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Ta có
\(x^3-6x^2+x^2y+9x-3y\\ =\left(x^3-6x^2+9x\right)+\left(x^2y-3y\right)\\ =x\left(x^2-3\right)^2+y\left(x^2-3\right)\)
=(x^2-3)(x+y)
( a - x )y3 - ( a - y )x3 + ( x - y )a3
= ay3 + a2y2 - ax2y - a2xy - a2y2 - a3y + a2x2 + a3x - xy3 - axy2 + x3y + ax2y + axy2 + a2xy - ax3 - a2x2
= ay( y2 +ay -x2 - ax ) - a2( y2 + ay -x2 -ax ) - xy( y2 + ay - x2 -ax ) + ax( y2 + ay -x2 -ax )
= ( y2 + ay - x2 - ax )( ay - a2 - xy + ax )
= ( y2 + xy +ay -xy -ax -x2 )[ ( y -a )a - x( y-a ) ]
= [ y( y +x +a ) - x( y + x + a )]( a - x )( a - y)
= ( y + x + a)( y -x )( a - x)( y - a)
\(\left(x-y\right)z^3+\left(y-z\right)x^3+\left(z-x\right)y^3\)
\(=\left(x-y\right)z^3-\left[\left(x-y\right)+\left(z-x\right)\right]x^3+\left(z-x\right)y^3\)
\(=\left(x-y\right)z^3-\left(x-y\right)x^3-\left(z-x\right)x^3+\left(z-x\right)y^3\)
\(=\left(x-y\right)\left(z^3-x^3\right)-\left(z-x\right)\left(x^3-y^3\right)\)
\(=\left(x-y\right)\left(z-x\right)\left(z^2+zx+x^2\right)-\left(z-x\right)\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=\left(x-y\right)\left(z-x\right)\left(z^2+zx+x^2-x^2-xy-y^2\right)\)
\(=\left(x-y\right)\left(z-x\right)\left[\left(x^2-x^2\right)+\left(zx-xy\right)+\left(z^2-y^2\right)\right]\)
\(=\left(x-y\right)\left(z-x\right)\left[x\left(z-y\right)+\left(z-y\right)\left(y+z\right)\right]\)
\(=\left(x-y\right)\left(z-x\right)\left(z-y\right)\left(x+y+z\right)\)
\(=-\left(x-y\right)\left(y-z\right)\left(z-x\right)\left(x+y+z\right)\)
a) \([(x-y)3 + (y-z)3]+ (z-x)3\)=\(\left(x-y+y-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2\right]-\left(x-z\right)^3\)
\(=\left(x-z\right)\left[\left(\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2-\left(x-z\right)^2\right)\right]\)
\(=\left(x-z\right)\left[\left(x-y\right)\left(x-y-y+z\right)+\left(y-z-x+z\right)\left(y-z+x-z\right)\right]=\left(x-z\right)\left[\left(x-2y+z\right)\left(x+z\right)-\left(x-y\right)\left(x+y-2z\right)\right]\)
\(=\left(x-z\right)\left(x-y\right)\left(x-2y+z-x-y+2z\right)=\left(x-z\right)\left(x-y\right)\left(z-y\right)3\)
b) \(=y^2\left(x^2y-x^3+z^3-z^2y\right)-z^2x^2\left(z-x\right)=y^2\left[-y\left(z^2-x^2\right)-\left(z^3-x^3\right)\right]-z^2x^2\left(z-x\right)\)
\(=y^2\left(z-x\right)\left(-yz-xy-z^2-zx-x^2\right)-z^2x^2\left(z-x\right)=\left(z-x\right)\left(-y^3z-xy^2-z^2y^2-xyz-x^2y^2-z^2x^2\right)\)
đến đây coi như là thành nhân tử rồi nha. em muốn gọn thì ráng ngồi nghĩ rồi tách nha. chỉ cần nhóm mấy cái có ngoặc giống nhau là đc. k khó đâu. chịu khó nghĩ để rèn luyện nha
c) \(x^8+2x^4+1-x^4=\left(x^4+1\right)^2-x^4=\left(x^4+1-x^2\right)\left(x^4+1+x^2\right)\)
\(\left(9a^3-6a^2\right)+\left(6a^2-4a\right)+\left(-9a+6\right)=3a^2\left(3a-2\right)+2a\left(3a-2\right)-3\left(3a-2\right)=\left(3a-2\right)\left(3a^2+2a-3\right)\)
d) em sửa đề đi. đề sai rồi. đồng nhất hệ số phải có dấu bằng nha.
có gì liên hệ chị. đúng nha ;)
bài a) bn trên đã dẫn link cho bn r
bài b)
Đặt x-y=a;y-z=b;z-x=c
\(=>a+b+c=x-y+y-z+z-x=0\)
\(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3=a^3+b^3+c^3\)
Theo câu a)\(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\) (do a+b+c=0)
\(=>a^3+b^3+c^3=3abc=>\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3=3\left(x-y\right)\left(y-z\right)\left(z-x\right)\)
a) Ta có :
\(a^3+b^3+c^3-3abc\)
\(\Rightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(\Rightarrow\left(a+b+c\right)\left[\left(a+b^2\right)-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
P/s tham khảo nha
hok tốt
\(\left(x+y\right)^3-\left(x-y\right)^3\)
\(=\left(x+y-x+y\right)\left[\left(x+y\right)^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)
\(=2y\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)\)
\(=2y\left(3x^2+y^2\right)\)
1) \(x^2-x-y^2-y=\left(x^2-y^2\right)-\left(x+y\right)=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)\)
\(x^2-2xy+y^2-z^2=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)
2)\(5x-5y+ax-ay=5\left(x-y\right)+a\left(x-y\right)=\left(x-y\right)\left(a+5\right)\)
\(a^3-a^2x-ay+xy=a^2\left(a-x\right)-y\left(a-x\right)=\left(a-x\right)\left(a^2-y\right)\)
`x^3+y^3+x+y`
`=(x+y)(x^2-xy+y^2)+x+y`
`=(x+y)(x^2-xy+y^2+1)`