\(x^2-7x+12=x^2-3x-4x+12=\left(x^2-3x\right)-\left(4x-12\right)=x\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x-4\right)\)

a) \(4x^2+12xy+9y^2\)

\(=\left(2x\right)^2+2.2x.3+\left(3y\right)^2\)

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

b) \(81x^2-18xy+y^2\)

\(=\left(9x\right)^2-2.9x.y+y^2\)

\(=\left(9x-y\right)^2\)

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

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

a. 25 - \(x^2\) = (5-x) (5+x)

b) -196 + 4\(x^2\) = 196 - 4\(x^2\) = (14- 2x) (14+2x)

c)\(5^4-81x^4\) = \(\left[\left(5^2\right)^2\right]-\left[\left(81x^2\right)^2\right]\) = (\(\left(5^2-81x^2\right)\left(5^2+81x^2\right)\)

\(a,25-e=\left(5-\sqrt{e}\right)\left(5+\sqrt{e}\right)\)

\(b,-196+g=-\left(196-g\right)=-\left(14-\sqrt{g}\right)\left(14+\sqrt{g}\right)\)

\(c,2^6-47^2=\left(2^3\right)^2-47^2=\left(2^3-47\right)\left(2^3+47\right)\)

\(d,5^4-81x^4=\left(5^2\right)^2-\left(9x^2\right)^2=\left(5^2-9x^2\right)\left(5^2+9x^2\right)=\left(25-9x^2\right)\left(25+9x^2\right)\)

\(i,\dfrac{25}{16}-9y^2=\left(\dfrac{5}{4}-3y\right)\left(\dfrac{5}{4}+3y\right)\)

\(A=\dfrac{4x^2\left(x+y\right)-9y^2\left(x+y\right)}{4x^2\left(x-y\right)-9y^2\left(x-y\right)}=\dfrac{\left(4x^2-9y^2\right)\left(x+y\right)}{\left(4x^2-9y^2\right)\left(x-y\right)}=\dfrac{x+y}{x-y}\)

Với x=2014, y=14:

\(A=\dfrac{2014+14}{2014-14}=\dfrac{2028}{2000}=\dfrac{507}{500}\)

Đặt $a=2x,\quad b=3y,\quad c=4z.$

Khi đó $a^2+b^2+c^2=1.$

Ta có $A=\dfrac{a}{1-a^2}+\dfrac{b}{1-b^2}+\dfrac{c}{1-c^2}.$

Xét hàm số $f(t)=\dfrac{t}{1-t^2}\quad(0<t<1).$

Ta có $f''(t)=\dfrac{2t(3+t^2)}{(1-t^2)^3}>0.$

Do đó $f(t)$ là hàm lồi trên $(0;1)$.

Theo bất đẳng thức Jensen:

$A\ge3f\left(\dfrac{a+b+c}{3}\right).$

Mặt khác, theo bất đẳng thức Cauchy:

$(a+b+c)^2\le3(a^2+b^2+c^2)=3$

$\Rightarrow\dfrac{a+b+c}{3}\le\dfrac1{\sqrt3}.$

Vì $f(t)$ đồng biến trên $(0;1)$ nên

$A\ge3f\left(\dfrac{a+b+c}{3}\right)$

$\ge3f\left(\dfrac1{\sqrt3}\right)$

$=3\cdot\dfrac{\frac1{\sqrt3}}{1-\frac13}$

$=3\cdot\dfrac{\frac1{\sqrt3}}{\frac23}$

$=\dfrac{3\sqrt3}{2}.$

$a=b=c=\dfrac1{\sqrt3}.$

Dấu '=' xảy ra khi:

\Leftrightarrow x=\dfrac1{2\sqrt3},\quady=\dfrac1{3\sqrt3},\quadz=\dfrac1{4\sqrt3}.

$A_{\min}=\dfrac{3\sqrt3}{2}.$

a) 4x²+4x+1

= (2x+1)²

b) x²-16x+64

= (x-8)²

c) 4x²-9y²

= (2x+3y)(2x-3y)

d) ( x-3).(x^2+3x+9)

Ta có:

\(\left(4x+9y+16z\right)\left(\frac{1}{x}+\frac{25}{y}+\frac{64}{z}\right)\ge\left(\sqrt{\frac{4x}{x}}+\sqrt{\frac{9y.25}{y}}+\sqrt{\frac{16z.64}{z}}\right)^2\)

\(\Leftrightarrow49\left(\frac{1}{x}+\frac{25}{y}+\frac{64}{z}\right)\ge\left(2+15+32\right)^2\)

\(\Leftrightarrow\frac{1}{x}+\frac{25}{y}+\frac{64}{z}\ge49\)

Dấu = xảy ra tại \(x=\frac{1}{2};y=\frac{5}{3};z=2\)

\(81x^2-6yz-9y^2-z^2\)

\(=\left(9x\right)^2-\left(6yz+9y^2+z^2\right)\)

\(=\left(9x\right)^2-\left(z+3y\right)^2\)

\(=\left(9x-z-3y\right)\left(9x+z+3y\right)\)

đề đúng chứ bạn????????

\(=81x^2-\left(z+3y\right)^2\)

\(=\left(9x+z+3y\right)\left(9x-z-3y\right)\)