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![](https://rs.olm.vn/images/avt/0.png?1311)
Câu a : Ta có : \(x^3+x^2z+y^2z-xyz+y^3=0\)
\(\Leftrightarrow\left(x^3+y^3\right)+\left(x^2z+y^2z-xyz\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(x^2-xy+y^2\right)+z\left(x^2-xy+y^2\right)=0\)
\(\Leftrightarrow\left(x^2-xy+y^2\right)\left(x+y+z\right)=0\)
\(\Leftrightarrow x+y+z=0\) ( đpcm )
Câu b : \(VT=\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)-a^3-b^3-c^3\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)=VP\)
Câu c : Ta có : \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow a+b+c=0\) ( đúng )
![](https://rs.olm.vn/images/avt/0.png?1311)
a)Đặt A=(x+y+z)3-x3-y3-z3
Xét (x+y+z)3=[(x+y)+z]3=(x+y)3+z3+3z(x+y)(x+y+z) =x3+y3+3xy(x+y)+z3+3z(x+y)(x+y+z)
=(x3+y3+z3)+3(x+y)(xy+xz+yz+z2)
=(x3+y3+z3)+3(x+y)[(xy+yz)+(xz+z2)]
=(x3+y3+z3)+3(x+y)[y(x+z)+z(x+z)]
=(x3+y3+z3)+3(x+y)(x+z)(y+z)
Từ đó suy ra A=(x3+y3+z3)+3(x+y)(x+z)(y+z)-x3-y3-z3=3(x+y)(x+z)(y+z)
![](https://rs.olm.vn/images/avt/0.png?1311)
bài 1
ab+bc+ca=0
=>ab+bc=-ca
ta có (a+b)(b+c)(c+a)/abc
=> (ab+ac+bc+b2)(c+a)/abc
=> (0+b2)(c+a)/abc
=>b2c+b2a/abc
=>b(ab+bc)/abc
=>b(-ac)/abc
=>-abc/abc=-1
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1 :
a) xy(x+y)+yz(y+z)+xz(x+z)+2xyz
= xy(x + y) + yz(y + z) + xyz + xz(x + z) + xyz
= xy(x + y) + yz(y + z + x) + xz(x + z + y)
= xy(x + y) + z(x + y + z)(y + x)
= (x + y)(xy + zx + zy + z²)
= (x + y)[x(y + z) + z(y + z)]
= (x + y)(y + z)(z + x)
b) \(x^3-x+3x^2y+3xy^2+y^3-x-y\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
Đã có kết quả
Bài 1,chữa phần a
xy(x+y)+yz(y+z)+xz(x+z)+2xyz
=[xy(x+y)+xyz]+[yz(y+z)+xyz]+xz(x+z)
=xy(x+y+z)+yz(x+y+z)+xz(x+z)
=y(x+y+z)(x+z)+xz(x+z)
=(x+z)(xy+y2+yz+xz)
=(x+z)(x+y)(y+z)
Chữa phần b
x3-x+3x2y+3xy2+y3-y
=(x+y)(x+y-1)(x+y+1)
Bài2
a3+b3+c3=(a+b)3-3ab(a+b)+c3=-c3-3ab(-c)+c3=3abc
Ai làm đúng như này ớ sẽ k
c, Ta có : a+b+c=0 ⇒ c=-(a+b)
⇒ a3+b3+c3= a3+b3-(a+b)3= x3+y3-(x3+3x2y+3xy2+y3)= x3+y3-x3-3x2y-3xy2-y3= -3x2y-3xy2= -3xy(x+y)= 3xyz(đpcm)
Câu a : Ta có :
\(x^3+x^2z+y^2z-xyz+y^3=0\)
\(\Leftrightarrow\left(x^3+y^3\right)+\left(x^2z-xyz+y^2z\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(x^2-xy+y^2\right)+z\left(x^2-xy+y^2\right)=0\)
\(\Leftrightarrow\left(x^2-xy+y^2\right)\left(x+y+z\right)=0\)
\(\Leftrightarrow x+y+z=0\)
Câu b : Khai triển VT ta có :
\(VT=\left(a+b+c\right)^3-a^3-b^3-c^3=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)-a^3-b^3-c^3=3\left(a+b\right)\left(b+c\right)\left(c+a\right)=VP\)
Câu c : Ta có :
\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-bc-ca+c^2\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
Luôn đúng vì \(a+b+c=0\)