P2 tách 1 hạng tử thành nhiều hạng tử
A) x2+4xy-21y2
b) 5x2 +6 xy+y2
c) x2+4xy-21y2
D) x2-4xy+10y2
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a: \(=4xy\left(1-5x^2y\right)\)
b: \(=\left(x-y\right)\left(x+y\right)+3\left(x-y\right)=\left(x-y\right)\left(x+y+3\right)\)
c: \(=x\left(x-a\right)+y\left(x-a\right)=\left(x-a\right)\left(x+y\right)\)
d: \(=\left(x+2y\right)^2-36=\left(x+2y+6\right)\left(x+2y-6\right)\)
c) \(5x^2+3y+15x+xy=5x\left(x+3\right)+y\left(x+3\right)=\left(x+3\right)\left(5x+y\right)\)
d) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x+3-y\right)\left(x+3+y\right)\)
e) \(x^2-y^2+2x+1=\left(x^2+2x+1\right)-y^2=\left(x+1\right)^2-y^2=\left(x+1-y\right)\left(x+1+y\right)\)
f) \(x^2-2xy-9+y^2=\left(x^2-2xy+y^2\right)-9=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\)
c: \(5x^2+15x+3y+xy\)
\(=5x\left(x+3\right)+y\left(x+3\right)\)
\(=\left(x+3\right)\left(5x+y\right)\)
d: \(x^2+6x+9-y^2\)
\(=\left(x+3\right)^2-y^2\)
\(=\left(x+3-y\right)\left(x+3+y\right)\)
e: \(x^2+2x+1-y^2\)
\(=\left(x+1\right)^2-y^2\)
\(=\left(x+1-y\right)\left(x+1+y\right)\)
f: \(x^2-2xy+y^2-9\)
\(=\left(x-y\right)^2-9\)
\(=\left(x-y-3\right)\left(x-y+3\right)\)
a, \(x^4+2x^2+1-x^2\)
= \(\left(x^2+1\right)^2-x^2\)
= \(\left(x^2+x+1\right)\left(x^2-x+1\right)\)
b, \(x^4+x^2+1\)
= \(x^4+2x^2+1-x^2\)
= .. ( như phần a )
c, \(y^4+64\)
= \(\left(y^2+8\right)\left(y^2-8\right)\)
d, \(4xy+3z-12y-xz\)
\(=4y\left(x-3\right)-z\left(x-3\right)\)
\(=\left(x-3\right)\left(4y-z\right)\)
e, \(x^2-4xy+4y^2-z^2+6z-9\)
\(=\left(x-2y\right)^2-\left(z-3\right)^2\)
g, \(x^2-4xy+5x+4y^2-10y\)
\(=\left(x^2-4xy+4y^2\right)+\left(5x-10y\right)\)
\(=\left(x-2y\right)^2+5\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x-2y+5\right)\)
h, \(x^2-7x+6\)
\(=x^2-6x-x+6\)
\(=x\left(x-6\right)-\left(x-6\right)\)
\(=\left(x-6\right)\left(x-1\right)\)
i, \(x^3+5x^2+6x+2\)
\(=x^3+x^2+4x^2+4x+2x+2\)
\(=x^2\left(x+1\right)+4x\left(x+1\right)+2\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+4x+2\right)\)
a) Áp dụng HĐT 1 thu được ( 2 x + y ) 2 .
b) Áp dụng HĐT 3 với A = 2x + l; B = x - l thu được
[(2x +1) + (x -1)] [(2x +1) - (x -1)] rút gọn thành 3x(x + 2).
c) Ta có: 9 - 6x + x 2 - y 2 = ( 3 - x ) 2 - y 2 = (3 - x - y)(3 -x + y).
d) Ta có: -(x + 2) + 3( x 2 - 4) = -{x + 2) + 3(x + 2)(x - 2)
= (x + 2) [-1 + 3(x - 2)] = (x + 2)(3x - 7).
a) Cách 1.
Ta có 2xy + 3z + 6y + xz = (2xy + xz) + (3z + 6y)
= x(2 y + z)+3(z + 2 y) = (z + 2y)(x + 3).
Cách 2.
Ta có 2xy + 3z + 6y + xz = (2x1/ + 6y) + (3z + xz)
= 2y(x + 3) + z(3 + x) = (z + 2y)(x + 3).
b) Biến đổi được a 4 - 9 rt 3 + a 2 -9a = (a- 9)a( a 2 +1).
c) Biến đổi được 3 x 2 + 5y - 3xy + (-5x) = (x - y)(3x - 5).
d) Biến đổi được x 2 - (a + b)x + ab = (x- a)(x - b).
e) Ta có 4 x 2 - 4xy + y 2 – 9 t 2 = ( 2 x - y ) 2 - ( 3 t ) 2
= (2x - y - 3t )(2x - y + 31).
g) Ta có x 3 - 3 x 2 y + 3 xy 2 - y 3 - z 3
= ( x - y ) 3 - z 3 = (x - y - z)( x 2 + y 2 + z 2 - 2xy + xz - yz).
h) Ta có x 2 - y 2 + 8x + 6y+ 7 = ( x 2 +8x + 16) - ( y 2 - 6y+ 9)
= ( x + 4 ) 2 - ( y - 3 ) 2 =(x-y + 7)(x + y + l).
a, \(x^2+4xy-21y^2\)
\(=x^2-3xy+7xy-21y^2\)
\(=x\left(x-3y\right)+7y\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x+7y\right)\)
b, \(5x^2+6xy+y^2\)
\(=5x^2+5xy+xy+y^2\)
\(=5x\left(x+y\right)+y\left(x+y\right)\)
\(=\left(x+y\right)\left(5x+y\right)\)
c, \(x^2+4xy-21y^2\)
(giống câu a)
d, \(x^2-4xy+10y^2\)
(bạn xem lại đề nhá)
a, \(x^2+4xy-21y^2\)
= \(\left(x-3y\right).\left(x+7y\right)\)
b, \(5x^2+6xy+y^2\)
= \(\left(x+y\right).\left(5x+y\right)\)
c, \(x^2+4xy-21y^2\)
= \(\left(x-3y\right).\left(x+7y\right)\)
d,\(x^2-4xy+10y^2\)
= \(x^2-2xy-2xy+10y^2\)
= \(x.\left(x-2y\right)-2y.\left(x+5y\right)\)
= ....................