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\(ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)\)
\(=ab\left(a-b\right)+bc\left(b-c\right)-ca\left(a-c\right)\)
\(=ab\left(a-b\right)+bc\left(b-c\right)-ca\left(a-b+b-c\right)\)
\(=ab\left(a-b\right)+bc\left(b-c\right)-ca\left(a-b\right)-ca\left(b-c\right)\)
\(=\left(a-b\right)\left(ab-ca\right)+\left(b-c\right)\left(bc-ca\right)\)
\(=\left(a-b\right)a\left(b-c\right)+\left(b-c\right)c\left(b-a\right)\)
\(=\left(a-b\right)a\left(b-c\right)-\left(b-c\right)c\left(a-b\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
mình làm vội, có chỗ nào sai bạn thông cảm nha
a/ Ta có : (x2 + x + 1)2 = [x2 + (x + 1)]2 = x4 + 2x2(x + 1) + (x + 1)2 Nên:
A = (x + 1)4 + (x2 + x + 1)2 = (x + 1)4 + x4 + 2x2(x + 1) + (x + 1)2 = [(x + 1)4 + (x + 1)2] + [x4 + 2x2(x + 1)]
= (x + 1)2(x2 + 2x + 2) + x2(x2 + 2x + 2) = (x2 + 2x + 2)(2x2 + 2x + 1).
b/ B = x10 + x5 + 1 Đặt \(|x^5|=t^2\) thì x10 = t4 Ta có B = t4 + t2 + 1 = (t2 + 1)2 - t2 = (t2 - t + 1)(t2 + t + 1)
Vậy : \(B=\left(x^5-\sqrt{|x|^5}+1\right)\left(x^5+\sqrt{|x|^5}+1\right).\)
c/ Nhân đa thức được: C = x2(x4 - 1)(x2 + 2) + 1 = (x6 - x2)(x2 + 2) + 1 = x6 (x2 + 2) - x2 (x2 + 2) + 1
C = x8 + 2x6 - x4 - 2x2 + 1 = x8 + 2x6 - 2x4 + x4 - 2x2 + 1 = (x4)2 + 2x4 (x2 - 1) + (x2 - 1)2
C = (x4 + x2 + 1)2 .
d/ D = 1 + ( a + b + c) + ab + bc + ca) + abc = (1 + a) + (abc + bc) + (b + ab) + (c + ca) = (1 + a) + bc(1 + a) + b(1 + a) + c(1 + a) =
= (1 + a)(1 + bc + b + c) = (1 + a)[(1 + b) + c(1 + b)] = (1 + a)(1 + b)(1 + c).
\(b,\)\(x^{10}+x^5+1\)
\(=x^{10}-x^7+x^7+x^5+x^3-x^3+1\)
\(=x^7\left(x^3-1\right)+x^3\left(x^4+x^2+1\right)-\left(x^3-1\right)\)
\(=x^7\left(x-1\right)\left(x^2+x+1\right)+x^3\left(x^4+2x^2+1-x^2\right)-\left(x-1\right)\left(x^2+x+1\right)\)
\(=x^7\left(x-1\right)\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)\left(x^2-x+1\right)\)\(-\left(x-1\right)\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^8-x^7+x^5-x^4+x^3-x+1\right)\)
\(d,\)\(1+\left(a+b+c\right)+\left(ab+bc+ca\right)+abc\)
\(=1+a+b+c+ab+bc+ca+abc\)
\(=\left(ab+b\right)+\left(abc+bc\right)+\left(ac+c\right)+\left(a+1\right)\)
\(=b\left(a+1\right)+bc\left(a+1\right)+c\left(a+1\right)+\left(a+1\right)\)
\(=\left(a+1\right)\left(b+bc+c+1\right)\)
\(=\left(a+1\right)\left[b\left(c+1\right)+\left(c+1\right)\right]\)
\(=\left(a+1\right)\left(b+1\right)\left(c+1\right)\)
a. \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(x^2+7x+11=t.\)Thay vào ta được :
\(\left(t+1\right)\left(t-1\right)-24\)
\(=t^2-1-24=t^2-25=\left(t+5\right)\left(t-5\right)\)
Thay \(t=x^2+7x+11\)Ta được :
\(\left(x^2+7x+11+5\right)\left(x^2+7x+11-5\right)\)
\(=\left(x^2+7x+16\right)\left(x^2+7x+6\right)\)