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DT
1
HN
22 tháng 8 2021
\(a^4\left(b-c\right)+b^4\left(c-a\right)+c^4\left(a-b\right)\)
\(=a^4\left(a+b-a-c\right)+b^4\left(c-a\right)+c^4\left(a-b\right)\)
\(=-a^4\left(c-a\right)-a^4\left(a-b\right)+b^4\left(c-a\right)+c^4\left(a-b\right)\)
\(=\left(b^4-a^4\right)\left(c-a\right)+\left(c^4-a^4\right)\left(a-b\right)\)
\(=\left(b^2+a^2\right)\left(b^2-a^2\right)\left(c-a\right)+\left(c^2-a^2\right)\left(c^2+a^2\right)\left(a-b\right)\)
\(=\left(b^2+a^2\right)\left(b-a\right)\left(b+a\right)\left(c-a\right)+\left(c-a\right)\left(c+a\right)+\left(c^2+a^2\right)\left(a-b\right)\)
\(=\left(b-a\right)\left(c-a\right)[\left(b^2+a^2\right)\left(a+b\right)-\left(c+a\right)\left(c^2+a^2\right)]\)
\(=\left(b-a\right)\left(c-a\right)\left(ab^2+a^3+b^3+a^2b-c^3-ac^2-a^3-a^2c\right)\)
\(=\left(b-a\right)\left(c-a\right)\left(ab^2+b^3+a^2b-c^3-ac^2-a^2c\right)\)
\(=\left(b-a\right)\left(c-a\right)[\left(ab^2-ac^2\right)+\left(a^2b-a^2c\right)+\left(b^3+c^3\right)]\)
\(=\left(b-a\right)\left(c-a\right)[a\left(b^2-c^2\right)+a^2\left(b-c\right)+\left(b-c\right)\left(b^2+bc+c^2\right)]\)
\(=\left(b-a\right)\left(c-a\right)\left(b-c\right)\left(ab+ac+a^2+b^2+c^2+bc\right)\)
BH
4 tháng 10 2016
a)\(a^4+a^3+a^3b+a^2b=\left(a^4+a^3b\right)+\left(a^3+a^2b\right)\)
\(=a^3\left(a+b\right)+a^2\left(a+b\right)\)
\(=\left(a^3+a^2\right)\left(a+b\right)\)
\(=a^2\left(a+1\right)\left(a+b\right)\)
b)\(\left(x-y+4\right)^2-\left(2x+3y-1\right)^2\)
\(=\left[\left(x-y+4\right)-\left(2x+3y-1\right)\right]\left[\left(x-y+4\right)+\left(2x+3y-1\right)\right]\)
\(=\left(x-y+4-2x-3y+1\right)\left(x-y+4+2x+3y-1\right)\)
\(=\left(-x-4y+5\right)\left(4x+2y+3\right)\)
c)\(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\)
\(=x^2\left(y-z\right)+y^2\left(z-y+y-x\right)+z^2\left(x-y\right)\)
\(=x^2\left(y-z\right)-y^2\left(y-z\right)-y^2\left(x-y\right)+z^2\left(x-y\right)\)
\(=\left(y-z\right)\left(x^2-y^2\right)-\left(x-y\right)\left(y^2-z^2\right)\)
\(=\left(y-z\right)\left(x-y\right)\left(x+y\right)-\left(x-y\right)\left(y-z\right)\left(y+z\right)\)
\(=\left(y-z\right)\left(x-y\right)\left(x+y-y-z\right)\)
\(=\left(y-z\right)\left(x-y\right)\left(x-z\right)\)
13 tháng 10 2019
\(a^4\left(b-c\right)+b^4\left(c-a\right)+c^4\left(a-b\right)\)
\(=a^4\left(b-c\right)+b^4[\left(c-b\right)-\left(a-b\right)]+c^4\left(a-b\right)\)
\(=a^4\left(b-c\right)+b^4\left(c-b\right)-b^4\left(a-b\right)+c^4\left(a-b\right)\)
\(=a^4\left(b-c\right)-b^4\left(b-c\right)-b^4\left(a-b\right)+c^4\left(a-b\right)\)
\(=\left(b-c\right)\left(a^4-b^4\right)-\left(a-b\right)\left(c^4-b^4\right)\)
\(=\left(b-c\right)\left(a^2-b^2\right)\left(a^2+b^2\right)-\left(a-b\right)\left(c^2-b^2\right)\left(c^2+b^2\right)\)
\(=\left(b-c\right)\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)+\left(a-b\right)\left(b-c\right)\left(c+b\right)\left(c^2+b^2\right)\)
\(=\left(b-c\right)\left(a-b\right)[\left(a+b\right)\left(a^2+b^2\right)+\left(c+b\right)\left(c^2+b^2\right)]\)
DM
3
DD
20 tháng 7 2017
\(a^4\left(b-c\right)+b^4\left(c-a\right)+c^4\left(a-b\right)\)
\(\Leftrightarrow\text{=a^4(b-c)-b^4[(b-c)+(a-b)]+c^4(a-b) =(b-c)(a^4-b^4)+(a-b)(c^4-b^4)}\)
\(\text{=(b-c)(a^2-b^2)(a^2+b^2)+(a-b)(c^2-b^2)... =(b-c)(a-b)(a+b)(a^2+b^2)-(a-b)(b-c)(b+... }\)
\(\text{=(b-c)(a-b)(a^3+ab^2+ba^2+b^3-bc^2-b^3-... mà ta có a^3+ab^2+ba^2-bc^2-c^3-cb^2 }\)
\(\text{=(a^3-c^3)+b^2(a-c)+b(a^2-c^2) =(a-c)(a^2+ac+c^2)+b^2(a-c)+b(a-c)(a+c) }\)
\(\text{=(a-c)(a^2+ac+c^2+b^2+ab+ac) } \)
\(\text{từ đó suy ra a^4(b-c)+b^4(c-a)+c^4(a-b) =(a-b)(b-c)(c-a)(a^2+b^2+c^2+ab+bc+ca)}\)
3 tháng 8 2018
a) \(\left(a+b-c\right)^2-\left(a-c\right)^2-2ab+2ac\)
\(=a^2+b^2+c^2+2ab-2bc-2ac-a^2+2ac-c^2-2ab+2ac\)
\(=b^2-2bc+2ac=b.\left(b-2c+2a\right)\)
b) \(x^4+2x^3+5x^2+4x-12\)
\(=x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12\)
\(=x^3.\left(x-1\right)+3x^2.\left(x-1\right)+8x.\left(x-1\right)+12.\left(x-1\right)\)
\(=\left(x-1\right)\left(x^3+3x^2+8x+12\right)\)
\(=\left(x-1\right)\left[\left(x^3+2x^2\right)+\left(x^2+2x\right)+\left(6x+12\right)\right]\)
\(=\left(x-1\right)\left[x^2.\left(x+2\right)+x.\left(x+2\right)+6.\left(x+2\right)\right]\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)
NB
3 tháng 8 2018
Pạn Khánh Châu ơi
Cái dòng thứ 2 đấy, dấu hiệu nhận biết là j vậy
Mà sao pạn phân tích hay vậy????
XO
24 tháng 6 2021
a(b3 - c3) + b(c3 - a3) + c(a3 - b3)
= ab3 - ac3 + bc3 - ba3 + ca3 - cb3
= b(c3 - a3 - cb2 + ab2) - ac(c2 - a2)
= b[(c - a)(a2 + ac + c2) - b2(c - a)] - ac(c - a)(c + a)
= (c - a)[a2b + abc + bc2 - b3 - ac2 - a2c]
= (c - a)[b(c2 - b2) - ac(c - b) - a2(c -b)]
= (c - a)(c - b)[b(b + c) - ac - a2] = (c - a)(c - b)(b2 + bc - ac - a2]
= (c - a)(c - b)[(b - a)(b + a) + c(b - a)]
= (c - a)(c - b)(b - a)(a + b + c)
XO
24 tháng 6 2021
(a + b)3 - (a - b)3
= (a + b - a + b)[(a + b)2 + (a + b)(a - b) + (a - b)2]
= 2b(a2 + 2ab + b2 + a2 - b2 + a2 - 2ab + b2]
= 2b(3a2 + b2]
x3 - 3x2 + 3x - 1 - y3
= (x - 1)3 - y3
= (x - y - 1)(x2 - 2x + 1 + xy - y + y2]
xm + 4 + xm + 3 - x - 1
= xm + 3(x +1) - (x + 1)
= (xm + 3 - 1)(x + 1)
= (x - 1)[xm + 2 + xm + 1 + .... + 1](x + 1)
JH
0
16 tháng 11 2016
a(b3-c3) -b(b3-c3+a3-b3)+c(a3-b3)
=a(b3-c3)-b(b3-c3)-b(a3-b3)+c(a3-b3)
=(b3-c3)(a-b)-(a3-b3)(b-c)
=(b-c)(b2+cb+c2)(a-b)-(a-b)(a2+ab+b2)(b-c)
=(b-c)(a-b)(b2+Cb+c2-a2-ab-b2)
=(b-c)(a-b)(c2+cb-ab-a2)
=(b-c)(a-b)[(c-a)(c+a)+b(c-a)]
=(b-c)(a-b)(c-a)(a+c+b)
(a-b) (c-a) (c-b) (c2+b c+a c+b2+a b+a2)