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a) a2 + b2 + 2ab + 2a + 2b + 1
= (a2 + b2 + 2ab) + (2a + 2b) + 1
= (a + b)2 + 2(a + b) + 1
= (a + b + 1)2
b) a3 - 3a + 3b - b3
= (a3 - b3) - (3a - 3b)
= (a - b)(a2 - ab + b2) - 3(a - b)
= (a - b)(a2 - ab + b2 - 3)
c) x2 + 2x - 15
= (x2 + 2x + 1) - 16
= (x + 1)2 - 16
= (x + 1 - 5)(x + 1 + 5)
= (x - 4)(x + 6)
d) a4 + 6a2b + 9b2 - 1
= (a2 + 3b)2 - 1
= (a2 + 3b - 1)(a2 + 3b + 1)
\(3y^2\left(a-3x\right)-a\left(a-3x\right)=\left(3y^2-a\right)\left(a-3x\right)\)
1)
b) \(\left(x-z\right)^2-y^2+2y-1\)
\(=\left(x^2-2xz+z^2\right)-\left(y-1\right)^2\)
\(=\left(y-z\right)^2-\left(y-1\right)^2\)
\(=\left[\left(x-z\right)+\left(y-1\right)\right]\cdot\left[\left(x-z\right)-\left(y+1\right)\right]\)
\(=\left(x-z+y-1\right)\cdot\left(x-z-y-1\right)\)
a. x2y2+1-x2-y2
=x2.(y2-1)-(y2-1)
=(y2-1).(x2-1)
=(y-1)(y+1)(x-1)(y-1)
c. x^3+3x^2-3x-1
= (x-1).(x^2+x+1)+3x.(x-1)
=(x-1)(x^2+x+1+x-1)
=(x-1)(x^2+2x)
=x(x-1)(x+2)
Phân tích đa thức thành nhân tử
a) (1-2x)(1+2x)-x(x+2)(x-2)
\(=1-4x^2-x\left(x^2-4\right)\)
\(=1-4x^2-x^3+4x\)
\(=\left(1-x^3\right)+\left(4x-4x^2\right)\)
\(=\left(1-x\right)\left(1+x+x^2\right)+4x\left(1-x\right)\)
\(=\left(1-x\right)\left(1+x+x^2+4x\right)\)
\(=\left(1-x\right)\left(x^2+5x+1\right)\)
\(a\left(a+2b\right)^3-b\left(2a+b\right)^3\)
\(=a\left(a^3+6a^2b+12ab^2+8b^3\right)-b\left(8a^3+12a^2b+6ab^2+b^3\right)\)
\(=a^4+6a^3b+12a^2b^2+8b^3a-8a^3b-12a^2b^2+6ab^3-b^4\)
\(=a^4+6a^3b+8b^3a-8a^3b-6ab^3-b^4\)
\(=\left(a^4-b^4\right)+\left(6a^3b-6ab^3\right)+\left(8b^3a-8a^3b\right)\)
\(=\left(a-b\right)\left(a^3+a^2b+ab^2+b^3\right)+6ab\left(a^2-b^2\right)+8ab\left(b^2-a^2\right)\)
\(=\left(a-b\right)\left(a^3+a^2b+ab^2+b^3\right)+6ab\left(a-b\right)\left(a+b\right)-8ab\left(a-b\right)\left(a+b\right)\)
\(=\left(a-b\right)\left(a^3+a^2b+ab^2+b^3+6a^2b+6ab^2-8a^2b-8ab^2\right)\)
\(=\left(a-b\right)\left(a^3-a^2b-ab^2+b^3\right)\)
\(=\left(a-b\right)\left[a^2\left(a-b\right)-b^2\left(a-b\right)\right]\)
\(=\left(a-b\right)^3\left(a+b\right)\)
\(\left(a-2b\right)^2-\left(3a+b\right)^2\)
\(\Rightarrow\left(a-2b-3a-b\right)\left(a-2b+3a+b\right)\)
\(\Rightarrow\left(-2a-2b\right)\left(4a-b\right)\)
\(\left(4a+3b\right)^2-\left(b-2a\right)^2\)
\(\Rightarrow\left(4a+3b-b+2a\right)\left(4a+3b+b-2a\right)\)
\(\Rightarrow\left(6a+2b\right)\left(2a+4b\right)\)
\(36\left(x-y\right)-25\left(2x-1\right)^2\)chịu
Phối hợp cả 3 phương phép để phân tích các đa thức sau thành phân tử:
a) 36 - 4a2 + 20ab - 25b2
= 36 - (4a2 - 20ab + 25b2)
= 62 - (2a - 5b)2
= (6 - 2a + 5b)(6 + 2a - 5b)
b) a3 + 3a2 + 3a + 1 - 27b3
= (a + 1)3 - (3b)3
= (a + 1 - 3b)[(a + 1)2 + 3b(a + 1) + 9b2]
= (a + 1 - 3b)(a2 + 2a + 1 + 3ab + 3b + 9b2)
c) x2 + 2xy + y2 - xz - yz
= (x + y)2 - z(x + y)
= (x + y)(x + y - z)
d) 5a3 - 10a2b + 5ab2 - 10a + 10b
= 5(a3 - 2a2b + ab2 - 2a + 2b)
= 5[a(a2 - 2ab + b2) - 2(a - b)]
= 5[a(a - b)2 - 2(a - b)]
= 5(a - b)(a2 - ab - 2)
Bài 1:
\(A=a^4-2a^3+3a^2-4a+5\)
\(=(a^4-2a^3+a^2)+2a^2-4a+5\)
\(=(a^4-2a^3+a^2)+2(a^2-2a+1)+3\)
\(=(a^2-a)^2+2(a-1)^2+3\)
\(=a^2(a-1)^2+2(a-1)^2+3=(a-1)^2(a^2+2)+3\)
Vì \((a-1)^2\geq 0,\forall a\in\mathbb{R}; a^2+2>0, \forall a\)
\(\Rightarrow A=(a-1)^2(a^2+2)+3\geq 0+3=3\)
Vậy \(A_{\min}=3\Leftrightarrow (a-1)^2=0\Leftrightarrow a=1\)
Bài 2:
a)
\(M=3xyz+x(y^2+z^2)+y(x^2+z^2)+z(x^2+y^2)\)
\(=3xyz+x^2y+x^2z+yx^2+yz^2+zx^2+zy^2\)
\(=(x^2y+xy^2+xyz)+(y^2z+yz^2+xyz)+(zx^2+z^2x+xyz)\)
\(=xy(x+y+z)+yz(y+z+x)+xz(z+x+y)\)
\(=(x+y+z)(xy+yz+xz)\)
b)
\(Q=(a+b+c)^3-a^3-b^3-c^3\)
\(=[(a+b)+c]^3-a^3-b^3-c^3\)
\(=(a+b)^3+c^3+3(a+b)^2c+3(a+b)c^2-a^3-b^3-c^3\)
\(=a^3+b^3+3ab^2+3a^2b+c^3+3(a+b)^2c+3(a+b)c^2-a^3-b^3-c^3\)
\(=3ab(a+b)+3(a+b)c(a+b+c)\)
\(=3(a+b)[ab+c(a+b+c)]=3(a+b)[a(b+c)+c(b+c)]\)
\(=3(a+b)(b+c)(a+c)\)
a, \(3a^2b^2-6a^2b^3+3a^2b^2\)
\(=6a^2b^2-6a^2b^3=6a^2b^2\left(1-b\right)\)
b, \(a^{n+1}-2a^{n-1}=a^2.a^{n-1}-2a^{n-1}=a^{n-1}\left(a^2-2\right)\)
c, \(3a^2b\left(a+b-2\right)-4ac^2-4bc^2+8c^2\)
\(=3a^2b\left(a+b-2\right)-4c^2\left(a+b-2\right)\)
\(=\left(3a^2b-4c^2\right)\left(a+b-2\right)\)
c, \(5a^n\left(a^2-ab+1\right)-2a^2b^n+2ab^{n+1}-2b^n\)
\(=5a^n\left(a^2-ab+1\right)-2a^2b^n+2ab^n.b-2b^n\)
\(=5a^n\left(a^2-ab+1\right)-2b^n\left(a^2-ab+1\right)\)
\(=\left(5a^n-2b^n\right)\left(a^2-ab+1\right)\)