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\(x^2+y^2-z^2=x^2+\left(y-z\right)\left(y+z\right)=x^2-x\left(y-z\right)=x\left(x-y+z\right)=x\left(-y-y\right)=-2xy\)
Tương tự \(x^2+z^2-y^2=-2xz;y^2+z^2-x^2=-2yz\)
Cộng VTV:
\(\Leftrightarrow\text{Biểu thức }=\dfrac{xy}{-2xy}+\dfrac{xz}{-2xz}+\dfrac{yz}{-2yz}=-\dfrac{1}{8}\)
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\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Leftrightarrow xy+yz+zx=0\)
\(\Rightarrow yz=-xy-zx\Rightarrow\dfrac{yz}{x^2+2yz}=\dfrac{yz}{x^2+yz-xy-zx}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)
Tương tự: \(\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(y-x\right)\left(y-z\right)}\) ; \(\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-z\right)\left(y-z\right)}\)
\(\Rightarrow A=\dfrac{-yz\left(y-z\right)-zx\left(z-x\right)-xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=1\)
Bài này dễ thôi:vv
Theo đề ta có: \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=0\Leftrightarrow\dfrac{xbc+yac+zab}{abc}=0\Leftrightarrow xbc+yac+zab=0\)
Lại có:\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=2\Rightarrow\left(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}\right)^2=4\)
=>\(\dfrac{a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}+2\left(\dfrac{ab}{xy}+\dfrac{bc}{yz}+\dfrac{ca}{xz}\right)=4\)
=>\(\dfrac{a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}+2\left(\dfrac{abz+bcx+cay}{xyz}\right)=4\)
=>\(\dfrac{a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}+2.0=4\Rightarrow\dfrac{a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}=2\)
Vậy...
Bài này ez thôi, làm mãi rồi.
Theo đề bài, ta có: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)
=>\(\dfrac{xy+yz+xz}{xyz}=0\)
=> xy+yz+zx=0
=> \(\left\{{}\begin{matrix}xy=-yz-zx\\yz=-xy-zx\\zx=-xy-yz\end{matrix}\right.\)
Ta có: x2+2yz=x2+yz-xy-zx=(x-y)(x-z)
y2+2xz=y2+xz-xy-yz=(x-y)(z-y)
z2+2xy=z2+xy-yz-xz=(x-z)(y-z)
=> \(\dfrac{yz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xz}{\left(x-y\right)\left(z-y\right)}+\dfrac{xy}{\left(x-z\right)\left(y-z\right)}=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\dfrac{\left(x-y\right)\left(x-z\right)\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=1\)
vi a/x + b/y + c/z =0 suy ra ayz/xyz + bxz/xyz + cxy/xyz =0 suy ra ayz+bxz+cxy /xyz =0 suy ra ayz + bxz + cxy =0
vi x/a + y/b =z/c =0 suy ra (x/a + y/b + z/c )^2 =0 suy ra x^2/a^2 +y^2/b^2 + z^2/c^2 + 2(xy/ab + xz/ac + yz/bc) =0
suy ra x^2/a^2 + y^2/b^2 + z^2/c^2 + 2(cxy+ bxz +ayz /abc) =0
suy ra x^2/a^2 + y^2/b^2 + z^2/c^2 =0
suy ra x^2/a^2 + y^2/b^2 + z^2/c^2 +2011 = 2011
Thay x=-y-z
suy ra x^2-y^2-z^2=(-y-z)^2-y^2-z^2=2yz
Vậy M=x^+y^+z^3/2xyz(quy đồng)
Tiếp tuc thay: x=-y-z
M=(-y-z)^3+y^3+z^3/2(-y-z)yz
M=-3x^2y-3xy^2/-2x^y-2xy^2=3/2
thanks bn nhiều