x^2 + 2x + 2 = x^2 + 2.x.1 + 1^2 +1 = (x + 1)^2 + 1 > 0

-x^2 + 4x - 4 = -(x^2 - 2.x.2 + 2^2) = -(x - 2)^2 <= 0

a) ta co ; x^2+ 2x+ 2= (x²+2x+1)+1=(x+1)²+1>0

vi (x+1)²>hoặc=0;1>0suy ra x^2+ 2x+ 2>0

b)ta co  -x²+4x-4=-(x²-4x+4)=-(x-2)²<0

      \(x^3-3x^2-4x+12=0\)   

\(\Rightarrow x^2\left(x-3\right)-4\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left(x^2-4\right)=0\)

\(\Rightarrow\left(x-3\right)\left(x-2\right)\left(x+2\right)=0\)

Tìm được x = 3, x = 2 và x = -2

      \(x^4+x^3-4x-4=0\)

\(\Rightarrow x^3\left(x+1\right)-4\left(x+1\right)=0\)

\(\Rightarrow\left(x+1\right)\left(x^3-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+1=0\\x^3=4\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=\sqrt[3]{4}\end{cases}}}\)

Chúc bạn học tốt.

Bài 1 : 

\(a)\)\(A=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)

\(A=\left(x^2+6x-x-6\right)\left(x^2+3x+2x+6\right)\)

\(A=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

\(A=\left(x^2+5x\right)^2-36\ge-36\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\left(x^2+5x\right)^2=0\)\(\Leftrightarrow\)\(x\left(x+5\right)=0\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)

Vậy GTNN của \(A\) là \(-36\) khi \(x=0\) hoặc \(x=-5\)

\(b)\)\(B=x^2-4x+y^2-8y+6\)

\(B=\left(x^2-4x+4\right)+\left(y^2-8y+16\right)-14\)

\(B=\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}\left(x-2\right)^2=0\\\left(y-4\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=4\end{cases}}}\)

Vậy GTNN của \(B\) là \(-14\) khi \(x=2\) và \(y=4\)

Chúc bạn học tốt ~ 

Bài 2 : 

\(a)\)\(0\le n\le5\)

\(b)\)\(n\ge2\)

\(c)\)\(\hept{\begin{cases}n\ge2\\n+1\ge5\end{cases}\Leftrightarrow\hept{\begin{cases}n\ge2\\n\ge4\end{cases}\Leftrightarrow}n\ge4}\)

\(d)\)\(\hept{\begin{cases}0\le n\le3\\0\le n\le2\\0\le n\le1\end{cases}\Leftrightarrow0\le n\le1}\)

Chúc bạn học tốt ~ 

       \(x^2-4x-5\)

\(=x^2-5x+x-5\)

\(=x\left(x-5\right)+\left(x-5\right)\)

\(=\left(x-5\right)\left(x+1\right)\)

      \(2x^2+7x+5\)

\(=2x^2+2x+5x+5\)

\(=2x\left(x+1\right)+5\left(x+1\right)\)

\(=\left(x+1\right)\left(2x+5\right)\)

1)

a)\(A=2013.2015=2013.\left(2014+1\right)=2013.2014+2013\)

\(B=2014^2=2014.\left(2013+1\right)=2014.2013+2014\)

Ta có: \(2014.2013+2014>2013.2014+2013\)

\(\Rightarrow2014^2>2013.2015\)

\(\Rightarrow B>A\)

Vậy \(B>A\)

b) \(A=4.\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)

\(\Rightarrow2A=2.4\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3-1\right).\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3^8-1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3^{16}-1\right)...\left(3^{64}+1\right)\)

\(\Rightarrow2A=3^{128}-1\)

\(\Rightarrow A=\frac{3^{128}-1}{2}< 3^{128}-1=B\)

\(\Rightarrow A< B\)

Vậy \(A< B\)

2)

a)\(9x^2-6x+3=\left(3x\right)^2-2.3x.1+1^2+2\)

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

Ta có: \(\left(3x-1\right)^2\ge0\forall x\)

\(\Rightarrow\left(3x-1\right)^2+2\ge2\forall x\)

\(\Rightarrow\left(3x-1\right)^2+2>0\forall x\)

                                đpcm

b)\(x^2+y^2+2x+6y+16\)

\(=\left(x^2+2x+1\right)+\left(y^2+2.y.3+3^2\right)+6\)

\(=\left(x+1\right)^2+\left(y+3\right)^2+6\)

Ta có: \(\hept{\begin{cases}\left(x+1\right)^2\ge0\forall x\\\left(y+3\right)^2\ge0\forall y\end{cases}\Rightarrow}\left(x+1\right)^2+\left(y+3\right)^2+6\ge6\forall x;y\)

\(\Rightarrow\left(x+1\right)^2+\left(y+3\right)^2+6>0\)

                                         đpcm

Tham khảo nhé~

1.

a) A = 2013.2015 = (2014 - 1)(2014 + 1) = 2014² - 1

Vì 2014² - 1 < 2014² => A < B

Vậy A < B

b) \(A=4\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3^8-1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)

\(\Rightarrow2A=3^{128}-1\Leftrightarrow A=\frac{3^{128}-1}{2}\)

\(\Rightarrow A< B\)

Vậy A < B

Bài 2:

a) \(9x^2-6x+2=\left(3x\right)^2-2.3x+1+2=\left(3x-1\right)^2+2\)

Vì \(\left(3x-1\right)^2\ge0\Rightarrow\left(3x-1\right)^2+2>0\)

=> 9x² - 6x + 2 luôn nhận giá trị dương với mọi x

b) \(x^2+y^2+2x+6y+16=\left(x^2+2x+1\right)+\left(y^2+6y+9\right)+6=\left(x+1\right)^2+\left(y+3\right)^2+6\)

Vì \(\left(x+1\right)^2\ge0;\left(y+3\right)^2\ge0\Rightarrow\left(x+1\right)^2+\left(y+3\right)^2\ge0\Rightarrow\left(x+1\right)^2+\left(y+3\right)^2+6>0\)

=> x² + y² + 2x + 6y + 16 luôn nhận giá trị dương với mọi x

a) Ta có \(2x^2-8x+13=2x^2-8x+8+5\)

\(=2\left(x^2-4x+4\right)+5\)

\(=2\left(x-2\right)^2+5\ge5\forall x\)

Giả sử trước khi làm nhé 

\(a)\)\(2x^2-8x+13>0\)

\(\Leftrightarrow\)\(4x^2-16x+26>0\)

\(\Leftrightarrow\)\(\left(4x^2-16+16\right)+10>0\)

\(\Leftrightarrow\)\(\left(2x-4\right)^2+10\ge10>0\) ( luôn đúng ) 

Vậy ... 

\(b)\)\(-2+2x-x^2< 0\)

\(\Leftrightarrow\)\(x^2-2x+2>0\)

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

\(\Leftrightarrow\)\(\left(x-1\right)^2+1\ge1>0\) ( luôn đúng ) 

Vậy ... 

Chúc bạn học tốt ~