VÌ \(\left(x-1\right)^{2012}\ge0\)

\(\left(y-2\right)^{2010}\ge0\)

\(\left(x-z\right)^{2008}\ge0\)

nên dấu \(=\)xảy ra khi \(\hept{\begin{cases}x=z\\x=1\\y=2\end{cases}\Leftrightarrow\hept{\begin{cases}x=z=1\\y=2\end{cases}}}\)

a.  x=1      y= -3

b.  x=5      y=7/2

c.  x= -1    y= -1/2

d.  x=1/4   y= 1/4

a) x = 1    

y = -3

b) x = 5

y = 7/2

c) x = -1

y = -1/2

d) x = 1/4 

y = 1/4

nha bn

Ta có | x + 2/5 | ≥ 0 ∀ x

         | 2y - 1/3 | ≥ 0 ∀ y

=> | x + 2/5 | + | 2y - 1/3 | ≥ 0 ∀ x, y

Đẳng thức xảy ra <=> \(\hept{\begin{cases}x+\frac{2}{5}=0\\2y-\frac{1}{3}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{2}{5}\\y=\frac{1}{6}\end{cases}}\)

Vậy x = -2/5 ; y = 1/6

\(\left|x+\frac{2}{5}\right|+\left|2y-\frac{1}{3}\right|=0\)

\(\orbr{\begin{cases}\left|x+\frac{2}{5}\right|=0\\\left|2y-\frac{1}{3}\right|=0\end{cases}}\)

\(\orbr{\begin{cases}x=0-\frac{2}{5}\\2y=0+\frac{1}{3}\end{cases}}\)

\(\orbr{\begin{cases}x=-\frac{2}{5}\\2y=\frac{1}{3}\end{cases}}\)

\(x=\frac{1}{3}:2\)

\(x=\frac{2}{3}\)

vậy \(\orbr{\begin{cases}x=-\frac{2}{5}\\x=\frac{2}{3}\end{cases}}\)

nhầm nha

\(\left|x+\frac{2}{5}\right|+\left|2y-\frac{1}{3}\right|=0\)

\(\orbr{\begin{cases}\left|x+\frac{2}{5}\right|=0\\\left|2y-\frac{1}{3}\right|=0\end{cases}}\)

\(\orbr{\begin{cases}x=0-\frac{2}{5}\\2y=0+\frac{1}{3}\end{cases}}\)

\(\orbr{\begin{cases}x=-\frac{2}{5}\\2y=\frac{1}{3}\end{cases}}\)

\(y=\frac{1}{3}:2\)

\(y=\frac{1}{6}\)

vậy \(\orbr{\begin{cases}x=-\frac{2}{5}\\y=\frac{1}{6}\end{cases}}\)

vậy 

CM : ( 3x - 2y )^2010 = 0 ,  / 5y - 6z /^2011 = 0 

=> 3x - 2y = 0    ,  5y - 6z = 0 

=> 3x = 2y     , 5y = 6z 

=> x/2 = y/3    , y/6 = z/5 

=> x/4 = y/6    , y/6 =z/5 

=> x/4 = y/6 = z/5 

=> 2x/ 8 , 5y/30 , 3z/15

Áp dụng tính chất DTSBN , ta có : 

2x/8 = 5y /30 = 3z / 15 = 2x - 5y + 3z / 8 - 30 + 15 = 54/-7 = -54 /7 

Rồi tính ra là xong 

áp dụng BDT AM-GM \(=>x+y\ge2\sqrt{xy}=>\left(x+y\right)^2\ge4xy\left(1\right)\)

mà \(x+y\le1=>\left(x+y\right)^2\le1\left(2\right)\)

(1)(2)\(=>4xy\le\left(x+y\right)^2\le1=>4xy\le1=>xy\le\dfrac{1}{4}\)

\(A=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\sqrt{1+x^2y^2}\ge2\sqrt{\dfrac{1+x^2y^2}{xy}}=2\sqrt{\dfrac{1}{xy}+xy}\)

\(=2\sqrt{\dfrac{1}{xy}+16xy-15xy}=2\sqrt{2\sqrt{16}-\dfrac{15}{4}}=\sqrt{17}\)

dấu"=" xảy ra<=>\(x=y=\dfrac{1}{2}\)

\(1\ge x+y\ge2\sqrt{xy}\Rightarrow xy\le\dfrac{1}{4}\Rightarrow\dfrac{1}{xy}\ge4\)

Ta có:

\(A\ge\dfrac{2}{\sqrt{xy}}.\sqrt{1+x^2y^2}=2\sqrt{\dfrac{1}{xy}+xy}=2\sqrt{\left(xy+\dfrac{1}{16xy}\right)+\dfrac{15}{16}.\dfrac{1}{xy}}\)

\(A\ge2\sqrt{2\sqrt{\dfrac{xy}{16xy}}+\dfrac{15}{16}.4}=\sqrt{17}\)

\(A_{min}=\sqrt{17}\) khi \(x=y=\dfrac{1}{2}\)

thông điệp nhỏ:

hay kkhi ko muốn k

\(\left\{{}\begin{matrix}x^3y^2+x^2y^3+x^3y+2x^2y^2+xy^3-30=0\\x^2y+xy^2+xy+x+y-11=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2y^2\left(x+y\right)+xy\left(x+y\right)^2-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)\left[xy+x+y\right]-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}xy\left(x+y\right)=u\\xy+x+y=v\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}uv-30=0\\u+v-11=0\end{matrix}\right.\)  \(\Rightarrow\left(u;v\right)=\left(6;5\right);\left(5;6\right)\)

TH1: \(\left\{{}\begin{matrix}xy\left(x+y\right)=6\\xy+x+y=5\end{matrix}\right.\)

Theo Viet đảo \(\Rightarrow\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(1;2\right);\left(2;1\right)\)hoặc \(\left\{{}\begin{matrix}x+y=2\\xy=3\end{matrix}\right.\)(vô nghiệm)

TH2: \(\left\{{}\begin{matrix}xy\left(x+y\right)=5\\xy+x+y=6\end{matrix}\right.\) 

\(\Rightarrow\left\{{}\begin{matrix}x+y=5\\xy=1\end{matrix}\right.\) \(\Rightarrow...\) hoặc \(\left\{{}\begin{matrix}x+y=1\\xy=5\end{matrix}\right.\) (vô nghiệm)

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