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\(a,=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\\ b,=x^2-5x+3x-15=\left(x-5\right)\left(x+3\right)\\ c,=5x^2y^3\left(1-5xy+2x\right)\\ d,=6y\left(2x^2-3xy-10y\right)\\ e,,=\left(x-y\right)\left(5-x\right)\\ f,=\left(2x+3\right)^2\)
1, x^3-2x^2+x
=x^3-x^2-x^2+x
=(x^3-x^2)-(x^2-x)
= x^2(x-1)-x(x-1)
=(x-1)(x^2-x)
=x(x-1)(x-1)
2, x^2-2x-15
=x^2-2x-9-6
= x^2-9-2x-6
=(x^2-9)-(2x+6)
=(x-3)(x+3)-2(x+3)
=(x+3)(x-3-2)=(x+3)(x-5)
3 , \(^{3x^3y^2-6x^2y^3+9x^2y^2}\)
= \(^{3x^2y^2\left(x-2y+3\right)}\)
4, \(^{5x^2y^3-25x^3y^4+10x^3y^3}\)
=\(^{5x^2y^2\left(y-5xy^2+2xy\right)}\)
5, \(^{12x^2y-18xy^2-30y^2}\)
=\(^{3y\left(4x^2-6xy-10y\right)}\)
Lời giải:
1. $x^3-2x^2+x=x(x^2-2x+1)=x(x-1)^2$
2. $x^2-2x-15=(x^2+3x)-(5x+15)=x(x+3)-5(x+3)=(x+3)(x-5)$
3. $3x^3y^2-6x^2y^3+9x^2y^2=3x^2y^2(x-2y+3)$
4. $5x^2y^3-25x^3y^4+10x^3y^3=5x^2y^3(1-5xy+2x)$
5. $12x^2y-18xy^2-30y^2=6y(2x^2-3xy-5y)$
Bạn cần viết đề bằng công thức toán để được hỗ trợ tốt hơn.
b: Ta có: \(xy-3x-2y+6\)
\(=x\left(y-3\right)-2\left(y-3\right)\)
\(=\left(y-3\right)\left(x-2\right)\)
a. 3xy( 4x + y - \(\dfrac{4}{3}\) )
b. 2x2( 3x + 1 )
c. (2x + 3 )( x - y )
d. xy( 1 - x )( x - 1 )
e. 6( 2x + 1 )( x + y )
\(8x^3+12x^2y+6xy^2+y^3-z^3\)
\(=\left(2x+y\right)^3-z^3\)
\(=\left(2x+y-z\right)\left[4x^2+z\left(2x+y\right)+z^2\right]\)
a, 8a3 - 36a2 +54ab2 - 27b3
=(8a3-36a2b +54ab2 - 27b3)
=(2a-3b)2
=(2a-3b)(2a-3b)(2a-3b)
b, 8x3 + 12x2y + 6xy2 + y3 - z 3
=(8x3 + 12x2y + 6xy2 + y3) - z3
=(2x + y)3 - y3
=(2x + y +z) . [ (2x + Y)2 + 2(2x + y)+ z2
= (2x + y + z)(4x2 + 4xy + y2 + 4x + 2y + z2
a) Áp dụng HĐT 5 thu được ( 2 a - 3 b ) 3 .
b) Ta có 8 x 3 + 12 x 2 y + 6 xy 2 + y 3 = ( 2 x + y ) 3 .
Áp dụng HĐT 7 với A = 2x + y; B = z
( 2 x + y ) 3 - z 3 = (2x + y - z)(4 x 2 + y 2 + z 2 + 4xy + 2xz + zy).
Ta có
8 x 3 + 12 x 2 y + 6 x y 2 + y 3 = ( 2 x ) 3 + 3 . ( 2 x ) 2 y + 3 . 2 x . y 2 + y 3 = ( 2 x + y ) 3
Đáp án cần chọn là: B
a: \(=x^2\left(2x+3\right)+\left(2x+3\right)\)
\(=\left(2x+3\right)\left(x^2+1\right)\)
b: \(=\left(x-4\right)\left(x+3\right)\)
e: =(x+3)(x-2)
a) \(=x^2\left(2x+3\right)+\left(2x+3\right)=\left(2x+3\right)\left(x^2+1\right)\)
b) \(=x\left(x-4\right)+3\left(x-4\right)=\left(x-4\right)\left(x+3\right)\)
c) \(=\left(2x\right)^2-\left(x^2+1\right)^2=\left(x^2-2x+1\right)\left(x^2+2x+1\right)=\left(x-1\right)^2\left(x+1\right)^2\)
d) \(=4xy\left(y-3x+2\right)\)
e) \(=x\left(x-2\right)+3\left(x-2\right)=\left(x-2\right)\left(x+3\right)\)
f) \(=x\left(x^2+2xy+y^2-4z^2\right)=x\left[\left(x+y\right)^2-4z^2\right]=x\left(x+y-2z\right)\left(x+y+2z\right)\)
g) \(=x\left(x^2-2xy+y^2-25\right)=x\left[\left(x-y\right)^2-25\right]=x\left(x-y-5\right)\left(x-y+5\right)\)
h) \(=x\left(x+1\right)-3\left(x+1\right)=\left(x+1\right)\left(x-3\right)\)
i) \(=x^2\left(x-3\right)-9\left(x-3\right)=\left(x-3\right)\left(x^2-9\right)=\left(x-3\right)^2\left(x+3\right)\)