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a ) \(\left(x+y+z\right)^2=x^2+y^2+z^{2^{ }}+2xy+2yz+2zx\)
Biến đổi vế trái ta được :
\(\left(x+y+z\right)^2=\left(x+y+z\right)\left(x+y+z\right)\)
\(=x^2+xy+xz+xy+y^2+yz+zx+zy+z^2\)
\(=x^2+y^2+z^{2^{ }}+2xy+2yz+2zx\)
Vậy \(\left(x+y+z\right)^2=x^2+y^2+z^{2^{ }}+2xy+2yz+2zx\)
a, \(x^3+y^3+z^3=3xyz\Rightarrow x^3+y^3+z^3-3xyz=0\)( 1 )
Nhận xét : \(\left(x+y\right)^3=x^3+y^3+3x^2y+3xy^2\Rightarrow x^3+y^3=\left(x+y\right)^3-3x^2-3xy^2\)
Thay vào ( 1 ) ta có :
\(\left(x+y\right)^3+c^3-3x^2y-3xy^2-3xyz\)
\(=\left(z+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(z+y+z\right)\left(z^2+2xy+y^2-xz-yz+z^2\right)-3xyz\left(z+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(z^2+x^2+y^2-xy-yz-xz\right)\)
Vì theo đầu bài ta có: \(x+y+z=0\)nên ta có ( DPCM ) ..... học cho tốt nhé!
\(a)x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow x^3+y^3+3x^2y+3xy^2-3x^2y-3xy^2+z^3-3xyz=0\)
\(\) \(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(\right.\) \(\left(x+y\right)^2-z\left(x+y\right)+z^2-3xy)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(\right.\) \(x^2+2xy+y^2-xz-yz+z^2-3xy)=0\)
Mà \(x+y+z=0\)
\(\Rightarrow0=0\left(đpcm)\right.\)
\(b)\left(x^2y^2+y^2z^2+x^2z^2+2\left.x^2yz+2xy^2z+2xyz^2\right)\right.=x^2y^2+y^2z^2+x^2z^2\)
\(\Leftrightarrow2\left(\right.\) \(x^2yz+xy^2z+xyz^2)=0\)
\(\Leftrightarrow2\left(x+y+z\right)\left(xyz\right)=0\)
Mà \(x+y+z=0\)
\(\Rightarrow0=0\left(đpcm\right)\)
\(c)\) Ta có:\(x+y+z=0\)
\(\Rightarrow\left(x+y+z\right)^2=0\)
\(\Rightarrow x^2+y^2+z^2+2\left(\right.\) \(x^2yz+xy^2z+xyz^2)=0\)
\(\Rightarrow2\left(\right.\) \(xy+yz+xz^{})=-\left(\right.\) \(x^2+y^2+z^2)\)
\(\Rightarrow4\left(\right.\) \(xy+yz+xz)^2=\) \(x^4+y^4+z^4+2\left(\right.\) \(x^2y^2+y^2z^2+x^2z^2)\left(1\right)\)
Mà ta có: \(\left(xy+yz+xz\right)^2=x^2y^2+y^2z^2+x^2z^2\) (theo câu b)
\(\Leftrightarrow2\left(xy+yz+xz\right)^2=2\left(\right.\) \(x^2y^2+y^2z^2+x^2z^2)\left(2\right)\)
\(\left(1\right)-\left(2\right)\Leftrightarrow2\left(xy+yz+xz\right)^2=x^4+y^4+z^4\left(đpcm\right)\)
1)x(x2 - 19 - 30)
2)x(x2 - 7 - 6)
3)x(x2 + 4x - 7 - 10)
( 4 tích mình làm tiếp 3 câu cuối)
a ) \(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3\)
\(=x^3-3x^2y+3xy^2-y^3+y^3-3y^2z+3yz^2-z^3+z^3-3z^2x+3zx^2-x^3\)
\(=-3x^2y+3xy^2-3y^2z+3yz^2-3z^2x+3zx^2\)
b)\(x\left(y^2-z^2\right)+z\left(x^2-y^2\right)+y\left(z^2-x^2\right)\)
=\(x\left(y^2-z^2\right)-\left(y^2-z^2+z^2-x^2\right)z+y\left(z^2-x^2\right)\)
=\(x\left(y^2-z^2\right)-z\left(y^2-z^2\right)-z\left(z^2-x^2\right)+y\left(z^2-x^2\right)\)
=\(\left(y^2-z^2\right)\left(x-z\right)+\left(z^2-x^2\right)\left(y-z\right)\)
=\(\left(y-z\right)\left(z-x\right)\left(-\left(y+z\right)+z+x\right)\)
=\(\left(y-z\right)\left(z-x\right)\left(x-y\right)\)