Ta có: \(x^2+y^2-z^2\)

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

\(=\left(x+y+z\right)\left(x+y-z\right)-2xy\)

\(=-2xy\)

Ta có: \(x^2+z^2-y^2\)

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

\(=\left(x+y+z\right)\left(x+z-y\right)-2xz\)

\(=-2xz\)

Ta có: \(y^2+z^2-x^2\)

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

\(=\left(x+y+z\right)\left(y+z-x\right)-2yz\)

\(=-2yz\)

Ta có: \(\dfrac{xy}{x^2+y^2-z^2}+\dfrac{xz}{x^2+z^2-y^2}+\dfrac{yz}{y^2+z^2-x^2}\)

\(=\dfrac{xy}{-2xy}+\dfrac{xz}{-2xz}+\dfrac{yz}{-2yz}\)

\(=\dfrac{1}{-2}+\dfrac{1}{-2}+\dfrac{1}{-2}\)

\(=\dfrac{-3}{2}\)

Ta có:

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)

\(\Rightarrow\dfrac{1}{z}=-\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)

\(\Rightarrow\left(\dfrac{1}{z}\right)^3=-\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^3\)

\(\Rightarrow\dfrac{1}{z^3}=-\left(\dfrac{1}{x^3}+3\cdot\dfrac{1}{x^2}\cdot\dfrac{1}{y}+3\cdot\dfrac{1}{x}\cdot\dfrac{1}{y^2}+\dfrac{1}{y^3}\right)\)

\(\Rightarrow\dfrac{1}{z^3}=-\dfrac{1}{x^3}-\dfrac{3}{x^2y}-\dfrac{3}{xy^2}-\dfrac{1}{y^3}\)

\(\Rightarrow\dfrac{1}{z^3}+\dfrac{1}{x^3}+\dfrac{1}{y^3}=-3\cdot\dfrac{1}{x}\cdot\dfrac{1}{y}\cdot\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)

\(\Rightarrow\dfrac{1}{z^3}+\dfrac{1}{x^3}+\dfrac{1}{y^3}=-3\cdot\dfrac{1}{x}\cdot\dfrac{1}{y}\cdot-\dfrac{1}{z}\)

\(\Rightarrow\dfrac{1}{z^3}+\dfrac{1}{x^3}+\dfrac{1}{y^3}=3\cdot\dfrac{1}{xyz}\)

\(\Rightarrow xyz\cdot\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=3\)

\(\Rightarrow\dfrac{xyz}{x^3}+\dfrac{xyz}{y^3}+\dfrac{xyz}{z^3}=3\)

\(\Rightarrow\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=3\)

Vậy \(A=3\)

Đề thiếu kìa :vv

⇔xy+yz+zx=0

=yz/(x−y)(x−z)

Tương tự: xy/z^2+2xy=xy/(x−z)(y−z)

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Leftrightarrow\dfrac{xy+yz+xz}{xyz}=0\Leftrightarrow xy+yz+xz=0\Leftrightarrow yz=-xy-xz\)

Ta có \(x^2+2yz=x^2+yz-xy-xz=\left(x-y\right)\left(x-z\right)\)

Tương tự \(y^2+2xz=\left(y-x\right)\left(y-z\right);z^2-2xy=\left(z-x\right)\left(z-y\right)\)

\(A=\dfrac{yz}{x^2+2yz}+\dfrac{xz}{y^2+2xz}+\dfrac{xy}{z^2+2xy}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xz}{\left(y-z\right)\left(y-x\right)}+\dfrac{xy}{\left(z-x\right)\left(z-y\right)}\\ A=\dfrac{-yz\left(y-z\right)-xz\left(z-x\right)-xy\left(x-z\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\\ A=\dfrac{-yz\left(y-z\right)+xz\left(y-z\right)+xz\left(x-y\right)-xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\\ A=\dfrac{\left(y-z\right)\left(xz-yz\right)+\left(x-y\right)\left(xz-xy\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\\ A=\dfrac{\left(x-y\right)\left(y-z\right)\left(z-x\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=1\)

⇒yz=−xy−zx⇒yz/x^2+2yz=yz/x^2+yz−xy−zx

=yz/(x−y)(x−z)

Tương tự: xy/z^2+2xy=xy/(x−z)(y−z)

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Leftrightarrow xy+yz+xz=0\)

\(\Leftrightarrow yz=-xy-xz\)\(\Leftrightarrow\dfrac{yz}{x^2+2yz}=\dfrac{yz}{x^2+yz-xy-xz}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)

Tương tự: \(\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(y-x\right)\left(y-z\right)}\)

                    \(\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-z\right)\left(y-z\right)}\)

\(\Rightarrow\dfrac{yz}{x^2+2yz}+\dfrac{xz}{y^2+2xz}+\dfrac{xy}{z^2+2xy}=\dfrac{-yz\left(y-z\right)-xz\left(z-x\right)-xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=1\)

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Leftrightarrow xy+yz+zx=0\)

\(\Rightarrow yz=-xy-zx\Rightarrow\dfrac{yz}{x^2+2yz}=\dfrac{yz}{x^2+yz-xy-zx}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)

Tương tự: \(\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(y-x\right)\left(y-z\right)}\) ; \(\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-z\right)\left(y-z\right)}\)

\(\Rightarrow A=\dfrac{-yz\left(y-z\right)-zx\left(z-x\right)-xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=1\)

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Rightarrow xy+yz+xz=0\)

A=\(xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}-\dfrac{3}{xyz}+\dfrac{3}{xyz}\right)=xyz.\dfrac{3}{xyz}=3\)

bạn tự chứng minh \(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}-\dfrac{3}{xyz}=0\) nha

đặt \(\dfrac{1}{x}=a;\dfrac{1}{y}=b;\dfrac{1}{z}=c\)

bài toán thành \(a^3+b^3+c^3-3abc=0\) nha