Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Xet ve phai :x^3+y^3+3x^2y+3xy^2-3x^2y-3xy^2+z^3
<=>x^3+y^3+z^3=ve trai
Xong
cho x,y,z nguyên và (x-y)*(y-z)*(z-x)=m. Chứng minh rằng: (x-y)^3 + (y-z)^3 + (z-x)^3 chia hết cho m
Một bài toán "lừa" người ta:
Đặt \(a=x-y,b=y-z,c=z-x\Rightarrow a+b+c=0\).
Ta có hằng đẳng thức \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\).
Trong trường hợp này thì \(a+b+c=0\) nên suy ra đpcm.
\(\left(x+y+z\right)^3-x^3-y^3-z^3\\ =x^3+y^3+z^3-x^3-y^3-z^3+3\left(x+y\right)\left(y+z\right)\left(z+x\right)\\ =3\left(x+y\right)\left(y+z\right)\left(z+x\right)\:\left(đpcm\right)\)
• \(VT=\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+3z\left(x+y\right)^2+3\left(x+y\right)z^2+z^3-x^3-y^3-z^3\)
\(=x^3+3x^2y+3xy^2+y^3+3z+\left(x+y\right)^2+3xz^2+3yz^2-x^3-y^3\)
\(=3x^2y+3xy^2+3z\left(x^2+2xy+y^2\right)+3xz^2+3yz^2\)
\(=3x^2y+3xy^2+3x^2z+6xyz+3y^2z+3xz^2+3yz^2\) (1)
• \(VP=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
\(=\left(3x+3y\right)\left(y+z\right)\left(z+x\right)\)
\(=\left(3xy+3xz+3y^2+3yz\right)\left(z+x\right)\)
\(=3xyz+3x^2y+3xz^2+3x^2z+3y^2z+3xy^2+3yz^2+3xyz\)
\(=6xyz+3x^2y+3xz^2+3x^2z+3y^2z+3xy^2+3yz^2\) (2)
Từ (1) và (2) suy ra \(VT=VP\) (đpcm)
a, x^4 - 5x^2 + 4
= x^4 - 4x^2- x+ 4
= x^2 . (x^2 - 4) - (x^2 - 4)
= (x^2 - 4) . (x^2 - 1)
= (x - 2) . (x + 2) . (x - 1) . (x + 1)
\(\left(x+y+z\right)^3=\left[\left(x+y\right)+z\right]^3=\left(x+y\right)^3+z^3+3\left(x+y\right)z\left(x+y+z\right)\)
\(=x^3+y^3+3xy\left(x+y\right)+c^3+3\left(x+y\right)z\left(x+y+z\right)\)
\(=x^3+y^3+z^3+3\left(x+y\right)\left[xy+z\left(x+y+z\right)\right]\)
\(=x^3+y^3+z^3+3\left(x+y\right)\left(xy+zx+zy+z^2\right)\)
\(=x^3+y^3+z^3+3\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)
\(=x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(z+x\right)\Rightarrow\left(dpcm\right)\)
Chúc bạn học tốt
T I C K nha cảm ơn bạn
x+y+z=6 = 1+2+3 <=> (x-1) +(y-2) +(z-3) = 0
mũ 3 lên ra pt cần CM
Đặt: x - 1 = a; y - 2 = b; c - 3 = z
=> a + b + c = 0
=> a + b = - c
=> (a + b)3 = - c3
a3 + b3 + c3
= a3 + b3 - (a + b)3
= a3 + b3 - a3 - 3ab(a + b) - b3
= - 3ab(a + b) = - 3ab(-c) = 3abc
Thay trở lại đc:
\(\left(x-1\right)^3+\left(y-2\right)^3+\left(z-3\right)^3=3\left(x-1\right)\left(y-2\right)\left(z-3\right)\)
\(VT=\left(x+y+z\right)^3-x^2-y^3-z^3\)
\(=x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(z+x\right)-x^3-y^3-z^3\)
\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)=VP\)
=> đpcm
=.= hok tốt!!
Đặt: \(A=\left(x+y+z\right)^3-x^3-y^3-z^3=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
Xét: \(\left(x+y+z\right)^3=\left[\left(x+y\right)+z\right]^3=\left(x+y\right)^3+z^3+3z\left(x+y\right)\left(x+y+z\right)\)
\(=x^3+y^3+3xy\left(x+y\right)+z^3+3z\left(x+y\right)\left(x+y+z\right)\)
\(=\left(x^3+y^3+z^3\right)+3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)
\(=\left(x^3+y^3+z^3\right)+3\left(x+y\right)\left[\left(xy+yz\right)+\left(xz+z^2\right)\right]\)
\(=\left(x^3+y^3+z^3\right)+3\left(x+y\right)\left[y\left(x+z\right)+z\left(x+z\right)\right]\)
\(=\left(x^3+y^3+z^3\right)+3\left(x+y\right)\left(x+z\right)\left(y+z\right)\)
=> ĐPCM
(x + y + z)3 = (x + y)3 + z3 + 3z(x + y)(x + y + z)
= x3 + y3 + z3 + 3xy(x + y) + 3z(x + y)(x + y + z)
= x3 + y3 + z3 + 3(x + y)[xy + z(x + y + z)]
= x3 + y3 + z3 + 3(x + y)(xy + xz + yz + z2)
= x3 + y3 + z3 + 3(x + y)[x(y + z) + z(y + z)]
= x3 + y3 + z3 + 3(x + y)(y + z)(x + z)
a³ - b³ = (a - b)³ + 3ab(a - b)
a³ + b³ = ( a + b )³ - 3ab( a + b )
= [ ( x + y + z )³ - x³ ] - ( y³ + z³ ) = ( y + z )³ + 3x( x + y + z )( y + z ) - [ ( y + z )³ - 3yz( y + z ) ]
= ( y + z )³ + 3x( x + y + z )( y + z ) - ( y + z )³ + 3yz( y + z )
= 3( y + z ) [ x ( x + y + z ) + yz ] = 3( y + z ) [ x ( x + y ) + xz + yz ) ]
= 3( y + z )[ x ( x + y ) + z( x + y ) ]
= 3( y + z ) ( x + y )( z + x )