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A = (3,1 – 2,5) – (-2,5 + 3,1) = 3,1 – 2,5 + 2,5 – 3,1 = 0
B = (5,3 – 2,8) – (4 + 5,3) = 5,3 – 2,8 – 4 – 5,3
A=(3,1-2,5)-(-2,5+3,1)=3,1-2,5+2,5-3,1=0
B=(5,3-2,8)-(4+5,3)=5,3-2,8-4-5,3=-6,8
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. \(C=2\left(3a+b-1\right)-3\left(2a+b-2\right)=6a+2b-2-6b-3b+6=4-3b\)
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. 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 \(A=2c\)
Bài 1: Phá dấu ngoặc rồi tính:
a. \(\left(a+b+c\right)-\left(a-b+c\right)\)
\(=a+b+c-a+b-c\)
\(=\left(a-a\right)+\left(b+b\right)+\left(c-c\right)\)
\(=2b\)
b. \(\left(4x+5y\right)-\left(5x-4y-1\right)\)
\(=4x+5y-5x+4y+1\)
\(=\left(4x-5x\right)+\left(5y+4y\right)+1\)
\(=-x+9y+1\)
a, (18+29) +(158-18-29)
= 18+29+158-18-29
= (18-18) +(29-29)+ 158
= 158
b, (13-135+49)- (13+49)
= 13-135+49-13-49
= (13-13) +(49-49) -135
= 135
a, \(\left(18+29\right)+\left(158-18-29\right)\)
\(=18+29+158-18-29\)
\(=\left(18-18\right)+\left(29-29\right)+158\)
\(=158\)
b,
\(\left(13-135+49\right)-\left(13+49\right)\)
\(=13-135+49-13-49\)
= \(\left(13-13\right)+\left(49-49\right)-135\)
= \(-135\)
a) (27+65)+(346-27-65)
= 27+65+346-27-65
= (27-27)+(65-65)+346
= 0 + 0 + 346
= 346
b) (42-69+17)-(42+17)
= 42-69+17 – 42 – 17
= (42-42)+(17-17)-69
= 0 + 0 – 69
= -69
a)(27+65)+(346-27-65)
=27+65+345-27-65
=(27-27)+(65-65)+346
=0+0+346
=346
b) (42-69+17)-(42+17)
=42-69+17-42-17
=(42-42)-69+(17-17)
=0-69+0
=-69
a) Ta có: \(\left(a+b\right).\left(a-b\right)=a^2-ab+ab-b^2=a^2-b^2\)
b) Ta có: \(\left(a-b\right).\left(a-b\right)=a^2-ab-ab+b^2=a^2-2ab+b^2\)