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Ta có b + c = (a + b) + (c – a) nên
A = ab(a + b) – bc[(a + b) + (c – a)] – ac(c – a)
= ab(a + b) – bc(a + b) – bc(c – a) – ac(c – a)
= b(a + b)(a – c) – c(c – a)(b + a)
= (a + b)(a – c)(b + c)
Đáp án cần chọn là: B
\(=a^2b-ab^2+b^2c-bc^2+ac^2-a^2c\)
\(=a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b-c\right)\left(b+c\right)\)
\(=\left(b-c\right)\left(a^2-bc-ab-ac\right)\)
\(=\left(b-c\right)\left[a\left(a-b\right)-c\left(a-b\right)\right]\)
\(ab\left(a-b\right)-ac\left(a+c\right)+bc\left(2a-b+c\right)\)
\(=a^2b-ab^2-a^2c-ac^2+2abc-b^2c+bc^2\)
\(=a^2b-ab^2-a^2c-ac^2+abc+abc-b^2c+bc^2\)
\(=\left(bc^2-ac^2+abc-a^2c\right)-\left(b^2c-abc-ab^2+a^2b\right)\)
\(=c\left(bc-ac+ab-a^2\right)-b\left(bc-ac-ab+a^2\right)\)
\(=\left(c-b\right)\left(bc-ac+ab-a^2\right)\)
\(=\left(c-b\right)\left[c\left(b-a\right)+a\left(b-a\right)\right]\)
\(=\left(c-b\right)\left(c+a\right)\left(b-a\right)\)
bc(a+d) 9b –c) – ac( b +d) (a-c) + ab(c+d) ( a-b)
= bc(a+d) [ (b-a) + (a-c)] – ac(a-c)(b+d) +ab(c+d)(a-b)
= -bc(a+d )(a-b) +bc(a+d)(a-c) –ac(b+d)(a-c) + ab(c+d)(a-b)
= b(a-b)[ a(c+d) –c(a+d)] + c(a-c)[ b(a+d) –a(b+d)]
= b(a-b). d(a-c) + c(a-c) . d(b-a)
= d(a-b)(a-c)(b-c)
\(=a^2b+ab^2-b^2c-bc^2-ac^2+a^2c\)
\(=a^2\left(b+c\right)+a\left(b-c\right)\left(b+c\right)-bc\left(b+c\right)\)
\(=\left(b+c\right)\left(a^2+ab-ac-bc\right)\)
\(=\left(b+c\right)\left[a\left(a+b\right)-c\left(a+b\right)\right]\)
\(=\left(b+c\right)\left(a+b\right)\left(a-c\right)\)