Gấp!!
Rút gọn:
a) (a+b+c)^2-(a+b)^2-(a+c)^2-(b+c)^2
b) (a+b+c)^2+(a-b+c)^2+(a+b-c)^2+(-a+b+c)^2
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Những câu hỏi liên quan
DD
13 tháng 7 2018
(a+b)\(^2\)-(b-a)\(^2\)
\(=a^2+2ab+b^2-b^2-2ba+a^2\)
\(=2a^2\)
19 tháng 7 2021
a) (2x+3)2-2(2x+3)(2x+5)+(2x+5)2
=4x2+12x+9-(4x+6)(2x+5)+4x2+20x+25
=4x2+12x+9-(8x2+12x+20x+30)+4x2+20x+25
=4x2+12x+9-8x2-12x-20x-30+4x2+20x+25
=4
b) (x2+x+1)(x2-x+1)(x2-1)
=((x2+1)2-x2)(x2-1)
=(x4+x2+1)(x2-1)
=x6+x4+x2-x4-x2-1
=x6-1
c)(a+b-c)2+(a-b+c)2-2(b-c)2
=a2+b2+c2+2ab-2ac-2bc+a2+b2+c2-2ab+2ac-2bc-2(b2-2bc+c2)
=2a2+2b2+2c2-4bc-2b2+4bc-2c2
=2a2
d) (a+b+c)2+(a-b-c)2+(b-c-a)2+(c-a-b)2
= a2+b2+c2+2ab+2ac+2bc+a2+b2+c2-2ab-2ac+2bc+a2+b2+c2+2bc-2ab+2ac+a2+b2+c2-2ac-2bc+2ab
=4a2+4b2+4c2+4ab+4bc
NL
Nguyễn Lê Phước Thịnh
CTVHS
1 tháng 3
Ta có: \(a^2b+b^2c+c^2a-ab^2-bc^2-a^2c\)
\(=a^2\left(b-c\right)+a\left(c^2-b^2\right)+bc\left(b-c\right)\)
\(=\left(b-c\right)\left(a^2+bc\right)-a\left(b-c\right)\left(b+c\right)\)
\(=\left(b-c\right)\left(a^2+bc-ab-ac\right)\)
\(=\left(b-c\right)\left(a^2-ab-ac+bc\right)\)
\(=\left(b-c\right)\left\lbrack a\left(a-b\right)-c\left(a-b\right)\right\rbrack=\left(a-b\right)\left(a-c\right)\left(b-c\right)\)
Ta có: \(a^3\left(b^2-c^2\right)+b^3\left(c^2-a^2\right)+c^3\left(a^2-b^2\right)\)
\(=a^3b^2-a^3c^2+b^3c^2-a^2b^3+c^3\left(a^2-b^2\right)\)
\(=a^2b^2\left(a-b\right)-c^2\left(a^3-b^3\right)+c^3\left(a-b\right)\left(a+b\right)\)
\(=\left(a-b\right)\left(a^2b^2+c^3a+c^3b\right)-c^2\left(a-b\right)\left(a^2+ab+b^2\right)\)
\(=\left(a-b\right)\left(a^2b^2+ac^3+bc^3-a^2c^2-abc^2-c^2b^2\right)\)
\(=\left(a-b\right)\left\lbrack a^2\left(b^2-c^2\right)+ac^2\left(c-b\right)+bc^2\left(c-b\right)\right\rbrack\)
\(=\left(a-b\right)\left(b-c\right)\left\lbrack a^2\left(b+c\right)-ac^2-bc^2\right\rbrack\)
\(=\left(a-b\right)\left(b-c\right)\left\lbrack a^2b+a^2c-ac^2-bc^2\right\rbrack=\left(a-b\right)\left(b-c\right)\cdot\left\lbrack b\left(a^2-c^2\right)+ac\left(a-c\right)\right\rbrack\)
=(a-b)(b-c)(a-c)\(\left\lbrack b\left(a+c\right)+ac\right\rbrack\)
=(a-b)(b-c)(a-c)(ab+bc+ac)
Ta có: \(C=\frac{a^2b+b^2c+c^2a-ab^2-bc^2-a^2c}{a^3\left(b^2-c^2\right)+b^3\left(c^2-a^2\right)+c^3\left(a^2-b^2\right)}\)
\(=\frac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)\left(ab+bc+ac\right)}\)
\(=\frac{1}{ab+bc+ac}\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
31 tháng 8 2021
Câu 3:
a: \(49^2=2401\)
b: \(51^2=2601\)
c: \(99\cdot100=9900\)
a)(a+b+c)2-(a+b)2-(a+c)2-(b+c)2
=a2+b2+c2+2ab+2bc+2ca-a2-2ab-b2-a2-2ac-c2-b2-2bc-c2
=-a2-b2-c2
=-(a2+b2+c2)
b)(a+b+c)2-(a-b+c)2+(a+b-c)2+(-a+b+c)2
=a2+2ab+b2+2bc+c2+2ac-a2-b2-c2+2ab+2bc-2ac+a2+b2+c2+2ab-2bc-2ac+a2+b2+c2-2ab-2ac+2bc
=2a2+2b2+2c2+4ab-4bc-4ac