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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
\(A=\left(-a-b+c\right)-\left(-a-b-c\right)\)
\(A=-a-b+c+a+b+c\)
\(A=\left(-a+a\right)+\left(b-b\right)+\left(c+c\right)\)
\(\Rightarrow A=2c\)
\(A=\left(-a-b+c\right)-\left(-a-b-c\right)\)
\(A=-a-b+c+a+b+c\)
\(A=2c\)
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. \(C=2\left(3a+b-1\right)-3\left(2a+b-2\right)=6a+2b-2-6b-3b+6=4-3b\)
a) Vế trái = a.(c + d) + b.( c+ d) - a.(b + c) - d.(b + c)
= a.[(c+ d) - (b + c)] + [b(c+d) - d.(b + c)]
= a.(d - b) + (bc + bd - db - dc) = a.(d - b) + c.(b - d) = a.(d - b) - c.(d - b) = (a - c).(d - b) = Vế phải
Vậy....
b) làm tương tự:
a) (a+b) (c+d) - (a+d) (b+c) = (ac + ad + bc + bd) - (ab + ac +bd + cd) = ac + ad + bc + bd - ab -ac - bd - cd
và bằng ad + bc - ab - cd = a( d-b ) + c( b-d ) = a (d-b) - c (d-b) = (a-c)(d-b) (dpcm)
p/s: ý B chứng minh tương tự.
a. VT:(x-y)-(x-z)
= x-y-x+z
= z-y
VP:(z+x)-(y+x)
=z+x-y-x
=z-y
=> VT=VP => đpcm.
b. VT:(x-y+z)-(y+z-x)-(x-y)
= x-y+z-y-z+x-x+y
= x-y
VP:(z-y)-(z-x)
= z-y-z+x
= x-y
=> VT=VP => đpcm.
c. VT: a(b+c)-b(a-c)
=ab+ac-ab+bc
= ac+bc
VP: (a+b)c
= ac+bc
=> VT=VP => đpcm.
d. VT: a(b-c)-a(b+d)
= ab-ac-ab-ad
= -ac-ad
VP: -a(c+d)
= -ac-ad
=> VT=VP => đpcm
tương tự...
\(\left(a+b\right)-\left(-a+b-c\right)+\left(c-a-b\right)\)
\(=a+b+a-b+c+c-a-b\)
\(=\)\(a-b+2c\)( 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
Ta có: \(A=\left(a+b-c\right)+\left(a-b\right)-\left(a-b\right)-\left(a-b-c\right).\)
\(\Rightarrow A=a+b-c+a+b+a+b+a+b+c\)
\(\Rightarrow A=a+b+a+b+a+b+a+b\)
\(\Rightarrow A=3.\left(a+b\right)\)
Rút gọn biểu thức :
A=(a+b−c)+(a−b)−(a−b)−(a−b−c)
Ta có: A=(a+b−c)+(a−b)−(a−b)−(a−b−c).
⇒A=a+b−c+a+b+a+b+a+b+c
⇒A=a+b+a+b+a+b+a+b
⇒A=3.(a+b)
nhé !