\(a,\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{7}{4}=0\\ \Leftrightarrow\left(x-y\right)^2+\left(x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}=0\\ \Leftrightarrow x,y\in\varnothing\left[\left(x-y\right)^2+\left(x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}>0\right]\\ b,\Leftrightarrow\left(x^2-2x+1\right)+\left(9y^2+12y+4\right)+\left(4z^2-4z+1\right)+14=0\\ \Leftrightarrow\left(x-1\right)^2+\left(3y+2\right)^2+\left(2z-1\right)^2+14=0\\ \Leftrightarrow x,y,z\in\varnothing\left[\left(x-1\right)^2+\left(3y+2\right)^2+\left(2z-1\right)^2+14\ge14>0\right]\)

\(c,\Leftrightarrow-\left(x^2-10xy+25y^2\right)-\left(y^2-20y+100\right)-50=0\\ \Leftrightarrow-\left(x-5y\right)^2-\left(y-10\right)^2-50=0\\ \Leftrightarrow x,y\in\varnothing\left[-\left(x-5y\right)^2-\left(y-10\right)^2-50\le-50< 0\right]\)

Bài 1:

$A=2x^2+y^2-2xy+x+2=(x^2+y^2-2xy)+(x^2+x+\frac{1}{4})+\frac{7}{4}$

$=(x-y)^2+(x+\frac{1}{2})^2+\frac{7}{4}$

Vì $(x-y)^2\geq 0; (x+\frac{1}{2})^2\geq 0$ với mọi $x,y$

$\Rightarrow A\geq 0+0+\frac{7}{4}=\frac{7}{4}$
Vậy $A_{\min}=\frac{7}{4}$. Giá trị này đạt được khi $x-y=x+\frac{1}{2}=0$

$\Leftrightarrow x=y=\frac{-1}{2}$

Bài 2:

$B=x^2+9y^2+4z^2-2x+12y-4z+20$

$=(x^2-2x+1)+(9y^2+12y+4)+(4z^2-4z+1)+14$

$=(x-1)^2+(3y+2)^2+(2z-1)^2+14$
Vì $(x-1)^2\geq 0; (3y+2)^2\geq 0; (2z-1)^2\geq 0$ với mọi $x,y,z$

$\Rightarrow B\geq 0+0+0+14=14$

Vậy $B_{\min}=14$. Giá trị này đạt được khi $x-1=3y+2=2z-1=0$

$\Leftrightarrow x=1; y=\frac{-2}{3}; z=\frac{1}{2}$

Bài 3:

$C=-x^2-26y^2+10xy-20y-150$
$-C=x^2+26y^2-10xy+20y+150$

$=(x^2+25y^2-10xy)+(y^2+20y+10^2)+50$

$=(x-5y)^2+(y+10)^2+50$
Vì $(x-5y)^2\geq 0; (y+10)^2\geq 0$ với mọi $x,y$

$\Rightarrow -C=(x-5y)^2+(y+10)^2+50\geq 0+0+50=50$

$\Rightarrow C\leq -50$

Vậy $C_{\max}=-50$. Giá trị này đạt được khi $x-5y=y+10=0$

$\Leftrightarrow y=-10; x=-50$

2) 

\(A=2x^2+2x+y^2-2xy=x^2-2xy+y^2+x^2+2x+1-1\)

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

Dấu \(=\)khi \(\hept{\begin{cases}x-y=0\\x+1=0\end{cases}}\Leftrightarrow x=y=-1\).

Vậy GTNN của \(A\)là \(-1\)đạt tại \(x=y=-1\).

\(B=2a^2+b^2+c^2-ab+ac+bc\)

\(2B=4a^2+2b^2+2c^2-2ab+2ac+2bc\)

\(=a^2-2ab+b^2+a^2+2ac+c^2+b^2+2bc+c^2+2a^2\)

\(=\left(a-b\right)^2+\left(a+c\right)^2+\left(b+c\right)^2+2a^2\ge0\)

Dấu \(=\)khi \(a=b=c=0\).

Vậy GTNN của \(B\)là \(0\)đạt tại \(a=b=c=0\).

1. 

a) \(2x^2+2x+1=x^2+x^2+2x+1=x^2+\left(x+1\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}x=0\\x+1=0\end{cases}}\)(vô nghiệm) 

suy ra đpcm

b) \(x^2+y^2+2xy+2y+2x+2=\left(x+y\right)^2+2\left(x+y\right)+1+1=\left(x+y+1\right)^2+1>0\)

c) \(3x^2-2x+1+y^2-2xy+1=x^2-2xy+y^2+x^2-2x+1+x^2+1\)

\(=\left(x-y\right)^2+\left(x-1\right)^2+x^2+1>0\)

d) \(3x^2+y^2+10x-2xy+26=x^2-2xy+y^2+x^2+10x+25+x^2+1\)

\(=\left(x-y\right)^2+\left(x+5\right)^2+x^2+1>0\)

Bài 3:

\(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{4}{xy}\)

\(\Leftrightarrow x^2y^2\left(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\ge\dfrac{4}{xy}.x^2y^2\)

\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2+y^2\ge4xy\)

\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2-2xy+y^2\ge2xy\)

\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2+\left(x-y\right)^2\ge2xy\)

\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2-2xy+\left(x-y\right)^2\ge0\)

\(\Leftrightarrow\left(\dfrac{xy}{x-y}-x+y\right)^2=0\) (luôn đúng)

-Tham khảo:

-Tham khảo:

\(2005^3-1=\left(2005-1\right)\left(2005^2+2005+1\right)=2004\times\left(2005^2+2005+1\right)⋮2004\left(\text{đ}pcm\right)\)

\(2005^3+125=\left(2005+5\right)\left(2005^2-2005\times5+5^2\right)=2010\times\left(2005^2-2005\times5+5^2\right)⋮2010\)

\(x^6+1=\left(x^2+1\right)\left(x^4-x^2+1\right)⋮x^2+1\left(\text{đ}pcm\right)\)

\(x^6-y^6=\left(x^2-y^2\right)\left(x^4+x^2y^2+y^2\right)=\left(x-y\right)\left(x+y\right)\left(x^4+x^2y^2+y^4\right)⋮x-y;x+y\left(\text{đ}pcm\right)\)

bài 4 í, có chắc đề đúng ko z

đề bài => 8x³ - y³ + 8x³ + y³ - 16x³ + 16xy = 32

=> 16xy = 32

=> xy = 2

=>\(\left[\begin{array}{nghiempt}x=1=>y=2\\x=-1=>y=-2\\x=2=>y=1\\x=-2=>y=-1\end{array}\right.\)

bài 3 í, đúng ra phải là y^2(y+5) chứ nhỉ