Phân tích đa thức thành nhân tử:a,27a^2b^2-18ab+3
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a, \(\left(y-2\right)\left(y+2\right)\left(y^2+4\right)-\left(y+3\right)\left(y-3\right)\left(y^2+9\right)\)
\(=\left(y^2-4\right)\left(y^2+4\right)-\left(y^2-9\right)\left(y^2+9\right)\)
\(=y^4-16-y^4+81=65\)
b, \(2\left(x^2-xy+y^2\right)\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)-2\left(x^6-y^6\right)\)
\(=2\left(x^3-y^3\right)\left(x^3+y^3\right)-2\left(x^6-y^6\right)\)
\(=2\left(x^6-y^6\right)-2\left(x^6-y^6\right)=0\)
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1) a) 2a + 2b = 2(a + b) ;
b) 9a - 9b = 9(a - b) ;
c) 3a - 6b - 9c = 3(a - 2b - 3c)
d) ab - ac = a(b - c)
e) 5a - 10ax - 15a = -10a - 10ax = 10a(x + 1)
f) 3a(ax - 2ay + 4)
g) 5a2(x - y) + 10a(x - y)
= 5a(x - y)(a + 2)
2) ax + ay + bx + by
= a(x + y) + b(x + y)
= (a + b)(x + y)
b) a2 - 49 = (a - 7)(a + 7)
c) 9a2 - 1 = (3a - 1)(3a + 1)
d) \(\frac{1}{4}a^2-b^2=\left(\frac{1}{2}a-b\right)\left(\frac{1}{2}a+b\right)\)
e) x2 + 14x + 49 = (x + 7)2
f) 4x2 + 20x + 25 = (2x + 5)2
g) 4x4 + 20x2 + 25 = (2x2 + 5)2
h) 2x3 + 8x2 + 8x = 2x(x2 + 4x + 4) = 2x(x + 2)2
i) 2x3 + 16 = 2(x3 + 8) = 2(x + 2)(x2 - 2x + 4)
3) x2 + 4x + 3 = x2 + x + 3x + 3 = x(x + 1) + 3(x + 1) = (x + 1)(x + 3)
x2 + 7x + 10 = x2 + 2x + 5x + 10 = x(x + 2) + 5(x + 2) = (x + 2)(x + 5)
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ta có 2(x^2+x+1)/x^2+1
=2x^2+2x+2/x^2+1
=1+(x+1)^2/x^2+1>=1 với mọi x
dấu bằng xảy ra khi x=-1
bạn tự kết luận nha
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![](https://rs.olm.vn/images/avt/0.png?1311)
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Trả lời:
\(27a^2b^2-18ab+3\)
\(=3\left(9a^2b^2-6ab+1\right)\)
\(=3\left[\left(3ab\right)^2-2.3ab.1+1^2\right]\)
\(=3\left(3ab-1\right)^2\)