Đề đây ạ:

Tìm các số nguyên x và y sao cho (x-3)(x+y)=7

Đặt \(\dfrac{x-y}{z}=m,\dfrac{y-z}{x}=n,\dfrac{z-x}{y}=p\), ta có:

\(\left(m+n+p\right)\left(\dfrac{1}{m}+\dfrac{1}{n}+\dfrac{1}{p}\right)=3+\dfrac{n+p}{m}+\dfrac{p+m}{n}+\dfrac{m+n}{p}\)

Tính \(\dfrac{n+p}{m}\) theo x, y, z ta được:

\(\dfrac{n+p}{m}=\dfrac{z}{x-y}.\dfrac{y^2-yz+xz-x^2}{xy}=\dfrac{z}{xy}\left(-x-y+x\right)\)

           \(=\dfrac{z}{xy}\left(-x-y-z+2z\right)=\dfrac{2x^2}{xy}\) vì \(\left(x+y+z\right)=0\)

Tương tự:    \(\dfrac{m+p}{n}=\dfrac{2x^2}{yz}.\dfrac{m+n}{p}=\dfrac{2y^2}{xz}\)

Vậy \(\left(m+n+p\right)\left(\dfrac{1}{m}+\dfrac{1}{n}+\dfrac{1}{p}\right)=3+\dfrac{2\left(x^3+y^3+z^3\right)}{xyz}=3+\dfrac{2.3xyz}{xyz}=3+6=9\)

tích mình đi

ai tích mình

mình ko tích lại đâu

thanks

tích mình đi

ai tích mình 

mình tích lại 

thanks

hs minh

Đặt \(P=\left(\dfrac{x-y}{z}+\dfrac{y-z}{x}+\dfrac{z-x}{y}\right)\left(\dfrac{z}{x-y}+\dfrac{x}{y-z}+\dfrac{y}{z-x}\right)=9\)

Đặt \(\left\{{}\begin{matrix}\dfrac{x-y}{z}=a\\\dfrac{y-z}{x}=b\\\dfrac{x-z}{y}=c\end{matrix}\right.\)

\(\Leftrightarrow P=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\\ =1+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{a}+1+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{c}{b}+1\\ =3+\dfrac{a+c}{b}+\dfrac{a+b}{c}+\dfrac{b+c}{a}\)

Ta có \(\dfrac{a+c}{b}=\dfrac{\dfrac{x-y}{z}+\dfrac{z-x}{y}}{\dfrac{y-z}{x}}=\dfrac{xy-y^2+z^2-xz}{yz}\cdot\dfrac{x}{y-z}\)

\(=\dfrac{\left(z-y\right)\left(y+z-x\right)x}{yz\left(y-z\right)}=\dfrac{x\left(x-y-z\right)}{yz}\)

Mà \(x+y+z=0\Leftrightarrow x=-y-z\)

\(\Leftrightarrow\dfrac{a+c}{b}=\dfrac{x\left(x+x\right)}{yz}=\dfrac{2x^2}{yz}\)

Cmtt ta được \(\dfrac{a+b}{c}=\dfrac{2y^2}{xz};\dfrac{b+c}{a}=\dfrac{2z^2}{xy}\)

Cộng vế theo vế

\(\Leftrightarrow P=\dfrac{2x^2}{yz}+\dfrac{2y^2}{xz}+\dfrac{2z^2}{xy}+3=\dfrac{2x^3+2y^3+2z^3}{xyz}+3\\ \Leftrightarrow P=\dfrac{2\left(x^3+y^3+z^3\right)}{xyz}+3\)

Lại có \(x+y+z=0\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)=0\)

\(\Leftrightarrow x^3+y^3+z^3-3xyz=0\\ \Leftrightarrow x^3+y^3+z^3=3xyz\)

Thế vào \(P\)

\(\Leftrightarrow P=\dfrac{2\cdot3xyz}{xyz}+3=6+3=9\)

Ta co: \(\hept{\begin{cases}x^2-y+\frac{1}{4}=0\\y^2-x+\frac{1}{4}=0\end{cases}}\)

\(\Rightarrow x^2-x+\frac{1}{4}+y^2-y+\frac{1}{4}=0\)

\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}x-\frac{1}{2}=0\\y-\frac{1}{2}=0\end{cases}\Rightarrow x=y=\frac{1}{2}}\)

Vậy \(x=y=\frac{1}{2}\)

Ta có: \(\hept{\begin{cases}x^2-y+\frac{1}{4}=0\\y^2-x+\frac{1}{4}=0\end{cases}}\)

\(\Rightarrow\left(x^2-x+\frac{1}{4}\right)+\left(y^2-y+\frac{1}{4}\right)=0\)

\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}x-\frac{1}{2}=0\\y-\frac{1}{2}=0\end{cases}\Rightarrow x=y=\frac{1}{2}}\)

Vậy \(x=y=\frac{1}{2}\)

\(P=\sum\dfrac{1}{x+y+1}\ge\dfrac{9}{2\left(x+y+z\right)+3}=\dfrac{9}{2.1+3}=\dfrac{9}{5}\)

Dấu \("="\Leftrightarrow x=y=z=\dfrac{1}{3}\)

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

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

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

Mà x;y;z>0 \(\Rightarrow x+y+z>0\)

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

hreury

Chọn đáp án A.