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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

1 tháng 10 2016

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

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

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

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

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

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

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

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

31 tháng 5 2018

a) a3 (b-c) + b3 (c-a) +c3 (a-b)

<=> a3b – a3c +b3c – b3a + c3a – c3b

<=>  b(a3 – c3) +c(a3 – b3) + a(b- c3)

(Tự áp dụng hằng đẳng thức)

b)

19 tháng 9 2018

Đặt:  \(a-b=x;\)\(b-c=y;\)\(c-a=z\)

thì:  \(x+y+z=0\)

Dễ dàng chứng minh đc:

\(x+y+z=0\)

thì   \(x^3+y^3+z^3=3xyz\)

đến đây bạn thay trở lại nhé