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a,\(\left(ab-1\right)^2+\left(a+b\right)^2\)
= \(a^2b^2-2ab+1+a^2+2ab+b^2\)
= \(\left(a^2b^2+a^2\right)+\left(b^2+1\right)\)
= \(a^2\left(b^2+1\right)+\left(b^2+1\right)\)
= \(\left(a^2+1\right)\left(b^2+1\right)\)
b, \(x^4+2x^3-4x-4\)
= \(\left(x^4-4\right)+\left(2x^3-4x\right)\)
= \(\left(x^2-2\right)\left(x^2+2\right)+2x\left(x^2-2\right)\)
=\(\left(x^2-2\right)\left(x^2+2x+2\right)\)
=
\(a^3+a^2c-abc+b^2c+b^3\)
\(=\left(a^3+b^3\right)+\left(a^2c+b^2c-abc\right)\)
\(=\left(a+b\right)\left(a^2-ab+b^2\right)+\)\(c\left(a^2+b^2-ab\right)\)
\(=\left(a^2+b^2-ab\right)\left(a+b+c\right)\)
\(a\left(a+2b\right)^3-b\left(2a+b\right)^3\)
\(=a\left(a ^3+3.a^22b+3.a2b^2+2b^3\right)-b\left(2a^3+3.2a^2.b+3.2a.b^2+b^3\right)\)
\(=a\left(a^3+6a^2b+6ab^2+8b^3\right)-b\left(8a^3+6a^2b+6ab^2+b^3\right)\)
\(=a^4+6a^3b+6a^2b^2+8ab^3-8a^3b-6a^2b^2-6ab^3-b^4\)
\(=a^4+6a^3b+8ab^3-8a^3b-6ab^3-b^4\)
\(a\left(a+2b\right)^3-b\left(2a+b\right)^3\)
\(=a\left(a^3+6a^2b+12ab^2+8b^3\right)-b\left(8a^3+12a^2b+6ab^2+b^3\right)\)
\(=a^4+6a^3b+12a^2b^2+8ab^3-8a^3b-12a^2b^2-6ab^3-b^4\)
\(=a^4-2a^3b+2ab^3-b^4\)
\(=\left(a^2-b^2\right)\left(a^2+b^2\right)-2ab\left(a^2-b^2\right)\)
\(=\left(a^2-b^2\right)\left(a^2-2ab+b^2\right)\)
\(=\left(a+b\right)\left(a-b\right)^3\)
a) a3+a2c-abc+b2c+b3 =(a3+b3)+(a2c-abc+b2c)=(a+b)(a2-ab+b2)+c(a2-ab+b2)=(a2-ab+b2)(a+b-c)
b) x3-7x-6 = x3+x2-x2-x-6x-6=x2(x+1)-x(x+1)-6(x+1)=(x+1)(x2-x-6)=(x+1)(x-3)(x+2)
c) x3-x2-14x+24=x3-2x2+x2-2x-12x+24=x2(x-2)+x(x-2)-12(x-2)=(x-2)(x2+x-12)=(x-2)(x+4)(x-3)
a(a+2b)3 -b(2a+b)3
\(=a\left(a^3+6a^2b+12ab^2+8b^3\right)-b\left(8a^3+12a^2b+6ab^2+b^3\right)\)
\(=a^4+6a^3b+12a^2b^2+8ab^3-8a^3b-12a^2b^2-6ab^3-b^4\)
\(=a^4-2a^3b+2ab^3-b^4\)
\(=\left[\left(a^2\right)^2+ \left(b^2\right)^2\right]-2ab\left(a^2-b^2\right)\)
\(=\left(a^2+b^2\right)\left(a^2-b^2\right)-2ab\left(a^2-b^2\right)\)
\(=\left(a^2-b^2\right)\left(a^2-2ab+b^2\right)\)
\(=\left(a-b\right)\left(a+b\right)\left(a-b\right)^2\)
\(=\left(a-b\right)^3\left(a+b\right)\)
\(a.\left(a+2b\right)^3-b.\left(2a+b\right)^3\)
\(=a.\left(a+20+b\right)^3-b.\left(20+a+b\right)^3\)
\(=\left(a-b\right).\left(a+20+b\right)^3\)
Thế này có phải là phân tích đa thức thành nhân tử k ạ
Chúc bạn học tốt
\(a\left(a+2b\right)^3-b\left(2a+b\right)^3\)
\(=\left(a^4+6a^3b+12a^2b^2+8ab^3\right)-\left(b^4+8a^3b+12a^2b^2+6ab^3\right)\)
\(=a^4-b^4-2a^3b+2ab^3\)
\(=\left(a^2-b^2\right)\left(a^2+b^2\right)-2ab\left(a^2-b^2\right)\)
\(=\left(a^2-b^2\right)\left(a^2-2ab+b^2\right)\)
\(=\left(a-b\right)^3\left(a+b\right)\)
OK ?
b)\(2a^2-3+5a\)
\(=\left(2a^2+6a\right)-\left(a+3\right)\)
\(=\left(a+3\right)\left(2a-1\right)\)
d)\(2a^2-5-3a\)
\(=\left(2a^2+2a\right)-\left(5a+5\right)\)
\(=\left(a+1\right)\left(2a-5\right)\)
a) \(a^2-3-2a\)
\(=a^2-2a+1-4\)
\(=\left(a^2-2a+1\right)-2^2\)
\(=\left(a-1\right)^2-2^2\)
\(=\left(a-1-2\right)\left(a-1+2\right)\)
\(=\left(a-3\right)\left(a+1\right)\)
c) \(4a+a^2+3\)
\(=a^2+4a+4-1\)
\(=\left(a^2+4a+4\right)-1^2\)
\(=\left(a+2\right)^2-1^2\)
\(=\left(a+2-1\right)\left(a+2+1\right)\)
\(=\left(a+1\right)\left(a+3\right)\)
Đặt \(a+b-2c=x,b+c-2a=y,c+a-2b=z\)
\(\Rightarrow x+y+z=0\)
Chắc bạn biết: \(x+y+z=0\Rightarrow x^3+y^3+z^3=3xyz\)
Vậy \(\left(a+b-2c\right)^3+\left(b+c-2a\right)^3+\left(c+a-2b\right)^3=3\left(a+b-2c\right)\left(b+c-2a\right)\left(c+a-2b\right)\)
Chúc bạn học tốt.
ban su dung hang dang thuc lop 8
chỉ rõ đi bạn ,ns vậy còn chung chung wa