Đặt: \(k=\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)

\(\Rightarrow k^3=\frac{xyz}{3.4.5}=\frac{1620}{60}=27\)

=> k = 3

Nên \(\frac{x}{3}=3\Rightarrow x=9\)

        \(\frac{y}{4}=3\Rightarrow y=12\)

         \(\frac{z}{5}=3\Rightarrow z=15\)

Vậy x = 9 , y = 12 , z = 15

a)

\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\Leftrightarrow x=3k;y=4k;z=5k\)và \(xyz=1620\)

\(\Rightarrow3k.4k.5k=1620\Leftrightarrow60k^3=1620\)

\(\Rightarrow k=\sqrt[3]{1620:60}=3\)

\(\hept{\begin{cases}\frac{x}{3}=3\Rightarrow x=3.3=9\\\frac{y}{4}=3\Rightarrow y=3.4=12\\\frac{z}{5}=3\Rightarrow z=3.5=15\end{cases}}\)

Vậy \(x=9;y=12;z=15\)

b) 

Ta có:

\(\frac{x}{2}=\frac{y}{3};\frac{y}{5}=\frac{z}{6}\Leftrightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{18}\) và \(x+y+z=334\)

Áp dụng tính chất của dãy tỉ số bằng nhau:

\(\frac{x}{10}=\frac{y}{15}=\frac{z}{18}=\frac{x+y+z}{10+15+18}=\frac{334}{43}\)

\(\hept{\begin{cases}\frac{x}{10}=\frac{334}{43}\Rightarrow x=\frac{334}{43}.10=\frac{3340}{43}\\\frac{y}{15}=\frac{334}{43}\Rightarrow y=\frac{334}{43}.15=\frac{5010}{43}\\\frac{z}{18}=\frac{334}{43}\Rightarrow z=\frac{334}{43}.18=\frac{6012}{43}\end{cases}}\)

Vậy \(x=\frac{3340}{43};y=\frac{5010}{43};z=\frac{6012}{43}\)

Ta có : A = 3 + 3² + 3³ + 3⁴ + ...... + 3¹⁰⁰

=> A = (3 + 3² + 3³ + 3⁴) + ...... + (3⁹⁷ + 3⁹⁸ + 3⁹⁹ + 3¹⁰⁰)

=> A = (3 + 3² + 3³ + 3⁴) + ...... + 3⁹⁶(3 + 3² + 3³ + 3⁴)

=> A = 120 + ..... + 3⁹⁶.120

=> A = 120(1 + .... + 3⁹⁶) chia hết cho 120

A=\(3+3^2+3^3+3^4+...+3^{100}\)

  =\(\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{99}+3^{100}\right)\)

  =\(\left(3+3^2\right)+3^2\left(3+3^2\right)+...+3^{98}\left(3+3^2\right)\)

  =\(\left(3+3^2\right)\left(1+3^2+3^4+...+3^{98}\right)\)

  =\(12\left(1+3^2+3^4+...+3^{98}\right)\)

Vì \(12⋮12\)=>\(12\left(1+3^2+3^4+...+3^{98}\right)⋮12\)

=>\(A⋮12\)

Vậy \(A⋮12\)

Ta thấy:
\(\left|3x+7\right|\ge0\) ( mọi x )
\(\Rightarrow4\left|3x+7\right|\ge0\) ( mọi x )
\(\Rightarrow4\left|3x+7\right|+3\ge3\) ( mọi x )
\(\Rightarrow\frac{15}{4\left|3x+7\right|+3}\le5\) ( mọi x )
\(\Rightarrow5+\frac{15}{4\left|3x+7\right|+3}\le10\) ( mọi x )
=> GTLN của \(A=5+\frac{15}{4\left|3x+7\right|+3}\) bằng 10 khi và chỉ khi:
\(\frac{15}{4\left|3x+7\right|+3}=5\)
\(\Rightarrow4\left|3x+7\right|+3=3\)
\(\Rightarrow4\left|3x+7\right|=0\)
\(\Rightarrow\left|3x+7\right|=0\)
\(\Rightarrow3x+7=0\)
\(\Rightarrow3x=-7\)
\(\Rightarrow x=\frac{-7}{3}\)
Vậy GTLN của \(A=5+\frac{15}{4\left|3x+7\right|+3}\) bằng 10 khi và chỉ khi \(x=\frac{-7}{3}\).