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\(4x^4-9x^2\)
\(=\left(2x^2\right)^2-\left(3x\right)^2\)
\(=\left(2x^2-3x\right)\left(2x^2+3x\right)\)
\(a,\left(x^2+y^2-5\right)^2-4x^2y^2-16xy-16\)
\(=\left(x^2+y^2-5\right)^2-4\left(x^2y^2-4xy-4\right)\)
\(=\left(x^2+y^2-5\right)^2-4\left(xy+2\right)^2\)
\(=\left(x^2+y^2-5\right)^2-\left[2xy+4\right]^2\)
\(=\left(x^2+y^2-5+2xy+4\right)\left(x^2+y^2-5-2xy-4\right)\)
\(=\left[\left(x^2+y^2+2xy\right)-1\right]\left[\left(x^2+y^2-2xy\right)-9\right]\)
\(=\left[\left(x+y\right)^2-1\right]\left[\left(x-y\right)^2-9\right]\)
\(=\left(x+y-1\right)\left(x+y+1\right)\left(x-y-3\right)\left(x-y+3\right)\)
\(b,x^3+5x^2+8x+4\)
\(=x^3+x^2+4x^2+8x+4\)
\(=x^2\left(x+1\right)+4\left(x^2+2x+1\right)\)
\(=x^2\left(x+1\right)+4\left(x+1\right)^2\)
\(=\left(x+1\right)\left[\left(x^2+4\right)\left(x+1\right)\right]\)
\(=\left(x+1\right)\left(x^2+4x+4\right)\)
\(=\left(x+1\right)\left(x+2\right)^2\)
\(c,x^3-6x^2-x+30\)
\(=x^3-5x^2-x^2+5x-6x+30\)
\(=x^2\left(x-5\right)-x\left(x-5\right)-6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2-x-6\right)\)
\(=\left(x-5\right)\left[x^2+2x-3x-6\right]\)
\(=\left(x-5\right)\left[x\left(x+2\right)-3\left(x+2\right)\right]\)
\(=\left(x-5\right)\left(x-3\right)\left(x+3\right)\)
\(d,125x^3-10x^2+2x-1\)
\(=\left(125x^3-1\right)-\left(10x^2-2x\right)\)
\(=\left(5x-1\right)\left(25x^2+5x+1\right)-2x\left(5x-1\right)\)
\(=\left(5x-1\right)\left(25x^2+5x+1-2x\right)\)
\(=\left(5x-1\right)\left(25x^2+3x+1\right)\)
1) \(x^2-10x+25=\left(x-5\right)^2\)
2) \(1-2xy+y^2=\left(1-y\right)^2\)
3) \(4x^2-8x+4=\left(2x-2\right)^2\)
4) \(x^2-16xy+64y^2=\left(x-8y\right)^2\)
\(-8x^2y^2-12xy^3-4xy^2\)
\(=-8x^2y^2-8xy^3-4xy^3-4xy\)
\(=-8xy\left(xy-y^2\right)-4xy\left(y^2-1\right)\)
\(=-8xy\left(y\left(x-y\right)\right)-4xy\left(y-1\right)\left(y+1\right)\)
\(=-4.2xy\left(y\left(x-y\right)\right)-4xy\left(y-1\right)\left(y+1\right)\)
\(=-4\left(2xy\left(y\left(x-y\right)\right)-xy\left(y-1\right)\left(y+1\right)\right)\)
Vậy thôi thành nhân tử là dc rồi
Ủng hộ nha
Thanks
a, = (x + y)5 - (x5 + y5)
= (x + y)5 - (x + y)(x4 - x3y + x2y2 - xy3 + y4)
= (x + y) [(x + y)4 - x4 + x3y - x2y2 + xy3 - y4]
= (x + y) (5x3y + 5x2y2 + 5xy3)
= 5xy(x + y)(x2 + xy + y2)
b, = x(x2 - 5xy - 14y2)
= x(x2 - 7xy + 2xy - 14y2)
= x(x + 2y)(x - 7y)
1.\(x^3+6x^2+12xy+8=x^3+3.2x^2+3.2^2x+2^3=\left(x+2\right)^3\)
3.\(x^4+2x^3+x^2-y^2=\left(x^2\right)^2+2x^2.x+x^2-y^2\)\(=\left(x^2+x\right)^2-y^2=\left(x^2+x-y\right)\left(x^2+x+y\right)\)
k mình nha bn !!!!!!! cái 2 bn xem lại đề đi, rồi mình giải cho
Bài làm
a) 2x2y - 4xy2 + 6xy
= 2xy( x - 2y + 3 )
b) 4x3y2 - 8x2y3 + 2x4y
= 2x2y( 2xy - 4y2 + x2 )
c) 9x2y3 - 3x4y2 - 6x3y2 + 18y4
= 3y2( 3x2y - x4 - 2x3 + 6y2 )
d) 7x2y2 - 21xy2z + 7xyz - 14xy
= 7xy( xy - 3yz + z - 2 )
# Học tốt #
\(=-\dfrac{6}{7}y-\dfrac{4}{7}xy+\dfrac{8}{7}\)