phân tích thành nhân tử: x3+y3+3xy
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x 3 + y 3 + z 3 – 3xyz = x + y 3 – 3xy(x + y) + z 3 – 3xyz
= [ x + y 3 + z 3 ] - [ 3xy.(x+ y) + 3xyz]
= [ x + y 3 + z 3 ] – 3xy(x + y + z)
= (x + y + z)[ x + y 2 – (x + y)z + z 2 ] – 3xy(x + y + z)
= (x + y + z)( x 2 + 2xy + y 2 – xz – yz + z 2 – 3xy)
= (x + y + z)( x 2 + y 2 + z 2 – xy – xz - yz)
![](https://rs.olm.vn/images/avt/0.png?1311)
x 3 – x + y 3 – y
= ( x 3 + y 3 ) − (x + y)
= (x + y)( x 2 – xy + y 2 ) − (x + y)
= (x + y)( x 2 – xy + y 2 − 1)
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x 3 - x + 3 x 2 y + 3 x y 2 + y 3 - y = x 3 + 3 x 2 y + 3 x y 2 + y 3 - x - y = x + y 3 - x - y = x + y x + y 2 - 1 = x + y x + y + 1 x + y - 1
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\(x^3+y^3+3y^2+3y+1\\ =x^3+\left(y+1\right)^3\\ =\left(x+y+1\right)\left[x^2-x\left(y+1\right)+\left(y+1\right)^2\right]\\ =\left(x+y+1\right)\left(x^2-xy-x+y^2+2y+1\right)\\ =\left(x+y+1\right)\left(x^2+y^2+2y+1-xy-x\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(Sửa:x^3+y^3+2x^2+2xy\\ =\left(x+y\right)\left(x^2-xy+y^2\right)+2x\left(x+y\right)\\ =\left(x+y\right)\left(x^2-xy+y^2+2x\right)\)
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\(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)
\(x^3-x+3x^2+3xy^2+y^3-y\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y+1\right)\left(x+y-1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x+y-z\right)^3-x^3-y^3+z^3\)
\(=\left[\left(x+y\right)-z\right]^3-x^3-y^3+z^3\)
\(=\left(x+y\right)^3-z^3-3\left(x+y\right)z\left(x+y-z\right)-x^3-y^3+z^3\)
\(=x^3+y^3-z^3+3xy\left(x+y\right)-3\left(x+y\right)z\left(x+y-z\right)-x^3-y^3+z^3\)
\(=3xy\left(x+y\right)-3z\left(x+y\right)\left(x+y-z\right)\)
\(=3\left(x+y\right)\left[xy-z\left(x+y-z\right)\right]\)
\(=3\left(x+y\right)\left(xy-zx-yz+z^2\right)\)
\(=3\left(x+y\right)\left[x\left(y-z\right)-z\left(y-z\right)\right]\)
\(=3\left(x+y\right)\left(y-z\right)\left(x-z\right)\)
#\(Urushi\text{☕}\)
Áp dụng (a+b)3 = a3+b3+3ab(a+b), ta có:
(x+y+z)3-x3-y3-z3
=[(x+y)+z]3-x3-y3-z3
=(x+y)3+z3+3z(x+y)(x+y+z)-x3-y3-z3
=x3+y3+3xy(x+y)+z3+3z(x+y)(x+y+z)-x3-y3-z3
=3(x+y)(xy+xz+yz+z2)
=3(x+y)[x(y+z)+z(y+z)]
=3(x+y)(y+z)(x+z)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn phải vt thêm dấu mũ vào mới giải đc chứ!! Để thế kia ai mà giải đc
vô đây Giúp tôi giải toán - Hỏi đáp, thảo luận về toán học - Học toán với OnlineMath
http://olm.vn/hoi-dap/question/163953.html bn vô đay tham khảo nhé