Trả lời

Từ giả thiết x+y+z=xyz <=> 1/xy + 1/yz + 1/zx = 1

Khi đó: x/1+x² = \(\frac{1}{\frac{x}{\left(\frac{1}{z}+\frac{1}{y}\right)\left(\frac{1}{x}+\frac{1}{z}\right)}}\)\(=\frac{xyz}{\left(x+y\right)\left(x+z\right)}\)

Tương tự cho 2 cái còn lại ta có:\(\frac{y}{1+y^2}=\frac{xyz}{\left(y+x\right)\left(y+z\right)}\)

\(\frac{z}{1+z^2}=\frac{xyz}{\left(z+x\right)\left(z+y\right)}\)

Suy ra VT=\(\frac{xyz\left(y+z\right)+2xyz\left(z+x\right)+3xyz\left(x+y\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)\(=\frac{xyz\left(5x+4y+3z\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)

ĐPCM

 Ta có:\(\frac{x}{1+x^2}=\frac{xyz}{yz+x^2yz}=\frac{xyz}{yz+x\left(xyz\right)}=\frac{xyz}{yz+x\left(x+y+z\right)}=\frac{xyz}{yz+x^2+xy+xz}=\frac{xyz}{y\left(x+z\right)+x\left(x+z\right)}\)

\(=\frac{xyz}{\left(x+z\right)\left(y+x\right)}\)

Chứng minh tương tự : \(\frac{2y}{1+y^2}=\frac{2xyz}{\left(y+z\right)\left(y+x\right)}\)

                                        \(\frac{3z}{1+z^2}=\frac{3xyz}{\left(x+z\right)\left(x+y\right)}\)

Khi đó VT \(=\frac{xyz}{\left(x+z\right)\left(y+x\right)}+\frac{2xyz}{\left(y+z\right)\left(y+x\right)}+\frac{3xyz}{\left(x+z\right)\left(z+y\right)}\)

\(=\frac{xyz\left[y+z+2\left(z+x\right)+3\left(x+y\right)\right]}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)

\(=\frac{xyz\left(5x+4y+3z\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\left(đpcm\right)\)

( mình đang vội nên làm hơi tắt mong bạn thông cảm )

chep bai vui khong may ban :)) 

\(x+y+z=0\)

⇔\(-x=y+z\)

⇔\(x^2=\left(y+z\right)^2\) 

⇔\(x^2=y^2+2yz+z^2\) 

⇔\(y^2+z^2-x^2=-2yz\)

Tương tự:

\(z^2+x^2-y^2=-2zx\)

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

➞ S = \(\dfrac{1}{-2xy}+\dfrac{1}{-2yz}+\dfrac{1}{-2zx}=\dfrac{x+y+z}{-2xyz}=0\) 

Vậy S = 0

Ta có:

\(x+y+z=0\)

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

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

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

Tương tự ta được:
\(S=\frac{1}{-2xy}+\frac{1}{-2yz}+\frac{1}{-2zx}=-\frac{1}{2}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)=-\frac{1}{2}\cdot\frac{x+y+z}{xyz}=0\)

Vậy S=0

Câu hỏi của Winkies:bạn tham khảo tại đây nhé!

Ta có:\(x^2+4y+4=0;y^2+4z+4=0;z^2+4x+4=0\)

\(\Leftrightarrow\left(x^2+4y+4\right)+\left(y^2+4z+4\right)+\left(z^2+4x+4\right)=0\)

\(\Leftrightarrow x^2+4x+4+y^2+4y+4+z^2+4z+4=0\)

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

Mà\(\left(x+2\right)^2\ge0;\left(y+2\right)^2\ge0;\left(z+2\right)^2\ge0\)

\(\Leftrightarrow\left(x+2\right)^2+\left(y+2\right)^2+\left(z+2\right)^2\ge0\)

Dấu "=" xảy ra\(\Leftrightarrow\hept{\begin{cases}x+2=0\\y+2=0\\z+2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-2\\y=-2\\z=-2\end{cases}\Leftrightarrow}x=y=z=-2}\)

Vậy\(x^{10}+y^{10}+z^{10}=x^{10}+x^{10}+x^{10}\)                         

                    \(=3\cdot x^{10}=3\cdot\left(-2\right)^{10}=3\cdot1024=3072\)

1/x + 1/y + 1/z = 13

<=> yz/x + xy/z + zx/y = 13

<=> xyz/x^2 + xyz/y^2 + xyz/z^2 = 13

<=> (x+y+z)(1/x^2 + 1/y^2 + 1/z^2) = 13

<=> 1/x^2 + 1/y^2 + 1/z^2 = 13/(x+y+z)

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