\(4+96:\left[\left(2^4.2+4\right):3^2\right]\)

\(=\)\(4+96:\left[\left(2^5+4\right):3^2\right]\)

\(=4+96:\left[\left(32+4\right):3^2\right]\)

\(=4+96:\left[36:3^2\right]\)

\(=4+96:\left[36:9\right]\)

\(=4+96:4\)

\(=4+24=28\)

28,0

\(x:\dfrac{3}{5}=\dfrac{1}{2}\)

\(x=\dfrac{1}{2}.\dfrac{3}{5}\)

\(x=\dfrac{3}{10}\)

\(x:\dfrac{3}{5}=\dfrac{1}{2}\)

\(x=\dfrac{1}{2}x\dfrac{3}{5}\)

\(x=\dfrac{3}{10}\)

a) \(180=2^2.3^2.5\)

\(234=2.3^2.13\)

\(UCLN\left(180;234\right)=2.3^2=18\)

\(UC\left(180;234\right)=\left\{1;2;3;6;9;18\right\}\) (các phần tử thuộc N)

$UC\left(180;234\right)=\left\{1;2;3;6;9;18\right\}$ 

\(63+27.97+18=63+18+27.97=81++27.97=27.3+27.97=27.\left(3+97\right)=27.100=2700\)

2700

Lời giải:

$-E=3x^2+x-2=3(x^2+\frac{x}{3})-2$

$=3[x^2+\frac{x}{3}+(\frac{1}{6})^2]-\frac{25}{12}$

$=3(x+\frac{1}{6})^2-\frac{25}{12}\geq \frac{-25}{12}$

$\Rightarrow E\leq \frac{25}{12}$

Vậy $E_{\max}=\frac{25}{12}$. Giá trị này đạt được khi $x+\frac{1}{6}=0\Leftrightarrow x=\frac{-1}{6}$

a/

\(\dfrac{2n+9}{n+1}=\dfrac{2\left(n+1\right)+7}{n+1}=2+\dfrac{7}{n+1}\)

\(\Rightarrow n+1=\left\{-7;-1;1;7\right\}\Rightarrow n=\left\{-8;-2;0;6\right\}\)

b/

\(\dfrac{3n+5}{n-1}=\dfrac{3\left(n-1\right)+8}{n-1}=3+\dfrac{8}{n-1}\)

\(\Rightarrow n-1=\left\{-8;-4;-2;-1;1;2;4;8\right\}\)

\(\Rightarrow n=\left\{-7;-3;-1;0;2;5;9\right\}\)