\(\dfrac{1}{4}\cdot\dfrac{1}{3}+\dfrac{1}{6}\cdot\dfrac{1}{4}+\dfrac{1}{4}\cdot\dfrac{1}{7}\\ =\dfrac{1}{4}\cdot\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{7}\right)\\ =\dfrac{1}{4}\cdot\dfrac{9}{14}\\ =\dfrac{9}{56}\)

3² \(\times\) ( \(x\) + 4) - 5² = 5.5²

9.(\(x\) + 4) - 25 = 125

9.(\(x\) + 4) = 150

    \(x\) + 4 =  \(\dfrac{50}{3}\)

     \(x\)       =  \(\dfrac{50}{3}\) - 4

     \(x\)      = \(\dfrac{38}{3}\)

\(3^2\cdot\left(x+4\right)-5^2=5\cdot5^2\\ \Leftrightarrow9\left(x+4\right)=6\cdot5^2\\ \Leftrightarrow x+4=\dfrac{50}{3}\\ \Leftrightarrow x=\dfrac{38}{3}\)

A = \(\dfrac{3^{123}+1}{3^{125}+1}\)  Vì 3¹²³ + 1 < 2¹²⁵ + 1 Nên A = \(\dfrac{3^{123}+1}{3^{125}+1}\)< \(\dfrac{3^{123}+1+2}{3^{125}+1+2}\)

A < \(\dfrac{3^{123}+3}{3^{125}+3}\) = \(\dfrac{3.\left(3^{122}+1\right)}{3.\left(3^{124}+1\right)}\) = \(\dfrac{3^{122}+1}{3^{124}+1}\) = B

Vậy A < B 

7x - 3³ = 2⁷ : 2⁴

7\(x\) - 27 = 2³

7\(x\)         = 8 + 27

7\(x\)         = 35

\(x\)            = 5

\(27-3\left(x+6\right)=6\\ \Leftrightarrow3\left(x+6\right)=27-6\\ \Leftrightarrow3\left(x+6\right)=21\\ \Leftrightarrow x+6=21:3\\ \Leftrightarrow x+6=7\\ \Leftrightarrow x=7-6\\ \Leftrightarrow x=1\)

\(3n+9⋮n+1\left(n\inℕ\right)\)

\(\Rightarrow3n+9-3\left(n+1\right)⋮n+1\)

\(\Rightarrow3n+9-3n-3⋮n+1\)

\(\Rightarrow6⋮n+1\)

\(\Rightarrow n+1\in\left\{1;2;3;6\right\}\)

\(\Rightarrow n\in\left\{0;1;2;5\right\}\left(n\inℕ\right)\)

\(\left(3n+9\right)⋮\left(n+1\right)\\ =>\left[3\left(n+1\right)+6\right]⋮\left(n+1\right)\\ =>6⋮\left(n+1\right)\\ =>n+1\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm6\right\}\\ =>n\in\left\{0;-2;1;-3;5;-7\right\}\)

Mà n là STN

Vậy \(n\in\left\{0;1;5\right\}\)

Bổ sung đề : Tìm n thuộc Z

+) \(11⋮\left(n-2\right)=>n-2\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\\ =>n\in\left\{3;1;13;-9\right\}\)

+) \(\left(n+7\right)⋮\left(n-3\right)\\ =>\left(n-3\right)+10⋮\left(n-3\right)\\ =>10⋮\left(n-3\right)\\ =>n-3\inƯ\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\\ =>n\in\left\{4;2;5;1;8;-2;13;-7\right\}\)