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4a2b2 + 36a2b3 + 6ab4
= 2ab2(2a + 18ab + 3b2)
4a2b3 - 6a3b2
= 2a2b2(2b - 3a)
\(3,\)Nhẩm nghiệm của đa thức trên ta đc : -1
Ta có lược đồ sau :
| 1 | 1 | -4 | -4 | |
| -1 | 1 | 0 | -4 | 0 |
Phân tích thành nhân tử ta có :\(\left(x+1\right)\left(x^2-4\right)\)
e) Ta có: \(a^3-a^2-a+1\)
\(=a^2\left(a-1\right)-\left(a-1\right)\)
\(=\left(a-1\right)\left(a^2-1\right)\)
\(=\left(a-1\right)^2\cdot\left(a+1\right)\)
f) Ta có: \(x^3-2xy-x^2y+2y^2\)
\(=x^2\left(x-y\right)-2y\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2-2y\right)\)
a) \(\left(a^2+b^2\right)^2-4a^2b^2=\left(a^2+b^2+2ab\right)\left(a^2+b^2-2ab\right)=\left(a+b\right)^2.\left(a-b\right)^2\)
b) \(3x^2-3xy-5x+5y=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)
c) \(-x^3+3x^2-3x+1=\left(1-x\right)^3\)
d) Đề sai ko ???
e) \(a^3-a^2-a+1=a^2\left(a-1\right)-\left(a-1\right)=\left(a-1\right)\left(a^2-1\right)=\left(a-1\right)^2\left(a+1\right)\)
f) \(x^3-2xy-x^2y+2y^2=x^2\left(x-y\right)-2y\left(x-y\right)=\left(x-y\right)\left(x^2-2y\right)\)
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)
\(=ab^3-ac^3+bc^3-a^3b+a^3c-b^3c\)
\(=a^3\left(c-b\right)+bc\left(c-b\right)\left(c+b\right)-a\left(c-b\right)\left(c^2+bc+b^2\right)\)
\(=a^3\left(c-b\right)+\left(c-b\right)\left(bc^2+b^2c\right)-\left(c-b\right)\left(ac^2+abc+ab^2\right)\)
\(=\left(c-b\right)\left(a^3+bc^2+b^2c-ac^2-abc-ab^2\right)\)
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)
\(=ab^3-ac^3+bc^3-a^3b+a^3c-b^3c\)
\(=a^3\left(c-b\right)+bc\left(c-b\right)\left(c+b\right)-a\left(c-b\right)\left(c^2+bc+b^2\right)\)
\(=a^3\left(c-b\right)+\left(c-b\right)\left(bc^2+b^2c\right)-\left(c-b\right)\left(ac^2+abc+ab^2\right)\)
\(=\left(c-b\right)\left(a^3+bc^2+b^2c-ac^2-abc-ab^2\right)\)
a)\(a^2+6a+8-b^2-2b=\left(a+3\right)^2-\left(b+1\right)^2=\left(a+3+b+1\right)\left(a+3-b-1\right)\)
\(=\left(a+b+4\right)\left(a-b+2\right)\)
b)\(a^2+6ax+8x^2-b^2-2bx\)
\(=\left(a+3x\right)^2-\left(b+x\right)^2\)
\(=\left(a+3x-b-x\right)\left(a+3x+b+x\right)=\left(a-b+2x\right)\left(a+b+4x\right)\)
a, 4a^2b^3 - 6a^3b^2 = 2a^2b^2(2b - 3a)
b, 5(a + b) +x( a + b ) = ( 5 + x )( a + b )
c, (a - b)^2 - ( b - a ) = ( a - b )^2 + ( a - b ) = (a - b) ( a - b + 1)