1,47 x 3,6 + 1,47 x 6,4

= 1,47 x ( 3,6 + 6,4 )

= 1,47 x 10

= 14,7

A=1-3+3^2-3^3+...-3^2003+3^2004

3A=3-3^2+3^3-3^4+...-3^2004+3^2005

3A+A=3-3^2+3^3-3^4+..-3^2004+3^2005+1-3+3^2-3^3+...-3^2003+3^2004

4A=3^2005+1

A=(3^2005+1):4

\(P=x^4-2x^2-3\left|x^2-1\right|-9\)

\(P=x^4-2x^2+1-3\left|x^2-1\right|-10\)

\(P=\left(x^2-1\right)^2-3\left|x^2-1\right|-10\)

\(P=\left|x^2-1\right|^2-3\left|x^2-1\right|-10\)

Đặt: \(\left|x^2-1\right|=t\ge0\) ta có:

\(pt\Leftrightarrow t^2-3t-10=t^2-3t+\dfrac{9}{4}-\dfrac{49}{4}\)

\(=\left(t-\dfrac{3}{2}\right)^2-\dfrac{49}{4}\ge\left(-\dfrac{3}{2}\right)^2-\dfrac{49}{4}=\dfrac{9}{4}-\dfrac{49}{4}=-10\)

Dấu "=" khi: \(t=0\Leftrightarrow\left|x^2-1\right|=0\Leftrightarrow x=\pm1\)

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

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

\(f\left(x\right)=x\left(2x-1\right)-3\left(2x-1\right)\)

\(f\left(x\right)=\left(x-3\right)\left(2x-1\right)\)

\(f\left(x\right)=0\Leftrightarrow\left(x-3\right)\left(2x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\)

\(\dfrac{a^3}{b+c}+\dfrac{b^3}{a+c}+\dfrac{c^3}{a+b}\)

\(=\dfrac{a^4}{ab+ac}+\dfrac{b^4}{ab+bc}+\dfrac{c^4}{ac+bc}\)

\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{2\left(ab+bc+ac\right)}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{2\left(a^2+b^2+c^2\right)}\)

\(=\dfrac{a^2+b^2+c^2}{2}=\dfrac{1}{2}\)

Dấu "=" xảy ra khi: \(a=b=c=\dfrac{1}{\sqrt{3}}\)

\(\dfrac{1}{3x^2+y^2}+\dfrac{2}{y^2+3xy}=\dfrac{1}{3x^2+y^2}+\dfrac{4}{2y^2+6xy}\)

\(\ge\dfrac{\left(1+2\right)^2}{3x^2+3y^2+6xy}=\dfrac{9}{3x^2+3y^2+6xy}\)

\(=\dfrac{9}{3\left(x^2+y^2+2xy\right)}=\dfrac{9}{3\left(x+y\right)^2}\ge\dfrac{9}{3}=3\)

Dấu "=" xảy ra khi: \(x=y=\dfrac{1}{2}\)

\(H=\dfrac{1}{1+xy}+\dfrac{1}{1+yz}+\dfrac{1}{1+xz}\ge\dfrac{\left(1+1+1\right)^2}{3+xy+yz+xz}=\dfrac{9}{3+xy+yz+xz}\)

Mặt khác,theo AM-GM: \(xy+yz+xz\le x^2+y^2+z^2=3\)

\(\Rightarrow\dfrac{9}{3+xy+yz+xz}\ge\dfrac{9}{3+3}=\dfrac{9}{6}=\dfrac{3}{2}\)

Dấu "=" khi: \(x=y=z=1\)

\(a^3+6=-3a-2a^2\)

\(\Rightarrow a^3+6+3a+2a^2=0\)

\(\Rightarrow a\left(a^2+3\right)+2\left(a^2+3\right)=0\)

\(\Rightarrow\left(a+2\right)\left(a^2+3\right)=0\)

Vì \(a^2+3>0\forall a\in R\) nên \(a+2=0\Leftrightarrow a=-2\)

\(A=\dfrac{a-1}{a+3}=\dfrac{-2-1}{-2+3}=\dfrac{-3}{1}=-3\)

\(a^2+b^2+c^2+d^2+e^2\ge a\left(b+c+d+e\right)\)

\(\Rightarrow a^2+b^2+c^2+d^2+e^2-ab-ac-ad-ae\ge0\)

\(\Rightarrow\left(\dfrac{a^2}{4}-ab+b^2\right)+\left(\dfrac{a^2}{4}-ac+c^2\right)+\left(\dfrac{a^2}{4}-ad+d^2\right)+\left(\dfrac{a^2}{4}-ae+e^2\right)\)

\(\Rightarrow\left(\dfrac{a}{2}-b\right)^2+\left(\dfrac{a}{2}-c\right)^2+\left(\dfrac{a}{2}-d\right)^2+\left(\dfrac{a}{2}-e\right)^2\ge0\) (đúng)

Dấu "=" xảy ra khi: \(\dfrac{a}{2}=b=c=d=e\)