\(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\le...">
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15 tháng 8 2018

Bài cuối hơi khó nhìn, bạn thông cảm nhé! ^^

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15 tháng 8 2018

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

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

\(=\left(a^2b-c^2b\right)-\left(a^2c-c^2a\right)-b^2\left(a-c\right)\)

\(=b\left(a^2-c^2\right)-ac\left(a-c\right)-b^2\left(a-c\right)\)

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

\(=\left(a-c\right)\left[b\left(a+c\right)-ac-b^2\right]\)

\(=\left(a-c\right)\left(ab+bc-ac-b^2\right)\)

\(=\left(a-c\right)\left[\left(ab-b^2\right)+\left(bc-ac\right)\right]\)

\(=\left(a-c\right)\left[b\left(a-b\right)+c\left(b-a\right)\right]\)

\(=\left(a-c\right)\left[b\left(a-b\right)-c\left(a-b\right)\right]\)

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

b) \(a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)\)

\(=a^3b-a^3c+c^3a-c^3b+b^3\left(c-a\right)\)

\(=\left(a^3b-c^3b\right)-\left(a^3c-c^3a\right)-b^3\left(a-c\right)\)

\(=b\left(a^3-c^3\right)-ac\left(a^2-c^2\right)-b^3\left(a-c\right)\)

\(=b\left(a-c\right)\left(a^2+ac+c^2\right)-ac\left(a-c\right)\left(a+c\right)-b^3\left(a-c\right)\)

\(=\left(a-c\right)\left[b\left(a^2+ac+c^2\right)-ac\left(a+c\right)-b^3\right]\)

\(=\left(a-c\right)\left(ba^2+abc+bc^2-a^2c-ac^2-b^3\right)\)

\(=\left(a-c\right)\left[\left(ba^2-a^2c\right)+\left(abc-ac^2\right)+\left(bc^2-b^3\right)\right]\)

\(=\left(a-c\right)\left[a^2\left(b-c\right)+ac\left(b-c\right)+b\left(c^2-b^2\right)\right]\)

\(=\left(a-c\right)\left[a^2\left(b-c\right)+ac\left(b-c\right)-b\left(b^2-c^2\right)\right]\)

\(=\left(a-c\right)\left[a^2\left(b-c\right)+ac\left(b-c\right)-b\left(b-c\right)\left(b+c\right)\right]\)

\(=\left(a-c\right)\left(b-c\right)\left[a^2+ac-b\left(b+c\right)\right]\)

\(=\left(a-c\right)\left(b-c\right)\left(a^2+ac-b^2-bc\right)\)

\(=\left(a-c\right)\left(b-c\right)\left[\left(a-b\right)\left(a+b\right)+c\left(a-b\right)\right]\)

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

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

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

nguồn câu hỏi tương tự

14 tháng 7 2019

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Trang 136 trong nâng cao phát triển có viết rồi mình cóp nó vô để mọi người dễ đọc nhé !

a: \(=ab\left(a+b\right)-bc\left(b+a\right)-bc\left(c-a\right)-ac\left(c-a\right)\)

\(=\left(a+b\right)\left(ab-bc\right)+\left(a-c\right)\left(bc-ac\right)\)

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

\(=\left(a-c\right)\left(ab+b^2+cb-ac\right)\)

b: \(=ab^2+ac^2+bc^2+a^2b+a^2c+b^2c+2abc\)

\(=ab\left(a+b\right)+c^2\left(a+b\right)+c\left(a+b\right)^2\)

\(=\left(a+b\right)\left(ab+c^2+ac+cb\right)\)

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

d: \(=a^3\left(b-c\right)-b^3\left(b-c+a-b\right)+c^3\left(a-b\right)\)

\(=a^3\left(b-c\right)-b^3\left(b-c\right)-b^3\left(a-b\right)+c^3\left(a-b\right)\)

\(=\left(b-c\right)\left(a-b\right)\left(a^2+ab+b^2\right)-\left(a-b\right)\left(b-c\right)\left(b^2+bc+c^2\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(a^2+ab+b^2-b^2-bc-c^2\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(a^2+ab-bc-c^2\right)\)

\(=\left(a-b\right)\left(b-c\right)\cdot\left[\left(a-c\right)\left(a+c\right)+b\left(a-c\right)\right]\)

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

a) \(A=\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(5x+5\right)^2\)

\(A=\left[\left(3x+1\right)-\left(5x+5\right)\right]^2\)

\(A=\left(-2x-4\right)^2\)

30 tháng 9 2017

A = (3x + 1)2 - 2(3x + 1)(5x + 5) + (5x + 5)2

= [(3x + 1)-(5x + 5)]2

= (3x + 1 - 5x - 5)2

= [(-2x) - 4]2

B = (3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 +1)(332 + 1)

=> (3 - 1)B = (3 - 1)(3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 +1)(332 + 1)

=>2B = (32 - 1)(32 + 1)(34 + 1)(38 + 1)(316 +1)(332 + 1)

= (34 - 1)(34 + 1)(38 + 1)(316 +1)(332 + 1)

= (38 - 1)(38 + 1)(316 +1)(332 + 1)

= (316 - 1)316 +1)(332 + 1)

= (332 - 1)(332 + 1)

= 364 - 1

vì 2B = 364 - 1

=> B = \(\dfrac{3^{64}-1}{2}\)

C = a2 + b2 + c2 + 2ab - 2ac - 2bc + a2 + b2 + c2 - 2ab + 2ac - 2bc - 2( b2 - 2bc + c2)

= 2a2 + 2b2 + 2c2 - 4bc - 2b2 + 4bc - 2c2

= 2a2

1 tháng 1 2022

a) \(x^7+x^5+1\)

\(=x^7-x+x^5-x^2+x^2+x+1\)

\(=x\left(x^6-1\right)+x^2\left(x^3-1\right)+\left(x^2+x+1\right)\)

\(=x\left(x^3+1\right)\left(x^3-1\right)+x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

\(=x\left(x^3+1\right)\left(x-1\right)\left(x^2+x+1\right)+x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)]

\(=\left(x^2+x+1\right)\left[x\left(x^3+1\right)\left(x-1\right)+x^2\left(x-1\right)+1\right]\)

\(=\left(x^2+x+1\right)\left[x\left(x^4-x^3+x-1\right)+x^3-x^2+1\right]\)

\(=\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+x^3-x^2+1\right)\)

\(=\left(x^2+x+1\right)\left(x^5-x^4+x^3-x+1\right)\)

b) \(x^5-x^4-1\)

\(=x^5-x^4+x^3-x^3+x^2-x-x^2+x-1\)

\(=x^3\left(x^2-x+1\right)-x\left(x^2-x+1\right)-\left(x^2-x+1\right)\)

\(=\left(x^2-x+1\right)\left(x^3-x-1\right)\)