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Áp dụng \(\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\)
\(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+z^3+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3-z^3\)
\(=x^3+y^3+3xy\left(x+y\right)+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3\)
\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)
\(=3\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)
\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
2) =((x+y)+z)^3-x^3-y^3-z^3
=(x+y)^3+3(x+y)^2z +3(x+y)z^2+z^3-x^3-y^3-z^3
=x^3+y^3+3xy(x+y)+3(x+y)^2z+3(x+y)z^2-x^3-y^3
=3xy(x+y)+3(x+y)^2z+3(x+y)z^2
=3(x+y)(xy+(x+y)z+z^2)
=3(x+y)(xy+xz+yz+z^2)
=3(x+y)(x(y+z)+z(y+z))
=3(x+y)(y+z)(x+z)
1) a^3-3a^2b+3ab^2-b^3+b^3-3b^2c+3bc^2-c^3+c^3-3c^2a+3ca^2-a^3
= -3(a^2b-ab^2+b^2c-bc^2+c^2a-ca^2)
=-3(ab(a-b)+c(b^2-a^2)-c^2(b-a))
= -3(ab(a-b)-c(a+b)(a-b)+c^2(a-b))
= -3(a-b)(ab-ac-bc+c^2)
= -3(a-b)(a(b-c)-c(b-c))
= -3(a-b)(b-c)(a-c)
( x + y + z )3 - x3 - y3 - z3
= [ ( x + y + z )3 - x3 ] - ( y3 + z3 )
= ( x + y + z - x )[ ( x + y + z )2 + ( x + y + z )x + x2 ] - ( y + z )( y2 - yz + z2 )
= ( y + z )( 3x2 + y2 + z2 + 2yz + 3zx + 3xy ) - ( y + z )( y2 - yz + z2 )
= ( y + z )( 3x2 + y2 + z2 + 2yz + 3zx + 3xy - y2 + yz - z2 )
= ( y + z )( 3x2 + 3yz + 3zx + 3xy )
= 3( y + z )( x2 + yz + zx + xy )
= 3( y + z )[ ( x2 + zx ) + ( xy + yz ) ]
= 3( y + z )[ x( x + z ) + y( x + z ) ]
= 3( y + z )( x + z )( x + y )
\(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left[\left(x+y+z\right)^3-x^3\right]-\left(y^3+z^3\right)\)
\(=\left(x+y+z-x\right).\left[\left(x+y+z\right)^2+\left(x+y+z\right).x+x^2\right]-\left(y+z\right)\left(y^2-yz+z^2\right)\)
\(=\left(y+z\right).\left(x^2+y^2+z^2+2xy+2yz+2xz+x^2+yx+zx+x^2\right)-\left(y+z\right)\left(y^2-yz+z^2\right)\)
\(=\left(y+z\right).\left[x^2+y^2+z^2+2xy+2yz+2xz+x^2+yx+zx+x^2-\left(y^2-yz+z^2\right)\right]\)
\(=\left(y+z\right).\left(x^2+y^2+z^2+2xy+2yz+2xz+x^2+yx+zx+x^2-y^2+yz-z^2\right)\)
\(=\left(y+z\right).\left(3x^2+3xy+3yz+3xz\right)\)
\(=\left(y+z\right).\left[\left(3x^2+3xy\right)+\left(3yz+3xz\right)\right]\)
\(=\left(y+z\right).\left[3x.\left(x+y\right)+3z.\left(y+x\right)\right]\)
\(=\left(y+z\right).\left(x+y\right).\left(3x+3z\right)\)
\(=3.\left(y+z\right).\left(x+y\right).\left(x+z\right)\)
\(x^3+y^3+z^3-3xyz=\left(x+y\right)^3-3x^2y-3xy^2+z^3-3xyz\)
\(=\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
19. 3x2-4x+1
= 3x2-3x-x+1
= (3x2-3x)-(x-1)
= 3x(x-1)-(x-1)
= (3x-1)(x-1)
20.3x2+4x-7
= 3x2+3x-7x-7
= (3x2+3x)-(7x+7)
= 3x(x+1)-7(x-1)
= (3x-7)(x-1)
21.3x2+7x-6
= 3x2+9x-2x-6
= (3x2+9x)-(2x+6)
= 3x(x+3)-2(x+3)
= (3x-2)(x+3)
22.3x2+3x-6
= 3x2+6x-3x-6
=(3x2+6x)-(3x+6)
= 3x(x+2)-3(x+2)
=(3x-3)(x+2)
= 3(x-1)(x+2)
23. 3x2-3x-6
=(3x2-6x)+(3x-6)
=3x(x-2)+3(x-2)
=(3x+3)(x-2)
= 3(x+1)(x-2)
24.6x2-13x+6
= 6x2-9x-4x+6
= (6x2-9x)-(4x-6)
=3x(2x-3)-2(2x-3)
=(3x-2)(2x-3)
25.6x2+13x+6
= 6x2+9x+4x+6
= (6x2+9x)+(4x+6)
=3x(2x+3)+2(2x+3)
=(3x+2)(2x+3)
26. 6x2+15x+6
= (6x2+12x)+(3x+6)
= 6x(x+2)+3(x+2)
=(6x+3)(x+2)
=3(2x+1)(x+2)
27. 6x2-15x+6
= (6x2-12x)-(3x-6)
= 6x(x-2)-3(x-2)
=(6x-3)(x-2)
=3(2x-1)(x-2)
28. 6x2+20x+6
= (6x2+18x)+(2x+6)
= 6x(x+3)+2(x+3)
= (6x+2)(x+3)
= 2(3x+1)(x+3)
29.6x2-20x+6
= (6x2-18x)-(2x-6)
= 6x(x-3)+2(x-3)
= (6x-2)(x-3)
= 2(3x-1)(x-3)
30.6x2+12x+6
= (6x2+6x)+(6x+6)
= 6x(x+1)+6(x+1)
= (6x+6)(x+1)
= 6(x+1)(x+1)
= 6(x+1)2
\(2x^2+5x-3=\left(2x^2-x\right)+\left(6x-3\right)\)\(=x\left(2x-1\right)+3\left(2x-1\right)=\left(x+3\right)\left(2x-1\right)\)
\(x^3+y^3+z^3-3xyz\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)+z^3-3x^2y-3xy^2-3xyz\)
\(=\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz+2xy\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)
Giỏi toán cần phải cọ xát nhiếu;
\(x^3+y^3+z^3-3xyz=\left(x^3+3x^2y+3xy^2+y^3\right)+z^3-3abc-3x^2y-3xy^2\)
Bạn thêm vào 2 hạng tử , sau đó bớt 2 hạng tử để biểu thức ko thay đổi nhé, ở đây xuất hiện 1 hằng đẳng thức:
\(=\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)\)
Ta thấy lại tiếp tục xuất hiên 1 hằng đẳng thức: a^3+b^3 nên ta có:
\(=\left(x+y+z\right)\left(\left(x+y\right)^2-\left(x+y\right)z+z^2\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
ủng hộ nha các bạn