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\(x^2+y^2+z^2=xy+yz+zx\\ \Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\\ \Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-z=0\\z-x=0\end{matrix}\right.\Leftrightarrow x=y=z\\ \text{Mà }x+y+z=-3\Leftrightarrow x=y=z=-1\\ \Leftrightarrow B=1-1+1=1\)
1, mk nhớ k lầm thì mk đã từng làm cho bn rồi ,kq=1/2
2,Dễ CM \(x^2+y^2+z^2\ge xy+yz+xz\) ,dấu "=" xảy ra <=>x=y=z
\(=>\left(x+y+z\right)^2\ge\left(xy+yz+xz\right)+2\left(xy+yz+xz\right)=3\left(xy+yz+xz\right)\)
\(=>9\ge3\left(xy+yz+xz\right)=>xy+yz+xz\le\frac{9}{3}=3\)
=>GTLN của xy+yz+xz=3
3)x3+y3+z3=3xyz
<=>x3+y3+z3-3xyz=0
<=>(x+y+z)(x2+y2+z2-xy-yz-xz)=0
<=>x+y+z=0 hoặc x2+y2+z2-xy-yz-xz=0
(+)x+y+z=0 thì x+y=-z;y+z=-x;x+z=-y
thế vô P =-1
(+)x2+y2+z2-xy-yz-xz=0
TH này thì x=y=z
thế vô P=2
Ta có:\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Leftrightarrow xy+yz+xz=0\)
Ta có: \(\left(xy+yz+xz\right)\left(x^2y^2+y^2z^2+x^2z^2-x^2yz-xy^2z-xyz^2\right)=0\)
\(\Leftrightarrow\left(xy\right)^3+\left(yz\right)^3+\left(xz\right)^3=3\left(xyz\right)^2\)
\(\Leftrightarrow\frac{xy}{z^2}+\frac{yz}{x^2}+\frac{zx}{y^2}=3\)
Từ đây ta có được K = 1
2: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
=>\(\frac{xy+yz+xz}{xyz}=0\)
=>xy+yz+xz=0
=>xy=-xz-yz; yz=-xy-xz; xz=-xy-yz
\(x^2+2yz=x^2+yz+yz\)
\(=x^2+yz-xy-xz=x\left(x-y\right)-z\left(x-y\right)=\left(x-y\right)\left(x-z\right)\)
\(y^2+2xz=y^2+xz+xz\)
\(=y^2+xz-xy-yz=y^2-xy+xz-yz\)
=y(y-x)+z(x-y)
=z(x-y)-y(x-y)=(x-y)(z-y)
\(z^2+2xy\)
\(=z^2+xy+xy\)
\(=z^2+xy-yz-xz\)
\(=z^2-xz+xy-yz=z\left(z-x\right)+y\left(x-z\right)=\left(x-z\right)\left(y-z\right)\)
\(A=\frac{yz}{x^2+2yz}+\frac{xz}{y^2+2xz}+\frac{xy}{z^2+2xy}\)
\(=\frac{yz}{\left(x-y\right)\left(x-z\right)}+\frac{xz}{\left(x-y\right)\left(z-y\right)}+\frac{xy}{\left(x-z\right)\left(y-z\right)}\)
\(=\frac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\frac{y^2z-yz^2-x^2z+xz^2+x^2y-xy^2}{\left(x-y\right)\cdot\left(x-z\right)\left(y-z\right)}\)
\(=\frac{z\left(y^2-x^2\right)+z^2\left(x-y\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\frac{\left(x-y\right)\left\lbrack-z\left(x+y\right)+z^2+xy\right\rbrack}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(=\frac{-xz-yz+z^2+xy}{\left(x-z\right)\left(y-z\right)}=\frac{z^2-yz-xz+xy}{\left(x-z\right)\left(y-z\right)}=\frac{z\left(z-y\right)-x\left(z-y\right)}{\left(x-z\right)\left(y-z\right)}=\frac{\left(z-x\right)\left(z-y\right)}{\left(z-x\right)\left(z-y\right)}\)
=1
2) \(\hept{\begin{cases}^{x^2-xy=y^2-yz}\left(1\right)\\^{y^2-yz=z^2-zx}\left(2\right)\\^{z^2-zx=x^2-xy}\left(3\right)\end{cases}}\)
lấy (2) - (1) suy ra\(2yz=2y^2+xy+xz-x^2-z^2\)
lấy (3) - (1) suy ra \(2xy=zx+yz-z^2+2x^2-y^2\)
lấy (3) - (2) suy ra \(2zx=xy+yz+2z^2-x^2-y^2\)
cộng lại đc \(yz+xz+xy=0\) do đó \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{yz+xz+xy}{xyz}=0\)
ta có: \(x+y+z=a\Rightarrow x^2+y^2+z^2+2\left(xy+yz+xz\right)=a^2\)
\(\Rightarrow b+2\left(xy+yz+xz\right)=a^2\Rightarrow xy+yz+xz=\frac{a^2-b}{2}\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{c}\Rightarrow\frac{xy+yz+xz}{xyz}=\frac{1}{c}\Rightarrow c\left(xy+yz+xz\right)=xyz\)
Ta có:\(x^3+y^3+z^3=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)+3xyz\)
\(=a\left(b-\frac{a^2-b}{2}\right)+\frac{3c\left(a^2-b\right)}{2}\)