\(\dfrac{18}{27}:2=\dfrac{18}{27}\times\dfrac{1}{2}\\ =\dfrac{9\times2\times1}{9\times3\times2}=\dfrac{1}{3}\)

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

Với mọi x thuộc R, ta luôn có:

\(\left(x+1\right)^2\ge0\\ \Rightarrow-3\left(x+1\right)^2\le0\\ \Rightarrow C\le-6< 0\)

Hay C luôn âm với mọi giá trị của x

\(B=4x-x^2-5=-\left(x^2-4x\right)-5\\ =-\left(x^2-4x+4\right)+4-5\\ =-\left(x-2\right)^2-1\)

Với mọi x thuộc R, ta luôn có:

 \(\left(x-2\right)^2\ge0\\ \Rightarrow-\left(x-2\right)^2\le0\\ \Rightarrow B=-\left(x-2\right)^2-1\le-1\)

Hay B luôn âm với mọi giá trị của x

Dấu "=" xảy ra khi: x-2=0 hay x=2

a) MSC: 9

\(\dfrac{7}{3}=\dfrac{7x3}{3x3}=\dfrac{21}{9};\dfrac{6}{9}\)

b) MSC: 18

\(\dfrac{13}{6}=\dfrac{13x3}{6x3}=\dfrac{39}{18};\dfrac{7}{9}=\dfrac{7x2}{9x2}=\dfrac{14}{18};\dfrac{17}{18}\)

\(25x199x4=\left(25x4\right)x199=100x199=19900\)

\(3046x39+3046x61=3046x\left(39+61\right)=3046x100=304600\)

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

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

\(2973780=2^2x3^3x5x5507\)

\(\dfrac{2}{13}< \dfrac{1}{y}< \dfrac{3}{7}\\ \dfrac{6}{39}< \dfrac{6}{6y}< \dfrac{6}{14}\\ 39>6y>14\\ \dfrac{39}{6}>y>\dfrac{14}{6}\)

Mà y là STN nên y nhận giá trị: 3;4;5;6

\(\dfrac{2}{13}< \dfrac{1}{y}< \dfrac{3}{7}\\ \dfrac{6}{39}< \dfrac{6}{y}< \dfrac{6}{14}\\ 39>y>14\)

y là STN nên y nhận giá trị: 15;16;17;18;...;37;38