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![](https://rs.olm.vn/images/avt/0.png?1311)
Vì bài dài nên mình sẽ tách ra nhé.
1a. Ta có:
$x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)=-2(xy+yz+xz)$
$x^3+y^3+z^3=(x+y+z)^3-3(x+y)(y+z)(x+z)=-3(x+y)(y+z)(x+z)$
$=-3(-z)(-x)(-y)=3xyz$
$\Rightarrow \text{VT}=-30xyz(xy+yz+xz)(1)$
------------------------
$x^5+y^5=(x^2+y^2)(x^3+y^3)-x^2y^2(x+y)$
$=[(x+y)^2-2xy][(x+y)^3-3xy(x+y)]-x^2y^2(x+y)$
$=(z^2-2xy)(-z^3+3xyz)+x^2y^2z$
$=-z^5+3xyz^3+2xyz^3-6x^2y^2z+x^2y^2z$
$=-z^5+5xyz^3-5x^2y^2z$
$\Rightarrow 6(x^5+y^5+z^5)=6(5xyz^3-5x^2y^2z)$
$=30xyz(z^2-xy)=30xyz[z(-x-y)-xy]=-30xyz(xy+yz+xz)(2)$
Từ $(1);(2)$ ta có đpcm.
1b.
$x^4+y^4=(x^2+y^2)^2-2x^2y^2=[(x+y)^2-2xy]^2-2x^2y^2$
$=(z^2-2xy)^2-2x^2y^2=z^4+2x^2y^2-4xyz^2$
$x^3+y^3=(x+y)^3-3xy(x+y)=-z^3+3xyz$
Do đó:
$x^7+y^7=(x^4+y^4)(x^3+y^3)-x^3y^3(x+y)$
$=(z^4+2x^2y^2-4xyz^2)(-z^3+3xyz)+x^3y^3z$
$=7x^3y^3z-14x^2y^2z^3+7xyz^5-z^7$
$\Rightarrow \text{VT}=7x^3y^3z-14x^2y^2z^3+7xyz^5$
$=7xyz(x^2y^2-2xyz^2+z^4)$
$=7xyz(xy-z^2)$
$=7xyz[xy+z(x+y)]^2=7xyz(xy+yz+xz)^2$
$=7xyz[x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)]$
$=7xyz(x^2y^2+y^2z^2+z^2x^2)$ (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 2/
\(\frac{1}{x^2\left(x^2+y^2\right)}+\frac{1}{\left(x^2+y^2\right)\left(x^2+y^2+z^2\right)}+\frac{1}{x^2\left(x^2+y^2+z^2\right)}=1\)
Điều kiện \(\hept{\begin{cases}x^2\ne0\\x^2+y^2\ne0\\x^2+y^2+z^2\ne0\end{cases}}\)
Xét \(x^2,y^2,z^2\ge1\)
Ta có: \(\hept{\begin{cases}x^2\ge1\\x^2+y^2\ge2\end{cases}}\)
\(\Rightarrow x^2\left(x^2+y^2\right)\ge2\)
\(\Rightarrow\frac{1}{x^2\left(x^2+y^2\right)}\le\frac{1}{2}\left(1\right)\)
Tương tự ta có: \(\hept{\begin{cases}\frac{1}{\left(x^2+y^2\right)\left(x^2+y^2+z^2\right)}\le\frac{1}{6}\left(2\right)\\\frac{1}{x^2\left(x^2+y^2+z^2\right)}\le\frac{1}{3}\left(3\right)\end{cases}}\)
Cộng (1), (2), (3) vế theo vế ta được
\(\frac{1}{x^2\left(x^2+y^2\right)}+\frac{1}{\left(x^2+y^2\right)\left(x^2+y^2+z^2\right)}+\frac{1}{x^2\left(x^2+y^2+z^2\right)}\le\frac{1}{2}+\frac{1}{6}+\frac{1}{3}=1\)
Dấu = xảy ra khi \(x^2=y^2=z^2=1\)
\(\Rightarrow\left(x,y,z\right)=?\)
Xét \(\hept{\begin{cases}x^2\ge1\\y^2=z^2=0\end{cases}}\) thì ta có
\(\frac{1}{x^4}+\frac{1}{x^4}+\frac{1}{x^4}=1\)
\(\Leftrightarrow x^4=3\left(l\right)\)
Tương tự cho 2 trường hợp còn lại: \(\hept{\begin{cases}x^2,y^2\ge1\\z^2=0\end{cases}}\) và \(\hept{\begin{cases}x^2,z^2\ge1\\y^2=0\end{cases}}\)
Bài 2/
Ta có: \(\frac{x}{y}+\frac{y}{z}+\frac{z}{t}+\frac{t}{x}\ge4\sqrt[4]{\frac{x}{y}.\frac{y}{z}.\frac{z}{t}.\frac{t}{x}}=4>3\)
Vậy phương trình không có nghiệm nguyên dương.
![](https://rs.olm.vn/images/avt/0.png?1311)
d)
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+.....+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)=\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+.....-\dfrac{1}{x+99}+\dfrac{1}{x+100}\)=\(\dfrac{1}{x}-\dfrac{1}{x+100}\)
=\(\dfrac{x+100}{x\left(x+100\right)}-\dfrac{x}{x\left(x+100\right)}\)
=\(\dfrac{x+100-x}{x\left(x+100\right)}=\dfrac{100}{x\left(x+100\right)}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1: \(=\dfrac{\left(x^2+2xy+y^2\right)-1}{\left(x^2+2x+1\right)-y^2}\)
\(=\dfrac{\left(x+y+1\right)\left(x+y-1\right)}{\left(x+1-y\right)\left(x+1+y\right)}=\dfrac{x+y-1}{x-y+1}\)
2: \(=\dfrac{\left(x^2-y^2\right)\left(x^2+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}\)
\(=\dfrac{\left(x-y\right)\left(x^2+y^2\right)}{x^2-xy+y^2}\)
3: \(=\dfrac{\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz}{2x^2+2y^2+2z^2-2xy-2yz-2xz}\)
\(=\dfrac{\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)}{2\left(x^2+y^2+z^2-xy-yz-xz\right)}\)
\(=\dfrac{x+y+z}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=\dfrac{1}{\left(x-y\right)\left(y-z\right)}-\dfrac{1}{\left(y-z\right)\left(x-z\right)}-\dfrac{1}{\left(x-y\right)\left(x-z\right)}\)
\(=\dfrac{x-z-x+y-y+z}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}=0\)
b: \(=\dfrac{1}{x\left(x-y\right)\left(x-z\right)}-\dfrac{1}{y\left(x-y\right)\left(y-z\right)}+\dfrac{1}{z\left(x-z\right)\left(y-z\right)}\)
\(=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{y^2z-yz^2-x^2z+xz^2+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{z\left(y^2-x^2\right)-z^2\left(y-x\right)-xy\left(y-x\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{\left(x-y\right)\left[-z\left(x+y\right)+z^2+xy\right]}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{-zx-zy+z^2+xy}{xyz\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{z\left(z-x\right)-y\left(z-x\right)}{xyz\left(y-z\right)\left(x-z\right)}=\dfrac{1}{xyz}\)
phương trình 1 có nhiều ẩn thế bn
Câu 2:, ta có
Xét x=1, ...
Xét x khác 1 ...
\(y=\frac{x^2+2}{x-1}=\frac{x^2-1+3}{x-1}=x+1+\frac{3}{x-1}\)
và y là số nguyên => x-1 llà ước của 3, đến đây tự giải nhé
^_^