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

\(=\dfrac{1}{4}x^2+xy+y^2+\dfrac{3}{4}x^2-3x-3\)

\(=\left(\dfrac{1}{2}x+y\right)^2+3\left(\dfrac{1}{4}x^2-x-1\right)\)

\(=\left(\dfrac{1}{2}x+y\right)^2+3\left(\dfrac{1}{4}x^2-x+1-2\right)\)

\(=\left(\dfrac{1}{2}x+y\right)^2+3\left(\dfrac{1}{2}x-1\right)^2-6>=-6\forall x,y\)

Dấu = xảy ra khi \(\left\{{}\begin{matrix}\dfrac{1}{2}x-1=0\\\dfrac{1}{2}x+y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-\dfrac{1}{2}x=-\dfrac{1}{2}\cdot2=-1\end{matrix}\right.\)

\(M=x^2-6x+25\\ \Rightarrow M=\left(x^2-6x+9\right)+16\\ \Rightarrow M=\left(x-3\right)^2+16\ge16\)

Dấu "=" xảy ra \(\Leftrightarrow x=3\)

Vậy \(M_{min}=16\Leftrightarrow x=3\)

GTNN của M = 16 khi x = 3

\(M=x^2-6x+25\)

\(=\left(x^2-6x+9\right)+16\)

\(=\left(x-3\right)^2+16\) ≥ \(16\)  ( Vì \(\left(x-3\right)^2\text{≥}0\) ∀ x)

Min M=16 ⇔ x=3

\(M=5xy+\dfrac{3}{2}x^2-6y^2+2xy-4y^2+3x^2\)

\(M=\left(5xy+2xy\right)+\left(\dfrac{3}{2}x^2+3x^2\right)+\left(-6y^2-4y^2\right)\)

\(M=7xy+\dfrac{9}{2}y^2-10x^2\)

Câu 2:  \(x^2-5x+1=0\Leftrightarrow x^2-2.x.\frac{5}{2}+\frac{25}{4}-\frac{25}{4}+1=0\)

\(\Leftrightarrow\left(x-\frac{5}{2}\right)^2-\frac{21}{4}=0\Leftrightarrow x-\frac{5}{2}=\pm\frac{\sqrt{21}}{2}\)\(\Leftrightarrow x=\pm\frac{\sqrt{21}+5}{2}\)

Thay vào biểu thức đó: 

\(\frac{x^2+1}{x^2}=1+\frac{1}{x^2}=1+\frac{1}{\frac{\left(\sqrt{21}+5\right)^2}{4}}\)

\(=1+\frac{1}{\frac{21+10\sqrt{21}+25}{4}}=1+\frac{4}{46+10\sqrt{21}}=\frac{50+10\sqrt{21}}{46+10\sqrt{21}}\)

\(=\frac{25+5\sqrt{10}}{23+5\sqrt{10}}\). ĐS...

\(\Leftrightarrow5^x=\dfrac{5^{2019}}{5^{2010}\cdot5^2}=5^7\)

hay x=7

Ta có:

\(N=\left(1+2\right)\left(2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{2008}+1\right)\)

\(\Leftrightarrow N=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{2008}+1\right)\)

\(\Leftrightarrow N=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{2008}+1\right)\)

\(\Leftrightarrow N=\left(2^8-1\right)...\left(2^{2008}+1\right)\)

\(\Leftrightarrow N=2^{4016}-1>2^{2016}=M\)

Ta có:

N=(1+2)(2−1)(22+1)(24+1)...(22008+1)N=(1+2)(2−1)(22+1)(24+1)...(22008+1)

⇔N=(22−1)(22+1)(24+1)...(22008+1)⇔N=(22−1)(22+1)(24+1)...(22008+1)

⇔N=(24−1)(24+1)...(22008+1)⇔N=(24−1)(24+1)...(22008+1)

⇔N=(28−1)...(22008+1)⇔N=(28−1)...(22008+1)

⇔N=24016−1>22016=M

Lời giải:
a.

$x^8+x^4+1=(x^4)^2+2x^4+1-x^4$
$=(x^4+1)^2-(x^2)^2=(x^4+1-x^2)(x^4+1+x^2)$

$=(x^4+1-x^2)[(x^2+1)^2-x^2]$

$=(x^4-x^2+1)(x^2+1-x)(x^2+1+x)$

b. 

$x^{12}-3x^6-1=(x^6-\frac{3}{2})^2-\frac{13}{4}$

$=(x^6-\frac{3}{2}-\frac{\sqrt{13}}{2})(x^6-\frac{3}{2}+\frac{\sqrt{13}}{2})$

c.

$3x^4+10x^2-25=(3x^4+15x^2)-(5x^2+25)$

$=3x^2(x^2+5)-5(x^2+5)=(x^2+5)(3x^2-5)$

$=(x^2+5)(\sqrt{3}x-\sqrt{5})(\sqrt{3}x+\sqrt{5})$

c.

$x^2-5y^2-y^4+2xy-9$

$=(x^2+2xy+y^2)-(y^4+6y^2+9)$
$=(x+y)^2-(y^2+3)^2$
$=(x+y+y^2+3)(x+y-y^2-3)$

\(a,x^8+x^4+1\\ =\left(x^8+2x^4+1\right)-x^4\\ =\left(x^4+1\right)^2-x^4\\ =\left(x^4-x^2+1\right)\left(x^4+x^2+1\right)\\ b,x^{12}-3x^6-1\\ =\left(x^{12}-2x^6+1\right)-x^6-2\\ =\left(x^6-1\right)^2-x^6-2\\ =\left(x^6-x^3-1\right)\left(x^6+x^3-1\right)-2???\\ c,3x^4+10x^2-25\\ =4x^4-\left(x^4-10x^2+25\right)\\ =4x^4-\left(x^2-5\right)^2\\ =\left(2x^2-x^2+5\right)\left(2x^2+x^2-5\right)\\ =\left(x^2+5\right)\left(3x^2-5\right)\\ d,x^2-5y^2-y^4+2xy-9\\ =\left(x^2+2xy+y^2\right)-\left(y^4+6y^2+9\right)\\ =\left(x+y\right)^2-\left(y^2+3\right)^2\\ =\left(x+y+y^2+3\right)\left(x+y-y^2-3\right)\)

a) x⁸+x⁴+1 = (x⁴+1)²-x⁴ = (x⁴-x²+1)(x⁴+x²+1)

b) x¹²-3x⁶-1 = (x⁶-1)²-x⁶ = (x⁶-x³-1)(x⁶+x³-1)

c) 3x⁴+10x²-25 = 4x⁴-(x⁴-10x²+25) = 4x⁴- (x²-5)² = (x²+5)(3x²-5)

d) x²-5y²-y⁴+2xy-9 = (x+y)²-(y²+3)² = (x+y-y²-3)(x+y+y²+3)

a) (x + 3)² - (x - 2)(x + 2) = 1

x² + 6x + 9 - x² + 4 - 1 = 0

6x + 12 = 0

6x = 0 - 12

6x = -12

x = -12/6

x = -2

b) M = x² - 6x

= x² - 6x + 9 - 9

= (x - 3)² - 9

Do (x - 3)² ≥ 0 với mọi x ∈ R

⇒ (x - 3)² - 9 ≥ -9

Vậy giá trị nhỏ nhất của M là -9 khi x = 3