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Ta có 1/x+1/y+1/z=0
=>1/x+1/y=-1/z
=>(1/x+1/y)^3= (-1/z)^3
=>1/x^3+1/y^3+3.1/x.1/y.(1/x+1/y) =-1/z^3
=>1/x^3+1/y^3+1/z^3= -3.1/x.1/y.(1/x+1/y) =3/(xyz) (vì 1/x+1/y=-1/z)
Mặt khác: 1/x+1/y+1/z=0
=>(xy+yz+zx)/(xyz)=0
=>xy+yz+zx=0
A=yz/x^2 +2yz + xz/y^2+ 2xz + xy/z^2+ 2 xy
=xyz/x^3+xyz/y^3+xyz/z^3 +2(xy+yz+zx) (vì x,y,z khác 0)
=xyz(1/x^3+1/y^3+1/z^3) (vì xy+yz+zx=0)
=xyz.3/(xyz) (vì 1/x^3+1/y^3+1/z^3=3/(xyz) )
=3
Vậy A=3.
\(\dfrac{21}{4x}+\dfrac{21}{4y}+\dfrac{21}{4z}=0\Leftrightarrow\dfrac{21}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=0\)
\(\Leftrightarrow\dfrac{xy+xz+yz}{xyz}=0\Leftrightarrow xy+xz+yz=0\) \(\Rightarrow\left\{{}\begin{matrix}xy=-xz-yz\\xz=-xy-yz\\yz=-xy-xz\end{matrix}\right.\)
Ta có:
\(x^2+2yz=x^2+yz+yz=x^2+yz-xy-xz=x\left(x-y\right)-z\left(x-y\right)=\left(x-y\right)\left(x-z\right)\)
\(\Rightarrow\dfrac{yz}{x^2+2yz}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)
Tương tự ta có \(\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(y-x\right)\left(y-z\right)}\) ; \(\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(z-x\right)\left(z-y\right)}\)
\(\Rightarrow A=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xz}{\left(y-x\right)\left(y-z\right)}+\dfrac{xy}{\left(z-x\right)\left(z-y\right)}\)
\(=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\dfrac{y^2z-yz^2-x^2z+xz^2+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(=\dfrac{z^2\left(x-y\right)-z\left(x^2-y^2\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\dfrac{\left(z^2-xz-yz+xy\right)\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(=\dfrac{\left(x\left(y-z\right)-z\left(y-z\right)\right)\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\dfrac{\left(x-z\right)\left(y-z\right)\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=1\)
mình chưa học đến lớp 8
theo bài ra ta có \(\frac{21}{4x}+\frac{21}{4y}+\frac{21}{4z}=0\)
\(\Leftrightarrow\frac{21xy+21yz+21xz}{4xyz}=0\)
\(\Leftrightarrow21\left(xy+yz+xz\right)=0\)
\(\Leftrightarrow xy+yz+xz=0\)
\(\Rightarrow\hept{\begin{cases}xy=-yz-xz\\yz=-xy-xz\\xz=-xy-yz\end{cases}}\)
thay vào P ta có
P=\(\frac{yz}{x^2-xy-xz+yz}+\frac{xz}{y^2-xy-yz+xz}+\frac{xy}{z^2-yz-xz+xy}\)
=\(\frac{yz}{\left(x-y\right)\left(x-z\right)}+\frac{xz}{\left(y-x\right)\left(y-z\right)}+\frac{xy}{\left(z-y\right)\left(z-x\right)}\)
=\(-\frac{yz}{\left(x-y\right)\left(z-x\right)}-\frac{xz}{\left(x-y\right)\left(y-z\right)}-\frac{xy}{\left(y-z\right)\left(z-x\right)}\)
=\(\frac{-yz\left(y-z\right)-xz\left(z-x\right)-xy\left(x-y\right)}{\left(x-y\right)\left(z-x\right)\left(y-z\right)}\)
=\(\frac{\left(y-z\right)\left(z-x\right)\left(x-y\right)}{\left(x-y\right)\left(z-x\right)\left(y-z\right)}\)(TỬ PHÂN TÍCH THÀNH NHÂN TỬ BẠN TỰ LÀM NHÉ)
=\(1\)
mà mình nghĩ cậu viết sai đề chỗ P= ........+\(\frac{xz}{y^2+2xz}\)+......nếu bạn ko phải sai đề thì mình giải sai rồi bạn xem lại nhé