\(T=-2\left(x^2+y^2+1-2xy+2x-2y\right)-2y^2+8y+2004\)

\(T=-2\left(x-y+1\right)^2-2\left(y-2\right)^2+2012\le2012\)

\(T_{max}=2012\) khi \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

\(2x^2+4y^2+4xy-6x+100=\left(x^2+4xy+4y^2\right)+\left(x^2-6x+9\right)+91=\left(x+2y\right)^2+\left(x-3\right)^2+91\ge91>0\)

\(2x^2-4xy+2y^2\\ =2\left(x^2-2xy+y^2\right)\\ =2\left(x-y\right)^2\)

a) 2x²-4xy+2y²
= 2x²-2xy-2xy+2y²
= 2x(x-y)-2y(x-y)
= (2x-2y)(x-y)
b) x²+4xy+4y²-9
= (x+2y)²-3²
= (x+2y-3)(x+2y+3)
c) x⁴-x³y+x-y
= x³(x-y)+(x-y)
= (x³+1)(x-y)

\(a,=2\left(x^2-2xy+y^2\right)=2\left(x-y\right)^2\\ b,=\left(x+2y\right)^2-9=\left(x+2y-3\right)\left(x+2y+3\right)\\ c,=x^3\left(x-y\right)+\left(x-y\right)\\ =\left(x+1\right)\left(x-y\right)\left(x^2-x+1\right)\)

tìm y nữa 

mình viết thiếu

\(A=\left(x^2+4y^2+1-4xy+2x-4y\right)+\left(x^2-4x+4\right)-3\)

\(A=\left(x-2y+1\right)^2+\left(x-2\right)^2-3\ge-3\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(2;\dfrac{3}{2}\right)\)

Lời giải:

$A=(x^2+4y^2+4xy)+x^2+5-8x-12y$

$=(x+2y)^2-6(x+2y)+x^2+5-2x$

$=(x+2y)^2-6(x+2y)+9+(x^2-2x+1)-5$

$=(x+2y-3)^2+(x-1)^2-5\geq 0+0-5=-5$

Vậy $A_{\min}=-5$. Giá trị này đạt được khi $x+2y-3=x-1=0$

$\Leftrightarrow x=1; y=1$

\(a,x^2-6x+5=0\\ \Rightarrow\left(x^2-5x\right)-\left(x-5\right)=0\\ \Rightarrow x\left(x-5\right)-\left(x-5\right)=0\\ \Rightarrow\left(x-1\right)\left(x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)

\(b,2x^2+4x-8=0\\ \Rightarrow x^2+2x-4=0\\ \Rightarrow\left(x^2+2x+1\right)-5=0\\ \Rightarrow\left(x+1\right)^2-\sqrt{5^2}=0\\ \Rightarrow\left(x+1+\sqrt{5}\right)\left(x+1-\sqrt{5}\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-1-\sqrt{5}\\x=-1+\sqrt{5}\end{matrix}\right.\)

\(c,4y^2-4y+1=0\\ \Rightarrow\left(2y-1\right)^2=0\\ \Rightarrow2y-1=0\\ \Rightarrow y=\dfrac{1}{2}\)

\(d,5x^2-x+2=0\)

Ta có:\(\Delta=\left(-1\right)^2-4.5.2=1-40=-39\)

Vì \(\Delta< 0\Rightarrow\) pt vô nghiệm

(Phần a mình lấy vế phải bằng 0 nha ^^)

a,

\(\left(5x-1\right)^2-\left(5x-4\right)\left(5x+4\right)+7=0\\ \Leftrightarrow25x^2-10x+1-\left(25x^2-16\right)+7=0\\ \Leftrightarrow25x^2-10x+1-25x^2+16+7=0\\ \Leftrightarrow-10x+24=0\\ \Leftrightarrow x=2,4\)

b,

\(5x^2+4xy+4y^2+4x+1=0\left(1\right)\\ \Leftrightarrow4x^2+4x+1+x^2+4xy+4y^2=0\\ \Leftrightarrow\left(2x+1\right)^2+\left(x+2y\right)^2=0\left(1a\right)\)

Do \(VT\ge0\) với \(\forall x,y\in R\) nên:

\(\left(1a\right)\Leftrightarrow\left\{{}\begin{matrix}2x+1=0\\x+2y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\frac{1}{2}\\y=\frac{1}{4}\end{matrix}\right.\)

c,

\(\left(x+2\right)^3-x\left(x-1\right)\left(x+1\right)=6x^2+21\\ \Leftrightarrow x^3+6x^2+12x+8-x\left(x^2-1\right)-6x^2-21=0\\ \Leftrightarrow x^3+12x+8-x^3+x-21=0\\ \Leftrightarrow13x-13=0\\ \Leftrightarrow x=1\)

Chúc bạn học tốt nha.

\(b)5x^2 + 4xy + 4y^2 + 4x + 1 = 0\)

\(\Leftrightarrow\) \(4x^2 + 4x + 1 + x^2 + 4xy + 4y^2 = 0\)

\(\Leftrightarrow\)\((2x + 1)^2 + (x + 2y)^2 = 0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+1=0\\x+2y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\frac{1}{2}\\y=\frac{1}{4}\end{matrix}\right.\)

\(c)(x+2)^3-x(x-1)(x+1)=6x^2+21\)

\(\Leftrightarrow x^3+6x^2+12x+8-x\left(x^2-1\right)=6x^2+21\\ \Leftrightarrow13x+8=21\\ \Leftrightarrow13x=21-8\\ \Leftrightarrow13x=13\\ \Leftrightarrow x=1\)