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theo đầu bài ta có\(\dfrac{x^2+y^2}{xy}=\dfrac{10}{3}\)=>\(3x^2+3y^2=10xy\)
A=\(\dfrac{x-y}{x+y}\)
=>\(A^2=\left(\dfrac{x-y}{x+y}\right)^2=\dfrac{x^2-2xy+y^2}{x^2+2xy+y^2}=\dfrac{3x^2-6xy+3y^2}{3x^2+6xy+3y^2}=\dfrac{10xy-6xy}{10xy+6xy}=\dfrac{4xy}{16xy}=\dfrac{1}{4}\)
=>A=\(\sqrt{\dfrac{1}{4}}=\dfrac{-1}{2}hoặc\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}\) (cộng trừ căn 1/4 nhé)
vì y>x>0=> A=-1/2
\(\Leftrightarrow\left(x^2-4xy+4y^2\right)+\left(y^2-6y+9\right)=5\)
\(\Leftrightarrow\left(x-2y\right)^2+\left(y-3\right)^2=5\)
\(\Leftrightarrow\left(x-2y\right)^2=5-\left(y-3\right)^2\) (1)
Do \(\left(x-2y\right)^2\ge0;\forall x;y\)
\(\Rightarrow5-\left(y-3\right)^2\ge0\Rightarrow\left(y-3\right)^2\le5\)
\(\Rightarrow\left[{}\begin{matrix}\left(y-3\right)^2=0\\\left(y-3\right)^2=1\\\left(y-3\right)^2=4\end{matrix}\right.\)
Thay vào (1):
- Với \(\left(y-3\right)^2=0\) \(\Rightarrow\left(x-2y\right)^2=5\) vô nghiệm do 5 ko phải SCP
- Với \(\left(y-3\right)^2=1\Rightarrow\left[{}\begin{matrix}y=4\\y=2\end{matrix}\right.\)
\(y=4\Rightarrow\left(x-8\right)^2=4\Rightarrow\left[{}\begin{matrix}x=10\\x=6\end{matrix}\right.\)
\(y=2\Rightarrow\left(x-4\right)^2=4\Rightarrow\left[{}\begin{matrix}x=6\\x=2\end{matrix}\right.\)
- Với \(\left(y-3\right)^2=4\Rightarrow\left[{}\begin{matrix}y=5\\y=1\end{matrix}\right.\)
\(y=5\Rightarrow\left(x-10\right)^2=1\Rightarrow\left[{}\begin{matrix}x=11\\x=9\end{matrix}\right.\)
\(y=1\Rightarrow\left(x-2\right)^2=1\Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
Em tự kết luận các cặp nghiệm
Chắc phải là cặp số nguyên chứ có vô số cặp x;y bất kì thỏa mãn pt này
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\(x^{10}+y^{10}+z^{10}=x^{10}+x^{10}+x^{10}\)
\(=3\cdot x^{10}=3\cdot\left(-2\right)^{10}=3\cdot1024=3072\)
Vì \(\left|x-1\right|\ge0\)
\(\left(y+2\right)^{2016}\ge0\)
=> \(\left|x-1\right|+\left(y+2\right)^{2016}=0\)
\(\Leftrightarrow\begin{cases}x-1=0\\y+2=0\end{cases}\)\(\Leftrightarrow\begin{cases}x=1\\y=-2\end{cases}\)
Có: \(2x^5-5y^3+2017=2\cdot1^5-5\cdot\left(-2\right)^3+2017=2059\)
3x^2+3y^2+4xy-2x+2y+2=0
=>2x^2+4xy+2y^2+x^2-2x+1+y^2+2y+1=0
=>x=1 và y=-1
M=(1-1)^2017+(1-2)^2018+(-1+1)^2015=1
Ta có : x2 - xy + y2 + 1
\(=x^2-2x.\frac{y}{2}+\frac{y^2}{4}+\frac{3y^2}{4}+1\)
\(=\left(x-\frac{y}{2}\right)^2+\left(\frac{3y}{2}\right)^2+1\)
Mà \(\left(x-\frac{y}{2}\right)^2\ge0\forall x\)
\(\left(\frac{3y}{2}\right)^2\ge0\forall x\)
Nên \(\left(x-\frac{y}{2}\right)^2+\left(\frac{3y}{2}\right)^2+1\ge1\forall x\)
Vậy \(\left(x-\frac{y}{2}\right)^2+\left(\frac{3y}{2}\right)^2+1>0\forall x\)
Hay : x2 - xy + y2 + 1 > 0 \(\forall x\)
\(x+y+z=0\)
⇔\(-x=y+z\)
⇔\(x^2=\left(y+z\right)^2\)
⇔\(x^2=y^2+2yz+z^2\)
⇔\(y^2+z^2-x^2=-2yz\)
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 \(x^2+3y^2=4xy\)
\(\Leftrightarrow x^2-xy-3xy+3y^2=0\)
\(\Leftrightarrow\left(x-y\right)\left(x-3y\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-y=0\\x-3y=0\end{cases}}\)
Vì x>y nên \(x-y\ne0\)\(\Rightarrow x-3y=0\Rightarrow x=3y\)
A= \(\frac{2x+5y}{x-2y}=\frac{11y}{y}=11\)
Thank you very much