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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) \(x^5-x^4-1\)
\(=x^5-x^4+x^3-x^3+x^2-x-x^2+x-1\)
\(=x^3\left(x^2-x+1\right)-x\left(x^2-x+1\right)-\left(x^2-x+1\right)\)
\(=\left(x^2-x+1\right)\left(x^3-x-1\right)\)
a) \(a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)\)
\(=a^3b-a^3c+b^3\left(c-a\right)+c^3a-c^3b\)
\(=\left(a^3b-c^3b\right)+\left(c^3a-a^3c\right)+b^3\left(c-a\right)\)
\(=-b\left(c^3-a^3\right)+ca\left(c^2-a^2\right)+b^3\left(c-a\right)\)
\(=-b\left(c-a\right)\left(c^2-ac+a^2\right)+ca\left(c+a\right)\left(c-a\right)+b^3\left(c-a\right)\)
\(=\left(c-a\right)\left(-c^2b+abc-a^2b\right)+\left(c-a\right)\left(c^2a+ca^2\right)+b^3\left(c-a\right)\)
\(=\left(c-a\right)\left(-c^2b+abc-a^2b+c^2a+ca^2+b^3\right)\)
a) a3 (b-c) + b3 (c-a) +c3 (a-b)
<=> a3b – a3c +b3c – b3a + c3a – c3b
<=> b(a3 – c3) +c(a3 – b3) + a(b3 - c3)
(Tự áp dụng hằng đẳng thức)
b)
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
Phân tích đa thức sau thành nhân tử :
\(a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)\)
\(a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)=a^3\left(b-c\right)+b^3c-b^3a+c^3a-c^3b\\ \)
\(\Rightarrow\)\(a^3\left(b-c\right)+bc\left(b^2-c^2\right)-a\left(b^3-c^3\right)\)
\(\Rightarrow\)\(a^3\left(b-c\right)+bc\left(b-c\right)\left(b+c\right)-a\left(b-c\right)\left(b^2+bc+c^2\right)\)
\(\Rightarrow\)\(\left(b-c\right)\left(a^3+bc\left(b+c\right)-a\left(b^2+bc+c^2\right)\right)\)
\(\Rightarrow\)\(\left(b-c\right)\left(a^3+b^2c+bc^2-ab^2-abc-ac^2\right)\)
\(\Rightarrow\)\(\left(b-c\right)\left(bc\left(c-a\right)+b^2\left(c-a\right)-a\left(c^2-a^2\right)\right)\)
\(\Rightarrow\)\(\left(b-c\right)\left(c-a\right)\left(bc+b^2-a\left(c+a\right)\right)\)
\(\Rightarrow\)\(\left(b-c\right)\left(c-a\right)\left(bc+b^2-ac-a^2\right)\)
\(\left(b-c\right)\left(c-a\right)\left(b^2-a^2+c\left(b-a\right)\right)=\left(b-c\right)\left(c-a\right)\left(b-a\right)\left(a+b+c\right)\)
a)hiệu hai lập phương
b) tổng hai lập phương
hằng đẳng thức có sẵn
bạn giải ra hộ mk đi