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

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

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

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

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: .... 

a: \(\Rightarrow10x^2+9x-\left(10x^2+15x-2x-3\right)=8\)

\(\Leftrightarrow10x^2+9x-10x^2-13x+3=8\)

=>-4x=5

hay x=-5/4

b: \(\Leftrightarrow21x-15x^2-35+25x+15x^2-10x+6x-4-2=0\)

=>42x=41

hay x=41/42

`a)(10x+9)x-(5x-1)(2x+3)=8`

`<=>10x^2+9x-10x^2-15x+2x+3=8`

`<=>-4x=5`

`<=>x=-5/4`     Vậy `S={-5/4}`

`b)(3x-5)(7-5x)+(5x+2)(3x-2)-2=0`

`<=>21x-15x^2-35+25x+15x^2-10x+6x-4-2=0`

`<=>42x=41`

`<=>x=41/42`       Vậy `S={41/42}`

a: 

=>-4x=5

hay x=-5/4

b: 

=>42x=41

hay x=41/42

\(\left(3x-1\right)^2.\left(x+5\right)^2=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-5\end{matrix}\right.\)

^2: Nếu trong ngoặc bằng 0 thì bình phương cũng bằng 0 và ngược lại

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 ...

a: =>3x+3=4x-4

=>-x=-7

hay x=7(nhận)

b: (x-1)(x-3)=0

=>x-1=0 hoặc x-3=0

=>x=1 hoặc x=3

c: 2(x-1)+x=0

=>2x-2+x=0

=>3x-2=0

hay x=2/3

a, ĐKXĐ : x ≠ 1 ; x ≠ -1

\(\Rightarrow3\left(x+1\right)=4\left(x-1\right)\)

\(\Leftrightarrow3x+3=4x-4\)

\(\Leftrightarrow-x=-7\)

\(\Leftrightarrow x=7\left(N\right)\)

b,

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

c,

\(\Leftrightarrow2x-2+x=0\)

\(\Leftrightarrow3x=2\)

\(\Leftrightarrow x=\dfrac{2}{3}\)

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)=\(\left(\dfrac{1}{5};-\dfrac{2}{5};3\right)\)