Thiếu đề. Nếu x = 1; y = 1 thì không thỏa mãn đề bài

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chịu

Đề sai rồi em, đề đúng phải là:

\(ab\left(x^2+y^2\right)+xy\left(a^2+b^2\right)=ab\)

Vế phải em thiếu a

\(\dfrac{x}{y}+\dfrac{y}{x}-2=\dfrac{\left(x-y\right)^2}{xy}\ge0\)

Ta có: \(\left\{{}\begin{matrix}x+y+z=0\\xy+yz+zx=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left(x+y+z\right)^2=0\\2\left(xy+yz+zx\right)=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x^2+y^2+z^2+2xy+2yz+2xz=0\\2xy+2yz+2xz=0\end{matrix}\right.\)

\(\Rightarrow x^2+y^2+z^2+2xy+2yz+2xz-2xy-2yz-2xz=0\)

\(\Rightarrow x^2+y^2+z^2=0\Rightarrow\left\{{}\begin{matrix}x^2\ge0\forall x\\y^2\ge0\forall y\\z^2\ge0\forall z\end{matrix}\right.\Rightarrow x^2+y^2+z^2\ge0\)

\("="\Leftrightarrow\left\{{}\begin{matrix}x^2=0\\y^2=0\\z^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\\z=0\end{matrix}\right.\)

\(\Rightarrow x=y=z=0\Rightarrow dpcm\)

\(x+y+z=0\Leftrightarrow\left(x+y+z\right)^2=0\)

\(\Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz=x^2+y^z+z^2+0=0\)

\(\Leftrightarrow x^2+y^2+z^2=0\Leftrightarrow x=y=z=0\)

b) Bằng chứ ^^
\(\left(x+y\right)^2=x^2+2xy+y^2=4xy\)

\(\Leftrightarrow x^2-2xy+y^2=0\Leftrightarrow\left(x-y\right)^2=0\Leftrightarrow x=y\)

a, Ta có: \(\left\{{}\begin{matrix}x+y+z=0\\xy+yz+zx=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x+y+z\right)^2=0\\2\left(xy+yz+zx\right)=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x^2+y^2+z^2+2\left(xy+yz+xz\right)=0\\2\left(xy+yz+zx\right)=0\end{matrix}\right.\)

\(\Rightarrow x^2+y^2+z^2=0\)

Vì \(\left\{{}\begin{matrix}x^2\ge0\\y^2\ge0\\z^2\ge0\end{matrix}\right.\) với \(\forall x,y,z\)

\(\Rightarrow x^2+y^2+z^2\ge0\)

Dấu ''='' xảy ra \(\Leftrightarrow x=y=z=0\)

Vậy ....

b, Ta có: \(\left(x-y\right)^2\ge0\)

\(\Rightarrow x^2-2xy+y^2\ge0\)

\(\Rightarrow x^2+2xy+y^2\ge4xy\)

\(\Rightarrow\left(x+y\right)^2\ge4xy\) (đpcm)

\(x^2+y^2-z^2>0\Rightarrow x^2+2xy+y^2-z^2>0\)

\(\Rightarrow\left(x+y\right)^2-z^2>0\)

\(\Rightarrow\left(x+y-z\right)\left(x+y+z\right)>0\)

Mà x;y;z>0 \(\Rightarrow x+y+z>0\)

\(\Rightarrow x+y-z>0\)

bạn cảm ơn ai vay có bn ấy có giup bn làm đau

mk chua hok den nen ko co bit lam

cảm ơn b nhé