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a) Ta có: (a+b)2 - (a-b)2
= (a+b+a-b)(a+b-a+b)
= 2a.2b
= 4ab
b) Ta có: (a+b)3 - (a-b)3 - 2b3
= a3 + 3a2b + 3ab2 + b3 - a3 + 3a2b - 3ab2 + b3 - 2b3
= 6a2b
c) Ta có: (x+y+z)2 - 2(x+y+z)(x+y) + (x+y)2
= (x+y+z-x-y)2
= z2
Ta có A=\(\left(ab+bc+ca\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-abc\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)
=\(2\left(a+b+c\right)+\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}-\frac{ab}{c}-\frac{bc}{a}-\frac{ca}{b}=2\left(a+b+c\right)\)
\(A=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2=a^2-ab+b^2+3ab\left(1-2ab\right)+6a^2b^2\)
=\(\left(a+b\right)^2-3ab+3ab-6a^2b^2+6a^2b^2=1\)
2) Ta có \(A=\left(a-1\right)\left(b-1\right)\left(c-1\right)=abc-ab-bc-ca+a+b+c-1=0\)
a)(x+y+z)2 - 2(x+y+z)(x+y)+(x+y)2
=[(x+y+z)-(x-y)]2
=(x+y+z-x-y)2
=z2
b) (a+b)3 - (a - b)3 - 2b3
=[(a+b)-(a-b)][(a+b)2+(a+b)(a-b)+(a-b)2]-2b3
=(a+b-a+b)(a2+2ab+b2+a2-b2+a2-2ab+b2)-2b3
=2b(3a2+b2)-2b3
=6a2b+2b3-2b3
=6a2b
c) (a + b)2 - (a - b)2=[a+b+(a-b)][a+b-(a-b)]=(a+b+a-b)(a+b-a+b)
=2a.2b=4ab
Bài 1:
\(A=a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\left(a^2+b^2\right)+6a^2b^2\)
\(=1^3-3ab+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\)
\(=1-3ab+3ab\left(1-2ab\right)+6a^2b^2=1\)
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: \(C=\frac{a^2b+b^2c+c^2a-ab^2-bc^2-a^2c}{a^3\left(b^2-c^2\right)+b^3\left(c^2-a^2\right)+c^3\left(a^2-b^2\right)}\)
\(=\frac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)\left(ab+bc+ac\right)}\)
\(=\frac{1}{ab+bc+ac}\)
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
(a+b)\(^2\)-(b-a)\(^2\)
\(=a^2+2ab+b^2-b^2-2ba+a^2\)
\(=2a^2\)