\(\left(3x+\dfrac{3}{5}\right)\left(\left|x\right|-\dfrac{1}{4}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{3}{5}=0\\\left|x\right|=\dfrac{1}{4}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=\dfrac{1}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)

Vậy \(x\in\left\{-\dfrac{1}{5};\dfrac{1}{4};-\dfrac{1}{4}\right\}\)

⇒\(\left\{{}\begin{matrix}3x+\dfrac{3}{5}=0\\\left|x\right|-\dfrac{1}{4}=0\end{matrix}\right.\)       ⇒\(\left\{{}\begin{matrix}3x=0-\dfrac{3}{5}=-\dfrac{3}{5}\\\left|x\right|=0+\dfrac{1}{4}=\dfrac{1}{4}\end{matrix}\right.\)

⇒\(\left\{{}\begin{matrix}x=-\dfrac{3}{5}:3=-\dfrac{1}{5}\\x=\dfrac{1}{4},-\dfrac{1}{4}\end{matrix}\right.\)

\(\left(3x+\dfrac{3}{5}\right).\left(\left|x\right|-\dfrac{1}{4}\right)=0\)

 \(\Rightarrow\left[{}\begin{matrix}3x+\dfrac{3}{5}=0\\\left|x\right|-\dfrac{1}{4}=0\end{matrix}\right.\) 

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{5}\\x=\dfrac{1}{4};x=\dfrac{-1}{4}\end{matrix}\right.\)

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

\(\Rightarrow x^3-3x^2-3x-1\) chia hết \(x^2+x+1\) khi \(3⋮x^2+x+1\)

\(\Rightarrow x^2+x+1=Ư\left(3\right)\) (1)

Mà x nguyên dương \(\Rightarrow x^2+x+1\ge1^2+1+1=3\) (2)

(1);(2) \(\Rightarrow x^2+x+1=3\)

\(\Rightarrow x=1\)

\(\left(\frac{2}{3}x-\frac{1}{2}\right).\frac{3}{4}-\frac{2}{5}x=4\frac{1}{4}\)

\(\frac{1}{2}x-\frac{3}{8}-\frac{2}{5}x=4\frac{1}{4}\)

\(\frac{1}{10}x=\frac{17}{4}+\frac{3}{8}\)

\(\frac{1}{10}x=\frac{37}{8}\)

\(x=\frac{185}{4}\)

\(\left(\frac{2}{3}x-\frac{1}{2}\right).\frac{3}{4}-\frac{2}{5}x=4\frac{1}{4}\)

\(\frac{1}{2}x-\frac{3}{8}-\frac{2}{5}x=\frac{17}{4}\)

\(\frac{1}{2}x-\frac{2}{5}x=\frac{17}{4}+\frac{3}{8}\)

\(\frac{1}{10}x=\frac{37}{8}\)

       \(x=\frac{37}{8}:\frac{1}{10}\)

      \(x=\frac{185}{4}\)

A = \(\dfrac{22-3x}{4-x}\)

A = \(\dfrac{3.\left(4-x\right)+10}{4-x}\)

A = 3 + \(\dfrac{10}{4-x}\)

A lớn nhất khi \(\dfrac{10}{4-x}\) lớn nhất. Vì 10 > 0; \(x\) \(\in\) Z nên \(\dfrac{10}{4-x}\) lớn nhất khi

 4 - \(x\) = 1 ⇒ \(x\) = 4 - 1 ⇒   \(x\) = 3

Vậy A_min  = 3 + \(\dfrac{10}{1}\) = 13 khi \(x\) =3

Kết luận giái trị lớn nhất của biểu thức là 13 xảy ra khi \(x\) = 3 

a. (-7)+14-(-7)+4.(-14)+12

= (-7)+14+7+(-56)+12

= (-7+7)+14-56+12

= 0 + 14+12-56

= -30

b. (3x-6).3=3⁴

=> 3x-6=3⁴:3

=> 3x-6=3³

=> 3x-6=27

=> 3x=27+6

=> 3x=33

=> x=33:3

Vậy x=11.

c. 96-3(x+1)=42

=> 3(x+1)=96-42

=> 3(x+1)=54

=> x+1=54:3

=> x+1=18

=> x=18-1

Vậy x=17.

a) \(\left(\frac{3}{5}x-\frac{2}{3}x-x\right).\frac{1}{7}=\frac{-5}{21}\)

\(\Rightarrow\left(\frac{3}{5}-\frac{2}{3}-1\right).x=\frac{-5}{21}:\frac{1}{7}=\frac{-5}{3}\)

\(\Rightarrow\frac{-16}{15}.x=\frac{-5}{3}\Rightarrow x=\frac{-5}{3}:\frac{-16}{15}=\frac{25}{16}\)

b) \(\left(x-\frac{1}{4}\right)^2=\frac{1}{36}\)

\(\Rightarrow\left(x-\frac{1}{4}\right)^2=\left(±\frac{1}{6}\right)^2\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{4}=\frac{1}{6}\\x-\frac{1}{4}=\frac{-1}{6}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{5}{12}\\x=\frac{1}{12}\end{cases}}\)

c) \(14-\left|\frac{3x}{2}-1\right|=9\)

\(\Rightarrow\left|\frac{3x}{2}-1\right|=14-9=5\)

\(\Rightarrow\orbr{\begin{cases}\frac{3x}{2}-1=5\\\frac{3x}{2}-1=-5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}\frac{3x}{2}=6\\\frac{3x}{2}=-4\end{cases}}\Rightarrow\orbr{\begin{cases}3x=12\\3x=-8\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=\frac{-8}{3}\end{cases}}\)