Phân tích đa thức sau thành nhân tử :
4(x^2y^2 + z^2t^2 + 2xyzt) – (x^2 + y^2 – z^2 – t^2)^2
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12 tháng 9 2021
\(4(x^2y^2+z^2t^2+2xyzt)-(x^2+y^2-z^2-t^2)^2\)
\(=[2(xy+zt]^2-(x^2+y^2-z^2-t^2)^2\)
\(=(2xy+2zt)^2-(x^2+y^2-z^2-t^2)^2\)
\(=(2xy+2zt-x^2-y^2+z^2+t^2)(2xy+2zt+x^2+y^2-z^2-t^2)^2\)
DN
3
2 tháng 9 2021
\(\left(x+y+z\right)^2+\left(x+y-z\right)^2-4z^2=\left(x+y+z\right)^2+\left(x+y-z-2z\right)\left(x+y-z+2z\right)=\left(x+y+z\right)^2+\left(x+y-3z\right)\left(x+y+z\right)=\left(x+y+z\right)\left(x+y+z+x+y-3z\right)=\left(x+y+z\right)\left(2x+2y-2z\right)=2\left(x+y+z\right)\left(x+y-z\right)\)
2 tháng 9 2021
Ta có:
(x + y + z)2 + (x + y – z)2 – 4z2
\(=\left(x+y-z\right)^2+\left(x+y-z\right)\left(x+y+3z\right)\)
\(=\left(x+y-z\right)\left(x+y+3z+x+y-z\right)\)
\(=2\left(x+y-z\right)\left(x+y+z\right)\)
TT
0
F
7 tháng 9 2020
\(a,4\left(2-x\right)^2+xy-2y\)
\(=4\left(2-x\right)^2-y\left(2-x\right)\)
\(=4-y\left(2-x\right)^2\left(2-x\right)\)
\(=\left(2-x\right)\left[\left(2-x\right)4-y\right]\)
\(=\left(2-x\right)\left(4x-8+y\right)\)
\(c,x^3+y^3+z^3-3xyz\)
\(=x^3+y^3+z^3+3x^2y-3x^2y+3xy^2-3xy^2-3xyz\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-3x^2y-3xy^2+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+1\right)+z^3-3xyz\)
\(=\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y\right)-3xyz\)
\(=\left[\left(x+y\right)+z\right]\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)
7 tháng 9 2020
a) 4(2 - x)2 + xy - 2y = 4(x - 2)2 + y(x - 2) = (4x - 8 + y)(x - 2)
b) 2(x - 1)3 - 5(x - 1)2 - (x - 1) = (x - 1)[2(x - 1)2 - 5(x - 1) - 1]
= (x - 1)(2x2 - 4x + 2 - 5x + 5 - 1) = (x - 1)(2x2 - 9x + 6)
c) x3 + y3 + z3 - 3xyz = (x + y)(x2 - xy + y2) + z3 - 3xyz
= (x + y)3 + z3 - 3xy(x + y) - 3xyz = (x + y + z)(x2 + 2xy + y2 - xz - yz + z2) - 3xy(x + y + z)
= (x + y + z)(x2 + y2 + z2 - xz - yz + 2xy - 3xy) = (x + y + z)(x2 + y2 + z2 - xy - yz - xz)
TT
0
DN
4
1 tháng 9 2021
\(x^2\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)\\ =\left(x+4\right)^2\left(x^2-1\right)-\left(x^2-1\right)\\ =\left(x^2-1\right)\left[\left(x+4\right)^2-1\right]\\ =\left(x-1\right)\left(x+1\right)\left(x+4-1\right)\left(x+4+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x+3\right)\left(x+5\right)\)
\(4\left(x^2y^2+z^2t^2+2xyzt\right)-\left(x^2+y^2-z^2-t^2\right)^2\)
\(=\left[2\left(xy+zt\right)\right]^2-\left(x^2+y^2-z^2-t^2\right)^2\)
\(=\left(2xy+2zt\right)^2-\left(x^2+y^2-z^2-t^2\right)^2\)
\(=\left(2xy+2zt-x^2-y^2+z^2+t^2\right)\left(2xy+2zt+x^2+y^2-z^2-t^2\right)^2\)
Ta có: \(4\left(x^2y^2+2xyzt+z^2t^2\right)-\left(x^2+y^2-z^2-t^2\right)^2\)
\(=\left(2xy+2tz\right)^2-\left(x^2+y^2-z^2-t^2\right)^2\)
\(=\left(2xy+2tz-x^2-y^2+z^2+t^2\right)\left(2xy+2tz+x^2+y^2-z^2-t^2\right)\)
\(=\left[-\left(x^2-2xy+y^2\right)+\left(z^2+2tz+t^2\right)\right]\left[\left(x^2+2xy+y^2\right)-\left(t^2-2tz+z^2\right)\right]\)
\(=\left(z+t-x+y\right)\left(z+t+x-y\right)\left(x+y-t+z\right)\left(x+y+t-z\right)\)