phân tích đa thức sau thành nhân tử : a^3(b^2-c^2)+b^3(c^2-a^2)+c^3(a^2-b^2)
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DM
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1 tháng 8 2019
\(B=\left(a^2+b^2\right)^3+\left(c^2-a^2\right)^3-\left(b^2+c^2\right)^3\)
\(=\left(a^2+b^2+c^2-a^2\right)\left[\left(a^2+b^2\right)^2-\left(c^2-a^2\right)\left(a^2+b^2\right)+\left(c^2-a^2\right)^2\right]-\left(b^2+c^2\right)^2\)
\(=\left(b^2+c^2\right)\left[\left(a^2+b^2\right)^2-\left(c^2-a^2\right)\left(a^2+b^2\right)+\left(c^2-a^2\right)^2\right]-\left(b^2+c^2\right)^2\)
\(=\left(b^2+c^2\right)\left(a^4+2a^2b^2+b^4-a^2c^2+a^4-b^2c^2+a^2b^2-b^4-2b^2c^2-c^4\right)\)
\(=\left(b^2+c^2\right)\left(2a^4-c^4+3a^2b^2-a^2c^2-3b^2c^2\right)\)
ko chắc
NH
1
29 tháng 6 2017
a) (a+b+c)^2 + (a+b-c)^2 - 4c^2
\(=\left(a+b+c\right)^2+\left[\left(a+b-c\right)^2-\left(2c\right)^2\right]\)
\(=\left(a+b+c\right)^2+\left(a+b-c+2c\right)\left(a+b-c-2c\right)\)
\(=\left(a+b+c\right)^2+\left(a+b+c\right)\left(a+b-3c\right)\)
\(=\left(a+b+c\right)\left(a+b+c+a+b-3c\right)\)
\(=\left(a+b+c\right)\left(2a+2b-2c\right)\)
\(=2\left(a+b+c\right)\left(a+b-c\right)\)
b) 4a^2b^2 - (a^2+b^2-c^2)^2
\(=\left(2ab\right)^2-\left(a^2+b^2-c^2\right)^2=\left(2ab+a^2+b^2-c^2\right)\left(2ab-a^2-b^2+c^2\right)\)
\(=\left[\left(a^2+2ab+b^2\right)-c^2\right]\left[c^2-\left(a^2-2ab+b^2\right)\right]\)
\(=\left[\left(a+b\right)^2-c^2\right]\left[c^2-\left(a-b\right)^2\right]\)
\(=\left(a+b+c\right)\left(a+b-c\right)\left(c+a-b\right)\left(c-a+b\right)\)
c) a(b^3-c^3) + b(c^3-a^3) + c(a^3-b^3)
\(=ab^3-ac^3+bc^3-a^3b+a^3c-b^3c\)
\(=a^3\left(c-b\right)+bc\left(c-b\right)\left(c+b\right)-a\left(c-b\right)\left(c^2+bc+b^2\right)\)
\(=a^3\left(c-b\right)+\left(c-b\right)\left(bc^2+b^2c\right)-\left(c-b\right)\left(ac^2+abc+ab^2\right)\)
\(=\left(c-b\right)\left(a^3+bc^2+b^2c-ac^2-abc-ab^2\right)\)
29 tháng 6 2017
a) (a+b+c)^2 + (a+b-c)^2 - 4c^2
\(=\left(a+b+c\right)^2+\left[\left(a+b-c\right)^2-\left(2c\right)^2\right]\)
\(=\left(a+b+c\right)^2+\left(a+b-c+2c\right)\left(a+b-c-2c\right)\)
\(=\left(a+b+c\right)^2+\left(a+b+c\right)\left(a+b-3c\right)\)
\(=\left(a+b+c\right)\left(a+b+c+a+b-3c\right)\)
\(=\left(a+b+c\right)\left(2a+2b-2c\right)\)
\(=2\left(a+b+c\right)\left(a+b-c\right)\)
b) 4a^2b^2 - (a^2+b^2-c^2)^2
\(=\left(2ab\right)^2-\left(a^2+b^2-c^2\right)^2=\left(2ab+a^2+b^2-c^2\right)\left(2ab-a^2-b^2+c^2\right)\)
\(=\left[\left(a^2+2ab+b^2\right)-c^2\right]\left[c^2-\left(a^2-2ab+b^2\right)\right]\)
\(=\left[\left(a+b\right)^2-c^2\right]\left[c^2-\left(a-b\right)^2\right]\)
\(=\left(a+b+c\right)\left(a+b-c\right)\left(c+a-b\right)\left(c-a+b\right)\)
c) a(b^3-c^3) + b(c^3-a^3) + c(a^3-b^3)
\(=ab^3-ac^3+bc^3-a^3b+a^3c-b^3c\)
\(=a^3\left(c-b\right)+bc\left(c-b\right)\left(c+b\right)-a\left(c-b\right)\left(c^2+bc+b^2\right)\)
\(=a^3\left(c-b\right)+\left(c-b\right)\left(bc^2+b^2c\right)-\left(c-b\right)\left(ac^2+abc+ab^2\right)\)
\(=\left(c-b\right)\left(a^3+bc^2+b^2c-ac^2-abc-ab^2\right)\)
M
31 tháng 10 2015
\(ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)\)
\(=ab\left(a-b\right)+bc\left(b-c\right)-ca\left(a-c\right)\)
\(=ab\left(a-b\right)+bc\left(b-c\right)-ca\left(a-b+b-c\right)\)
\(=ab\left(a-b\right)+bc\left(b-c\right)-ca\left(a-b\right)-ca\left(b-c\right)\)
\(=\left(a-b\right)\left(ab-ca\right)+\left(b-c\right)\left(bc-ca\right)\)
\(=\left(a-b\right)a\left(b-c\right)+\left(b-c\right)c\left(b-a\right)\)
\(=\left(a-b\right)a\left(b-c\right)-\left(b-c\right)c\left(a-b\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
mình làm vội, có chỗ nào sai bạn thông cảm nha