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Lizy
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meme
12 tháng 9 2023 lúc 14:11

(a(b-c)^2 + b(c-a)^2 + c(a-b)^2) - (a^3 + b^3 + c^3) + 4abc

= a(b^2 - 2bc + c^2) + b(c^2 - 2ac + a^2) + c(a^2 - 2ab + b^2) - (a^3 + b^3 + c^3) + 4abc

= ab^2 - 2abc + ac^2 + bc^2 - 2abc + ba^2 + ca^2 - 2abc + cb^2 - a^3 - b^3 - c^3 + 4abc

= ab^2 + ac^2 + bc^2 + ba^2 + ca^2 + cb^2 - a^3 - b^3 - c^3 + 4abc - 6abc

= a(b^2 + c^2 + a^2) + b(a^2 + c^2 + b^2) + c(a^2 + b^2 + c^2) - (a^3 + b^3 + c^3) - 2abc

= a^3 + b^3 + c^3 + a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 - a^3 - b^3 - c^3 - 2abc

= a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 - 2abc

= ab(a + b) + ac(a + c) + bc(b + c) - 2abc

= (a + b)(ab - ac + bc) - 2abc

Vậy, ta có thể viết bài toán dưới dạng nhân tử là: (a + b)(ab - ac + bc) - 2abc.

Huỳnh Minh Long
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Nguyễn Nam
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Park Chanyeol
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Nguyễn Văn Nam
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Vỹ Ly
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Nguyễn Huệ Lam
29 tháng 6 2017 lúc 16:34

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)

\(=ab^3-ac^3+bc^3-a^3b+a^3c-b^3c\)

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

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

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

Nguyễn Huệ Lam
29 tháng 6 2017 lúc 16:35

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)

\(=ab^3-ac^3+bc^3-a^3b+a^3c-b^3c\)

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

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

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

quynh le
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quynh le
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Roy Wang
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Nhã Hy
15 tháng 7 2017 lúc 21:58

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

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

\(D=a^3\left(b-c\right)\left(b^3c-ab^3+ac^3-bc^3\right)\)

\(D=a^3\left(b-c\right)\left[\left(b^3c-bc^3\right)-\left(ab^3-ac^3\right)\right]\)

\(D=a^3\left(b-c\right)\left[bc\left(b^2-c^2\right)-a\left(b^3-c^3\right)\right]\)

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

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

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

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

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

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

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

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

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

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

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

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

Chúc bạn học tốt.