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a: Ta có: \(\left(x+y\right)^2\)
\(=x^2+2xy+y^2\)
\(\Leftrightarrow x^2+y^2=\dfrac{\left(x+y\right)^2}{2xy}\ge\dfrac{\left(x+y\right)^2}{2}\forall x,y>0\)
Có x+y+z=0
<=>(x+y+z)+(x+y+z)=0
<=>x+y+z+x+y+z=0
<=>2x+2y+2z=0
<=>(2x+2y+2z).2=0(1)
Tương tự có :(4x+4y+4z).2=0(2)
Từ (1)và(2) có (x2+y2+z2).2=2.(x4+y4+z4)
Chúc bạn học tốt nha
Bài 3:
\(\left(x-3\right)\left(x-1\right)\left(x+1\right)\left(x+3\right)+15\)
\(=\left(x^2-9\right)\left(x^2-1\right)+15\)
\(=x^4-10x^2+9+15\)
\(=x^4-10x^2+24\)
\(=\left(x^2-4\right)\left(x^2-6\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(x^2-6\right)\)
Ta có:
\(x^4\ge0\); \(y^4\ge0\) ;\(z^4\ge0\)
\(\Rightarrow x^4+y^4+z^4\ge0\)
Ta cũng có:
\(x^2\ge0\); \(y^2\ge0\) ;\(z^2\ge0\)
\(\Rightarrow x^2+y^2+z^2\ge0\)
Mà: \(x^4>x^2;y^4>x^2;z^4>z^2\)
\(\Rightarrow x^4+y^4+z^4\ge\left(x^2+y^2+z^2\right):3\) (đpcm)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\\ \Rightarrow\left\{{}\begin{matrix}1+\dfrac{x}{y}+\dfrac{x}{z}=0\\\dfrac{y}{x}+1+\dfrac{y}{z}=0\\\dfrac{z}{x}+\dfrac{z}{y}+1=0\end{matrix}\right.\\ \Rightarrow\dfrac{x}{y}+\dfrac{x}{z}+\dfrac{y}{x}+\dfrac{y}{z}+\dfrac{z}{x}+\dfrac{z}{y}=-3\)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\\ \Rightarrow\dfrac{yz+xz+xy}{xyz}=0\\ \Rightarrow yz+xz+xy=0\)
\(\Rightarrow\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)\left(xy+xz+yz\right)=0\\ \Rightarrow\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}+\dfrac{x}{y}+\dfrac{x}{z}+\dfrac{y}{x}+\dfrac{y}{z}+\dfrac{z}{x}+\dfrac{z}{y}=0\\ \Rightarrow\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=3\)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)
\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{-1}{z}\)
\(\Rightarrow\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^3=\left(\dfrac{-1}{z}\right)^3\)
\(\Leftrightarrow\dfrac{1}{x^3}+3\dfrac{1}{x^2}\dfrac{1}{y}+3\dfrac{1}{x}\dfrac{1}{y^2}+\dfrac{1}{y^3}=\dfrac{-1}{z^3}\)
\(\Leftrightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=-3.\dfrac{1}{x}\dfrac{1}{y}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
\(\Leftrightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=-3\dfrac{1}{x}\dfrac{1}{y}\dfrac{-1}{z}\)
\(\Leftrightarrow\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)xyz=3\dfrac{1}{x}\dfrac{1}{y}\dfrac{1}{z}.xyz\)
\(\Leftrightarrow\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=3\)
a)(x-y)3+(y-z)3+(z-x)3
=3(x-y+y-z+z-x)=3
b)nhân vào là rồi đối trừ là hết luôn ( nhưng là mũ 2 hay nhân 2 v mk là theo nhân 2 nhé]
x2+y2−z22xy−y2+z2−x22yz+z2+x2−y22xz=1x2+y2−z22xy−y2+z2−x22yz+z2+x2−y22xz=1
Tính P = x + y + z
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
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)\(\Rightarrow A+B+C\Rightarrow X=-\left(Y+Z\right)\Rightarrow X^2=\left(Y+Z\right)^2\)
\(\Leftrightarrow X^2=Y^2+2YZ+Z^2\)
\(\Leftrightarrow X^2-Y^2-Z^2=2YZ\)
\(\Leftrightarrow\left(X^2-Y^2-Z^2\right)^2=4Y^2Z^2\)
\(\Leftrightarrow X^4+Y^4+Z^4=2X^2Y^2+2Y^2Z^2+2Z^2X^2\)
\(\Leftrightarrow2\left(X^4+Y^4+Z^2\right)=\left(X^2+Y^2+Z^2\right)^2=A^4\)
\(\Rightarrow X^4+Y^4+Z^4=\frac{A^4}{2}\)