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x3 + y3 + z3 - 3xyz
= [ (x3+y3)+z3] -3xyz
=[ (x+y)3 +z3 +3xy (x+y) ] -3xyz
=[(x+y)3 +z3] +3x2y+3xy2-3xyz
=(x+y+z)[(x+y)2 -z(x+y)+z2] -3xy(x+y+z)
=(x+y+z) [x2+2xy +y2 - xz-yz +z2 -3xy ]
=(x+y+z) (x2+y2+z2 -xy-yz-zx)
x3 + y3 + z3 - 3xyz = (x + y)3 – 3x2y – 3xy2 + z3 – 3xyz
= (x + y)3 + z3 – 3x2y – 3xy2 - 3xyz
= (x + y +z)[(x + y)2 – (x + y)z + z2)] - 3xy(x + y + z)
= (x + y + z)(x2 +2xy + y2 – xz – yz +z2 – 3xy)
= (x + y + z)(x2 + y2 +z2 – xy - yz – xz)
\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3x^2y-3xy^2+z^3-3xyz\)
\(=\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+2xy-xz-yz+z^2-3xyz\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xz-yz-xy\right)\)
= (x + y)3 + z3 – 3x2y – 3xy2 - 3xyz
= (x + y +z)[(x + y)2 – (x + y)z + z2)] - 3xy(x + y + z)
= (x + y + z)(x2 +2xy + y2 – xz – yz +z2 – 3xy)
= (x + y + z)(x2 + y2 +z2 – xy - yz – xz)
x3 - y3 - z3 +3xyz
= (x3 - 3x2y +3xy2 -y^3) +3x2y-3xy2 - z3 +3xyz
= [(x-y)3 -z3] + 3x2y -3xy2 +3xyz
= (x-y-z)(x2 + 2xy+y2 +zx+zy + z2) + 3xy( x-y+z)
x3−y3−z3+3xyz=(x+y+z)(xy+yz+xz−x2−y2−z2) =-(x^3+y^3+y^3-3xyz)$
Ta tính x3+y2+z3−3xyz trước
ta có:
x3+y3+z3−3xyz=(x+y)3+z3−3xy(x+y)−3xyz=(x+y+z)(x2+y2+z2−xy−yz−xz)
=>x3−y3−z3+3xyz=(x+y+z)(xy+zy+xz−x2−y2−z2)
a) Ta có:
x³ + y³ + z³ - 3xyz = (x+y)³ - 3xy(x-y) + z³ - 3xyz
= [(x+y)³ + z³] - 3xy(x+y+z)
= (x+y+z)³ - 3z(x+y)(x+y+z) - 3xy(x-y-z)
= (x+y+z)[(x+y+z)² - 3z(x+y) - 3xy]
= (x+y+z)(x² + y² + z² + 2xy + 2xz + 2yz - 3xz - 3yz - 3xy)
= (x+y+z)(x² + y² + z² - xy - xz - yz).
\(a,x^3+y^3+z^3-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2+2xy-xz-yz\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
b) \(f\left(x\right)=12x^3-32x^2+25x-6\)
Thấy \(x=\frac{3}{2}\) là một nghiệm.Vậy đa thức có chứa nhân tử \(\left(x-\frac{3}{2}\right)\)
Ta có: \(f\left(x\right)=\left(x-\frac{3}{2}\right)\left(\frac{12x^3-32x^2+25x-6}{x-\frac{3}{2}}\right)\)
\(=\left(x-\frac{3}{2}\right)\left(12x^2-14x+4\right)\)
\(=\left(x-\frac{3}{2}\right)\left[\left(12x^2-6x\right)-\left(8x-4\right)\right]\)
\(=\left(x-\frac{3}{2}\right)\left(x-\frac{2}{3}\right)\left(12x-6\right)\)
\(=12\left(x-\frac{3}{2}\right)\left(x-\frac{2}{3}\right)\left(x-\frac{1}{2}\right)\)
\(=\left(x^3+y^3\right)+z^3-3xyz=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz\)
\(=\left(x+y+z\right)^3-3z\left(x+y\right)\left(x+y+z\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(\left(x+y+z\right)^2-3z\left(x+y\right)-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy+yz+zx\right)\)
\(x^3+y^3+z^3-3xyz=\left(x^3+y^3\right)-3xyz+z^3\)
\(=\left(x+y\right)^3-3xy.\left(x+y\right)-3xyz+z^3\)
\(=\left[\left(x+y\right)^3+z^3\right]-\left[3xy.\left(x+y\right)+3xyz\right]\)
\(=\left(x+y+z\right).\left(x^2+2xy+y^2-zx-zy+z^2\right)-3xy.\left(x+y+z\right)\)
\(=\left(x+y+z\right).\left(x^2+y^2+z^2-zx-zy+2zy-3xy\right)\)
\(=\left(x+y+z\right).\left(x^2+z^2+y^2-zx-zy-xy\right)\)
Vừa làm xong . Chúc bạn học tốt !
\(=\left(x+y\right)^3+z^z-3x^2y-3xy^2-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
(x+y+z)(x²+y²+z²-xy-yz-xz)
Có thể xem trên wikipedia với từ khóa 7 HDT đáng nhớ mục liên quan nhé bạn
\(x^3+y^3+z^3-3xyz\)
\(=x^3+y^3+z^3-xyz-xyz-xyz\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
x3 + y3 + z3 - 3xyz
= [ (x3+y3)+z3] -3xyz
=[ (x+y)3 +z3 +3xy (x+y) ] -3xyz
=[(x+y)3 +z3] +3x2y+3xy2-3xyz
=(x+y+z)[(x+y)2 -z(x+y)+z2] -3xy(x+y+z)
=(x+y+z) [x2+2xy +y2 - xz-yz +z2 -3xy ]
=(x+y+z) (x2+y2+z2 -xy-yz-zx)