c+c-a-a+b+c
=(a-a-a)+(b-b-b)+(c+c+c)
=-a-b+3c
c. C=(-a+b+c)-(a-b+c)-(-a+b-c)
=-a+b+c-a+b-c+a-b+c
=(a-a-a)+(b+b-b)+(c+c-c)
=-a+b+c
a) A= ( a-b) + (a+b-c) - ( a-b-c)
= a-b+a+b-c-a+b+c
= ( a +a -a) -( b-b-b) - (c-c)
= a - (-b) - 0
= a +b
b) B= ( a -b) - (b-c) + (c-a) -( a-b-c)
= a - b - b +c +c - a - a +b +c
= ( a - a -a) - (b+b -b) + ( c+c +c)
= - a - b + 3c
c) C= (-a +b+c ) - ( a-b+c) - (-a +b -c)
= -a+b+c -a+b-c +a -b+c
= (-a-a+a) + (b+b-b) + ( c-c+c)
= -a + b + c
Ta có: \(A=\left(a+b-c\right)+\left(a-b\right)-\left(a-b\right)-\left(a-b-c\right).\)
\(\Rightarrow A=a+b-c+a+b+a+b+a+b+c\)
\(\Rightarrow A=a+b+a+b+a+b+a+b\)
\(\Rightarrow A=3.\left(a+b\right)\)
\(A=\left(-a-b+c\right)-\left(-a-b-c\right)\)
\(A=-a-b+c+a+b+c\)
\(A=\left(-a+a\right)+\left(b-b\right)+\left(c+c\right)\)
\(\Rightarrow A=2c\)
\(A=\left(-a-b+c\right)-\left(-a-b-c\right)\)
\(A=-a-b+c+a+b+c\)
\(A=2c\)
Vậy \(A=2c\)
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
a) \(A=\left(a-2b+c\right)-\left(a-2b-c\right)\)
\(A=a-2b+c-a+2b+c=2c\)
b) \(B=\left(-x-y+3\right)-\left(-x+2-y\right)\)
\(B=-x-y+3+x-2+y=1\)
c) \(C=2\left(3a+b-1\right)-3\left(2a+b-2\right)\)
\(C=6a+2b-2-6a-3b+6=4-b\)
a. \(A=\left(a-2b+c\right)-\left(a-2b-c\right)=a-2b+c-a+2b+c=0\)
b. \(B=\left(-x-y+3\right)-\left(-x+2-y\right)=-x-y+3+x-2+y=1\)
c. \(C=2\left(3a+b-1\right)-3\left(2a+b-2\right)=6a+2b-2-6b-3b+6=4-3b\)
\(\left(a+b\right)-\left(-a+b-c\right)+\left(c-a-b\right)\)
\(=a+b+a-b+c+c-a-b\)
\(=\)\(a-b+2c\)( đpcm )
\(a\left(b-c\right)-a\left(b+d\right)\)
\(=a\left(b-c-b-d\right)\)
\(=\)\(a\left(-c-d\right)\)
\(=-a\left(c+d\right)\)( đpcm )
học tốt
#)Giải :
a) \(A=\frac{4^5.9^4-2^6.6^9}{2^{10}.3^8+6^8.20}=\frac{2^{10}.3^8-2^{10}.3^8.3}{2^{10}.3^8+2^8.3^8.2^2.5}=\frac{2^{10}.3^8-2^{10}.3^8.3}{2^{10}.3^8+2^{10}.3^8.5}=\frac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=-\frac{1}{3}\)
\(a,A=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)
\(=\frac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=\frac{-1}{3}\)
Học tốt!!!!!!!!!!!!!
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
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