Question

Bỏ ngoặc rồi rút gọn biểu thức:

 a) - ( - a + c – d ) – ( c – a + d )                 ; b) – ( a + b  - c + d ) + ( a – b – c –d )

 c)  a( b – c – d ) – a ( b + c – d )                ; d)(a+ b) ( c + d) – ( a + d ) ( b + c )

 e)( a + b ) ( c – d ) – ( a – b ) ( c + d )      ; f) ( a + b ) ² – ( a – b ) ²

Answer 1

a) - ( - a + c – d ) – ( c – a + d ) 

= a - c - d - c + a + d

= (a + a) + (-c - c) + (-d + d)

= 2a - 2c               

b) – ( a + b  - c + d ) + ( a – b – c –d )

= - a - b + c - d + a - b - c - d

= (-a + a) + (-b - b) + (c - c) + (-d - d)

= -2b - 2d

Answer 2

a) - ( - a + c - d) - ( c - a + d )

= a - c + d - c + a - d 

= 2a

b) - ( a+ b - c + d ) + ( a -b -c -d )

= - a-b+c-d+a-b-c-d

=-2d -2b

c) a(b-c-d) - a(b+c-d)

= a(b-c-d-b-c+d)

= ab-ac-ad-ab-ac+ad

= -2ab-2ac

d) (a+b)(c+d)-(a+d)(b+c)

= ac+ad+bc+bd - (ab+ac+bd+cd)

= ac+ad+bc+bd-ab-ac-bd-cd

=ad+bc-ab-cd

Answer 3

a) - ( - a + c - d) - ( c - a + d )

= a - c + d - c + a - d 

= 2a

b) - ( a+ b - c + d ) + ( a -b -c -d )

= - a-b+c-d+a-b-c-d

=-2d -2b

c) a(b-c-d) - a(b+c-d)

= a(b-c-d-b-c+d)

= ab-ac-ad-ab-ac+ad

= -2ab-2ac

d) (a+b)(c+d)-(a+d)(b+c)

= ac+ad+bc+bd - (ab+ac+bd+cd)

= ac+ad+bc+bd-ab-ac-bd-cd

=ad+bc-ab-cd

e)(a+b)(c-d)-(a-b)(c+d)

= ac-ad+bc-bd-ac-ad+bc+bd

= 2bc-2ad

f) ( a + b )² – ( a – b )²

= a²+2ab+b² - (a²+2ab-b²)

=a²+2ab+b² - a²-2ab+b²

=2b²

Answer 4

Giải:

a) \(-\left(-a+c-d\right)-\left(c-a+d\right)\) 

\(=a-c+d-c+a-d\) 

\(=\left(a+a\right)+\left(-c-c\right)+\left(d-d\right)\) 

\(=2a+-2c+0\) 

\(=2a-2c\) 

b) \(-\left(a+b-c+d\right)+\left(a-b-c-d\right)\) 

\(=-a-b+c-d+a-b-c-d\) 

\(=\left(-a+a\right)+\left(-b-b\right)+\left(c-c\right)+\left(-d-d\right)\) 

\(=0+-2b+0+-2d\) 

\(=-2b-2d\) 

c) \(a.\left(b-c-d\right)-a.\left(b+c-d\right)\) 

\(=ab-ac-ad-ab-ac+ad\) 

\(=\left(ab-ab\right)+\left(-ac-ac\right)+\left(-ad+ad\right)\) 

\(=0+-2ac+0\) 

\(=-2ac\) 

d) \(\left(a+b\right).\left(c+d\right)-\left(a+d\right).\left(b+c\right)\) 

\(=ac+bc+ad+bd-ab-db-ac-dc\) 

\(=\left(ac-ac\right)+\left(bd-db\right)+b.\left(c-d\right)+d.\left(b-c\right)\) 

\(=0+0+bc-bd+db-dc\) 

\(=\left(-bd+db\right)+c.\left(b-d\right)\) 

\(=0+c.\left(b-d\right)\) 

\(=c.\left(b-d\right)\) 

e) \(\left(a+b\right).\left(c-d\right)-\left(a-b\right).\left(c+d\right)\) 

\(=ac+bc-ad+bd-ac+bc-ad+bd\) 

\(=\left(ac-ac\right)+\left(bc+bc\right)+\left(-ad-ad\right)+\left(bd+bd\right)\) 

\(=0+2bc+-2ad+2bd\) 

\(=2bc-2ad+2bd\) 

\(=2.\left(bc-ad+bd\right)\) 

f) \(\left(a+b\right)^2-\left(a-b\right)^2\) 

\(=\left(a+b\right).\left(a+b\right)-\left(a-b\right).\left(a-b\right)\) 

\(=aa+ba+ab+bb-aa+ba-ab+bb\) 

\(=\left(aa-aa\right)+\left(ba+ba\right)+\left(ab-ab\right)+\left(bb+bb\right)\) 

\(=0+2ba+0+2bb\) 

\(=2ba+2bb\) 

\(=2b.\left(a+b\right)\)

Answer 5

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