Phân tích đa thức thành nhân từ bằng phương pháp dùng hàng đẳng thức:
a) x^2 - 2xy + y^2 - 4m^2 + 4mn - n^2
b) x^2 - 4x^2y^2 + y^2 + 2xy
c) x6 - y6
d) 25 - a^2 + 2ab - b^2
e)4b^2c^2 - (b^2+c^2-a^2)^2
f) (a+b+c)^2 + (a-b+c)^2 - 4c^2
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a) Ta có: \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2\)
\(=\left(4x^2-3x-18-4x^2-3x\right)\left(4x^2-3x-18+4x^2+3x\right)\)
\(=\left(-6x-18\right)\left(8x^2-18\right)\)
\(=-6\left(x+3\right)\cdot2\left(4x^2-9\right)\)
\(=-12\left(x+3\right)\left(2x-3\right)\left(2x+3\right)\)
b) Ta có: \(9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)
\(=\left(3x+3y-3\right)^2-\left(4x+6y+2\right)^2\)
\(=\left(3x+3y-3-4x-6y-2\right)\left(3x+3y-3+4x+6y+2\right)\)
\(=-\left(x+3y+5\right)\left(7x+9y-1\right)\)
c) Ta có: \(-4x^2+12xy-9y^2+25\)
\(=-\left(4x^2-12xy+9y^2-25\right)\)
\(=-\left[\left(2x-3y\right)^2-25\right]\)
\(=-\left(2x-3y-5\right)\left(2x-3y+5\right)\)
d) Ta có: \(x^2-2xy+y^2-4m^2+4mn-n^2\)
\(=\left(x^2-2xy+y^2\right)-\left(4m^2-4mn+n^2\right)\)
\(=\left(x-y\right)^2-\left(2m-n\right)^2\)
\(=\left(x-y-2m+n\right)\left(x-y+2m-n\right)\)
a) \(x^4-y^4\)
\(=\left(x^2\right)^2-\left(y^2\right)^2\)
\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\)
\(=\left(x+y\right)\left(x-y\right)\left(x^2+y^2\right)\)
b) \(x^2-3y^2\)
\(=x^2-\left(y\sqrt{3}\right)^2\)
\(=\left(x-y\sqrt{3}\right)\left(x+y\sqrt{3}\right)\)
c) \(\left(3x-2y\right)^2-\left(2x-3y\right)^2\)
\(=\left(3x-2y+2x-3y\right)\left(3x-2y-2x+3y\right)\)
\(=\left(5x-5y\right)\left(x+y\right)\)
\(=5\left(x-y\right)\left(x+y\right)\)
d) \(9\left(x-y\right)^2-4\left(x+y\right)^2\)
\(=\left[3\left(x-y\right)+2\left(x+y\right)\right]\left[3\left(x-y\right)-2\left(x+y\right)\right]\)
\(=\left(3x-3y+2x+2y\right)\left(3x-3y-2x-2y\right)\)
\(=\left(5x-y\right)\left(x-5y\right)\)
e) \(\left(4x^2-4x+1\right)-\left(x+1\right)^2\)
\(=\left(2x-1\right)^2-\left(x+1\right)\)
\(=\left(2x-1+x+1\right)\left(2x-1-x-1\right)\)
\(=3x\left(x-2\right)\)
f) \(x^3+27\)
\(=x^3+3^3\)
\(=\left(x+3\right)\left(x^2-3x+9\right)\)
g) \(27x^3-0,001\)
\(=\left(3x\right)^3-\left(0,1\right)^3\)
\(=\left(3x-0,1\right)\left(9x^2+0,3x+0,01\right)\)
h) \(125x^3-1\)
\(=\left(5x\right)^3-1^3\)
\(=\left(5x-1\right)\left(25x^2+5x+1\right)\)
a: =x^2+2xy+y^2-4x^2y^2
=(x+y)^2-(2xy)^2
=(x+y+2xy)(x+y-2xy)
b: =49-(a^2-2ab+b^2)
=49-(a-b)^2
=(7-a+b)(7+a-b)
c: =\(a^2-\left(b^2-4bc+4c^2\right)\)
\(=a^2-\left(b-2c\right)^2=\left(a-b+2c\right)\left(a+b-2c\right)\)
d:
\(=\left(bc\right)^2-\left(b^2+c^2-a^2\right)^2\)
\(=\left(bc-b^2-c^2+a^2\right)\left(bc+b^2+c^2-a^2\right)\)
e: \(=\left(a+b\right)^2+2c\left(a+b\right)+c^2+\left(a+b\right)^2-2c\left(a+b\right)+c^2-4c^2\)
=2(a+b)^2-2c^2
=2[(a+b)^2-c^2]
=2(a+b-c)(a+b+c)
1) x^2-4x^2y^2+y^2+2xy
=x2+2xy+y2-4x2y2
=(x+y)2-4x2y2
=(x+2xy+y)(x-2xy+y)
2) 25-a^2+2ab-b^2
=25-(a2-2ab+b2)
=25-(a-b)2
=[5-(a-b)][5+(a-b)]
=(5-a+b)(5+a-b)
Bổ sung nha :
(x - y)2 - (2m - n)2
= (x - y -2m - n) . (x - y + 2m - n) .
Chúc bạn học tốt !
x2-2xy+y2-4m2+4mn-n2 mới đúng tui giải cho
<=> (x-y)2-(4m-n)2< Áp dụng hằng đẳng thức số 2 >
<=> (x-y-4m-2).(x-y+4m-2) < HĐT số 3 >
a) x2 - 2xy + y2 - 4m2 + 4mn - n2 = (x - y)2 - [(2m)2 - 2.2m.n + n2] = (x - y)2 - (2m - n)2
= [(x - y) - (2m - n)][(x - y) + (2m - n)] = (x - y - 2m + n)(x - y + 2m - n)
b) x2 - 4x2y2 + y2 + 2xy = x2 + 2xy + y2 - 4x2y2 = (x + y)2 - (2xy)2 = (x + y - 2xy)(x + y + 2xy)
c) x6 - y6 = (x3)2 - (y3)2 = (x3 - y3)(x3 + y3) = (x - y)(x2 + xy + y2)(x + y)(x2 - xy - y2)
d) 25 - a2 + 2ab - b2 = 25 - (a2 - 2ab + b2) = 52 - (a - b)2 = (5 - a + b)(5 + a - b)
xin lỗi các bạn, đề mink có vấn đề: ý c phải là: x^6 - y^6