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) giúp tớ giải mấy câu này đi

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

Answer 1

a) \(-\left(-a+c-d\right)-\left(c-a+d\right)\\ =a-c+d-c+a-d\\ =\left(a+a\right)-\left(c+c\right)+\left(d-d\right)\\ =2a-2c\)

b) \(-\left(a+b-c+d\right)+\left(a-b-c-d\right)\\ =-a-b+c-d+a-b-c-d\\ =\left(a-a\right)-\left(b+b\right)+\left(c-c\right)-\left(d+d\right)\\ =-2b-2d\)

c) \(a\left(b-c-d\right)-a\left(b+c-d\right)\\ =ab-ac-ad-ab-ac+ad\\ =\left(ab-ab\right)-\left(ac+ac\right)+\left(ad-ad\right)\\ =-2ac\)

d) \(\left(a+b\right)\left(c+d\right)-\left(a+d\right)\left(b+c\right)\\ =a\left(c+d\right)+b\left(c+d\right)-\left[a\left(b+c\right)+d\left(b+c\right)\right]\\ =ac+ad+bc+bd-\left(ab+ac+bd+dc\right)\\ =ac+ad+bc+bd-ab-ac-bd-dc\\ =\left(ac-ac\right)+ad+bc+\left(bd-bd\right)-dc\\ =ad+bc-dc\)

e) \(\left(a+b\right)\left(c-d\right)-\left(a-b\right)\left(c+d\right)\\ =a\left(c-d\right)+b\left(c-d\right)-\left[a\left(c+d\right)-b\left(c+d\right)\right]\\ =ac-ad+bc-bd-\left(ac+ad-bc-bd\right)\\ =ac-ad+bc-bd-ac-ad+bc+bd\\ =\left(ac-ac\right)-\left(ad+ad\right)+\left(bc+bc\right)+\left(bd-bd\right)\\ =-2ad+2bc\)

f) \(\left(a+b\right)^2-\left(a-b\right)^2\\ =\left(a+b\right)\left(a+b\right)-\left(a-b\right)\left(a-b\right)\\ =a\left(a+b\right)+b\left(a+b\right)-\left[a\left(a-b\right)-b\left(a-b\right)\right]\\ =a^2+ab+ab+b^2-\left(a^2-ab-ab+b^2\right)\\ =a^2+2ab+b^2-\left(a^2-2ab+b^2\right)\\ =a^2+2ab+b^2-a^2+2ab-b^2\\ =\left(a^2-a^2\right)+\left(2ab+2ab\right)+\left(b^2-b^2\right)\\ =4ab\)