Áp dụng tính chất của dãy tỉ số = nhau ta có:

\(\frac{y-2x+4z}{2x}=\frac{z-2y+4x}{2y}=\frac{x-2z+4y}{2z}=\)\(=\frac{\left(y-2x+4z\right)+\left(z-2y+4x\right)+\left(x-2z+4y\right)}{2x+2y+2z}=\frac{3\left(x+y+z\right)}{2\left(x+y+z\right)}=\frac{3}{2}\)

\(\Rightarrow\left\{\begin{matrix}2\left(y-2x+4z\right)=6x\\2\left(z-2y+4x\right)=6y\\2\left(x-2z+4y\right)=6z\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}y-2x+4z=3x\\z-2y+4x=3y\\x-2z+4y=3z\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}y+4z=5x\\z+4x=5y\\x+4y=5z\end{matrix}\right.\)

\(P=\left(2+\frac{x}{2y}\right)\left(2+\frac{y}{2z}\right)\left(2+\frac{z}{2x}\right)\)

\(P=\frac{4y+x}{2y}.\frac{4z+y}{2z}.\frac{4x+z}{2x}=\frac{5z}{2y}.\frac{5x}{2z}.\frac{5y}{2x}=\frac{125}{8}\)

Ta có:

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

\(\Leftrightarrow\frac{yz+zx+xy}{xyz}=0\) (Quy đồng)

\(\Rightarrow yz+zx+xy=0\)

Vì:

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

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

Nên.....(tự kết luận nha)

giải chi tiết ( vì sao ) đoạn dưới đây = 0 hộ mk vs :

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

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

-Ta có:

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

\(\Leftrightarrow xy+yz+zx=0\)

Đặt \(xy=a,yz=b,zx=c\) thì bài toán thành

Cho \(a+b+c=0\)chứng minh \(\left(a^2+b^2+c^2\right)^2=2\left(a^4+b^4+c^4\right)\)

Ta có:

\(\left(a^2+b^2+c^2\right)^2-2\left(a^4+b^4+c^4\right)\)

\(=2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4\)

\(=c^2\left(a+b\right)^2+c^2\left(a-b\right)^2-\left(a^2-b^2\right)^2-c^4\)

\(=c^2\left[\left(a+b\right)^2-c^2\right]+\left(a-b\right)^2\left[c^2-\left(a+b\right)^2\right]\)

\(=c^2\left(a+b+c\right)\left(a+b-c\right)+\left(a-b\right)^2\left(a+b+c\right)\left(c-a-b\right)\)

\(=\left(a+b+c\right)\left(a+b-c\right)\left[c^2-\left(a-b\right)^2\right]=0\)

Vậy \(\left(a^2+b^2+c^2\right)^2=2\left(a^4+b^4+c^4\right)\)

Từ giả thiết \(x+y+z=xyz\Leftrightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1\)

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

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

Trần Việt Linh vào giúp bạn này đi

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 :)) 

Đề bài thực chất thiếu điều kiện \(xyz\ne0.\) Bây giờ ta sẽ giải bài toán với thêm điều kiện bổ sung này:

Theo giả thiết \(x+y+z=xyz\Leftrightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1.\)

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

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

Từ đó suy ra vế trái bằng \(\frac{xyz\left(y+z\right)+2xyz\left(z+x\right)+3xyz\left(x+y\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{xyz\left(5x+4y+3z\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}.\)   (ĐPCM).

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

\(=\frac{xyz}{xy\left(\frac{1}{x}+\frac{1}{y}\right)zx\left(\frac{1}{z}+\frac{1}{x}\right)}=\frac{xyz}{\left(x+y\right)\left(z+x\right)}\)

Tương tự, ta cũng có: \(\frac{2y}{1+y^2}=\frac{2xyz}{\left(x+y\right)\left(y+z\right)}\)\(;\)\(\frac{3z}{1+z^2}=\frac{3xyz}{\left(y+z\right)\left(z+x\right)}\)

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

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