\(x^2+2y^2-2xy+4y+3< 0\)

\(\Rightarrow x^2-2xy+y^2+y^2+4y+4-1< 0\)  

\(\Rightarrow\left(x^2-2xy+y^2\right)+\left(y^2+4y+4\right)-1< 0\)

\(\Rightarrow\left(x-y\right)^2+\left(y+2\right)^2-1< 0\)

Mà: \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\forall x,y\\\left(y+2\right)^2\ge0\forall y\end{matrix}\right.\) 

\(\Rightarrow\left(x-y\right)^2+\left(y+2\right)^2-1\ge-1\forall x,y\)

Mặt khác: \(\left(x-y\right)^2+\left(y+2\right)^2-1< 0\)

Dấu "=" xảy ra:

\(\left\{{}\begin{matrix}x-y=0\\y+2=0\end{matrix}\right.\)

\(\Rightarrow x=y=-2\)

Vậy: .... 

Cảm ơn anh/chị/bạn nhiều ạ!

bn ko nói là tìm x hay y hay xy à

`xy-x-y=0`

`<=>xy-x-y+1=1`

`<=> x(y-1)-(y-1)=1`

`<=> (y-1)(x-1)=1`

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y-1=1\\x-1=1\end{matrix}\right.\\\left\{{}\begin{matrix}y-1=-1\\x-1=-1\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=2\\x=2\end{matrix}\right.\\\left\{{}\begin{matrix}y=0\\x=0\end{matrix}\right.\end{matrix}\right.\)

Vì \(\left\{{}\begin{matrix}\left|2x-27\right|^{2011}\text{≥0,∀x}\\\left(3y+10\right)^{2012}\text{≥0,∀y}\end{matrix}\right.\)

⇒ \(\left|2x-27\right|^{2011}+\left(3y+10\right)^{2012}\text{≥0,∀x},y\)

Dấu "=" ⇔ \(\left\{{}\begin{matrix}2x-27=0\\3y+10=0\end{matrix}\right.\)

⇒ \(\left\{{}\begin{matrix}x=\dfrac{27}{2}\\y=-\dfrac{10}{3}\end{matrix}\right.\)

Vậy ...

Cảm ơn a

Ta có: \(\left(x-\dfrac{1}{5}\right)^{2004}\ge0\forall x\)

\(\left(y+0.4\right)^{100}\ge0\forall y\)

\(\left(z-3\right)^{678}\ge0\forall z\)

Do đó: \(\left(x-\dfrac{1}{5}\right)^{2004}+\left(y+0.4\right)^{100}+\left(z-3\right)^{678}\ge0\forall x,y,z\)

Dấu '=' xảy ra khi

\(\left\{{}\begin{matrix}x-\dfrac{1}{5}=0\\y+0.4=0\\z-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=-\dfrac{2}{5}\\z=3\end{matrix}\right.\)

Vậy: (x,y,z)=\(\left(\dfrac{1}{5};-\dfrac{2}{5};3\right)\)

Vì |3,7 - x| > 0

=> |3,7 - x| + 2,5 > 2,5

=> A > 2,5

Dấu "=" xảy ra 

<=> |3,7 - x| = 0

<=> 3,7 - x = 0

<=> x = 3,7

KL: A_min = 2,5 <=> x = 3,7

Vì |x + 1,5| > 0

=> |x + 1,5| - 4,5 > - 4,5

=> B > - 4,5

Dấu "=" xảy ra

<=> |x + 1,5| = 0

<=> x+1,5 = 0

<=> x = -1,5

KL: B_min = -4,5 <=> x = -1,5

Do \(\left|x\right|,\left|x^2+x\right|\ge0\forall x\)

\(\Rightarrow\left\{{}\begin{matrix}x=0\\x^2+x=0\end{matrix}\right.\)

\(\Rightarrow x=0\)

x = 0

Vì bất cứ số nào nhân hay chia với o đều = 0

-2,5 + |3x + 5| = -1,5

|3x + 5| = -1,5 + 2,5

|3x + 5| = 1

Với x -5/3 ta có:

3x + 5 = 1

3x = 1 - 5

3x = -4

x = -4/3 (nhận)

Với x < -5/3 ta có:

3x + 5 = -1

3x = -1 - 5

3x = -6

x = -6/3

x = -2 (nhận)

Vậy x = -2; x = -4/3

a, Ta có

 \(\left|x-1,7\right|=2,3\\ \Rightarrow\left[{}\begin{matrix}x-1,7=2.3\\x-1.7=-2,3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-0,6\end{matrix}\right.\)

   Vậy....

b, Ta có :

\(\left|x+\dfrac{3}{4}\right|-\dfrac{1}{3}=0\\ \Rightarrow\left|x+\dfrac{3}{4}\right|=\dfrac{1}{3}\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=\dfrac{1}{3}\\x+\dfrac{3}{4}=-\dfrac{1}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{12}\\x=-\dfrac{13}{12}\end{matrix}\right.\)

    Vậy...

=> 1,5-x = 0 và 2,5-x = 0

x = 1,5-0 và x = 2,5-0

x = 1,5 và x = 2,5

=> không có x thỏa mãn