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 :)) 

a) \(x\left(x^2-2x\right)+\left(x-2x\right)=x^2\left(x-2\right)+x\left(x-2\right)=\left(x-2\right)\left(x^2+x\right)⋮x-2\forall x,y\in Z\)

b) \(x^3y^2-3yx^2+xy=xy\left(x^2y-3x+1\right)⋮xy\forall x,y\in Z\)

c) \(x^3y^2-3x^2y^3+xy^2=xy^2\left(x^2-3xy+1\right)⋮\left(x^2-3xy+1\right)\forall x,y\in Z\)

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

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

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 ) 

Ta có: \(x+y+z=1\) nên:

\(\Rightarrow y+z=1-x\)

Thay \(y+z=1-x\) và áp dụng BĐT \(\left(a+b\right)^2\ge4ab\) ta được:

\(4\left(1-x\right)\left(1-y\right)\left(1-z\right)=4\left(y+z\right)\left(1-y\right)\left(1-z\right)\le\left[\left(y+z\right)+\left(1-z\right)\right]^2\left(1-y\right)\)

\(\Rightarrow4\left(y+z\right)\left(1-y\right)\left(1-z\right)\le\left(1+y\right)^2\left(1-y\right)=\left(1+y\right)\left(1-y^2\right)\le1+y\)

\(\Rightarrow4\left(1-x\right)\left(1-y\right)\left(1-z\right)\le1+y=x+2y+z\left(đpcm\right)\)

Áp dụng BĐT AM - GM, ta có:

\(4\left(1-x\right)\left(1-y\right)\left(1-z\right)\)

\(=4\left(y+z\right)\left(x+z\right)\left(x+y\right)\)

\(\le\frac{\left(x+y+y+z\right)^2}{4}\times4\left(x+z\right)\)

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

\(\le\left(x+2y+z\right)\times\frac{\left(x+2y+z+x+z\right)^2}{4}\)

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

\(=x+2y+z\left(\text{đ}pcm\right)\)

Dấu "=" xảy ra khi a = b = c = \(\frac{1}{3}\)

Dấu = xảy ra:\(\hept{\begin{cases}x=z=\frac{1}{2}\\y=0\end{cases}}\)

`@` `\text {Ans}`

`\downarrow`

 Viết các biểu thức sau dưới dạng hiệu chứ ạ?

`e,`

`(x+1)(x-1)`

`= x(x-1) + x - 1`

`= x^2 - x + x - 1`

`= x^2 - 1`

`f,`

`(x-2y)(x+2y)?`

`= x(x+2y) - 2y(x+2y)`

`= x^2 + 2xy - 2xy - 4y^2`

`= x^2 - 4y^2`

`g,`

`(x+y+z)(x-y-z)`

`= x(x-y-z) + y(x-y-z) + z(x-y-z)`

`= x^2 - xy - xz + xy - y^2 - yz + xz - yz - z^2`

`= x^2 - y^2 - z^2 - 2yz`

`h,`

`(x-y+z)(x+y+z)`

`= x(x+y+z) - y(x+y+z) + z(x+y+z)`

`= x^2 + xy + xz - xy - y^2 - yz + xz + yz + z^2`

`= x^2 - y^2 + z^2 + 2xz`

Câu này c xem lại đề.