Question

1, cho x²+y²+z²= xy+yz+zx. chứng minh x=y=z

2, cho x+y+z =0. chứng minh (x² +y²+z²)² = 2(x⁴+y⁴+z⁴)

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Answer 1

1,

\(x^2+y^2+z^2=xy+yz+zx\)

\(\Leftrightarrow x^2+y^2+z^2-xy-yz-zx=0\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=2.0=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

<=> x - y = 0

y - z = 0

z - x =0 

<=> x=y

y=z

z=x

<=> x=y=z

Answer 2

1)VD:\(X=Y=Z\Leftrightarrow XY+YZ+ZX=X^2+Y^2+Z^2\)

\(\Leftrightarrow X^2+Y^2+Z^2=XY+YZ+ZX\left(1\right)\)

VD:\(X^2+Y^2+Z^2=XY+YZ+ZX\Leftrightarrow2X^2+2Y^2+2Z^2=2XY+2YZ+2ZX\)

\(\Leftrightarrow2X^2+2Y^2+2Z^2-2XY-2YZ-2ZX=0\)

\(\Leftrightarrow\left(X-Y\right)^2+\left(Y-Z\right)^2+\left(Z-X\right)^2=0\left(HĐT\right)\)

\(\Rightarrow X=Y=Z\left(2\right)\)

\(1\&2\Rightarrow X^2+Y^2+Z^2=XY+YZ+ZX\)

\(\Leftrightarrow X=Y=Z\)

2)\(\Rightarrow A+B+C\Rightarrow X=-\left(Y+Z\right)\Rightarrow X^2=\left(Y+Z\right)^2\)

\(\Leftrightarrow X^2=Y^2+2YZ+Z^2\)

\(\Leftrightarrow X^2-Y^2-Z^2=2YZ\)

\(\Leftrightarrow\left(X^2-Y^2-Z^2\right)^2=4Y^2Z^2\)

\(\Leftrightarrow X^4+Y^4+Z^4=2X^2Y^2+2Y^2Z^2+2Z^2X^2\)

\(\Leftrightarrow2\left(X^4+Y^4+Z^2\right)=\left(X^2+Y^2+Z^2\right)^2=A^4\)

\(\Rightarrow X^4+Y^4+Z^4=\frac{A^4}{2}\)

Answer 3

1.Ta có: x²+y²+z²=xy + yz + zx <=>2x^2+2y^2+2z^2-2xy-2yz-2xz=0 <=> (x-y)^2 + (y-z)^2 + ( x-z )^2 = 0 

Vì (x-y)^2>=0,(y-z)^2>=0,(x-z)^2>=0 

Dấu "=" xảy ra <=> x-y=0,y-z=0,x-z=0 <=> x=y=z