Lời giải:

$45-5(x+4)=10$

$5(x+4)=45-10=35$

$x+4=35:5=7$

$x=7-4=3$

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$(7x-15):3-6=17$

$(7x-15):3=17+6=23$
$7x-15=23\times 3=69$

$7x=69+15=84$

$x=84:7=12$

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$3^{x+2}-5.3^x=108$

$3^x(3^2-5)=108$

$3^x.4=108$

$3^x=108:4=27=3^3$

$\Rightarrow x=3$

\(45-5\left(x+4\right)=10\)

\(\Rightarrow5\left(x+4\right)=45-10\)

\(\Rightarrow5\left(x+4\right)=35\)

\(\Rightarrow x+4=35:5\)

\(\Rightarrow x+4=7\)

\(\Rightarrow x=7-4\)

\(\Rightarrow x=3\)

______________

\(\left(7x-15\right):3-6=17\)

\(\Rightarrow\left(7x-15\right):3=17+6\)

\(\Rightarrow\left(7x-15\right):3=23\)

\(\Rightarrow7x-15=23\cdot3\)

\(\Rightarrow7x-15=69\)

\(\Rightarrow7x=69+15\)

\(\Rightarrow7x=84\)

\(\Rightarrow x=12\)

______________

\(3^{x+2}-5\cdot3^x=108\)

\(\Rightarrow3^x\cdot\left(3^2-5\right)=108\)

\(\Rightarrow3^x\cdot\left(9-5\right)=108\)

\(\Rightarrow3^x\cdot4=108\)

\(\Rightarrow3^x=108:4\)

\(\Rightarrow3^x=27\)

\(\Rightarrow3^x=3^3\)

\(\Rightarrow x=3\)

a) 17 + x = -5 + 2x

=> x - 2x = -5 - 17

=> -x = -22

=> x = 22

b) 15 - 3x = -22 - 4x 

=> -3x + 4x = -22 - 15

=> x = -37

c) 47 - (4x + 5) = -(3x - 2)

=> 47 - 4x - 5 = -3x + 2

=> -4x + 3x = 2 - 47 + 5

=> -x = -40

=> x = 40

d) -58 + (3 - 7x) = 15 - 8x

=> -58 + 3 - 7x = 15 - 8x

=> -7x + 8x =15 + 58 - 3

=> x = 70

a) Ta có: \(3x\left(7x-2\right)-14x+4=0\)

\(\Leftrightarrow3x\left(7x-2\right)-2\left(7x-2\right)=0\)

\(\Leftrightarrow\left(7x-2\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}7x-2=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}7x=2\\3x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{7}\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{2}{7};\dfrac{2}{3}\right\}\)

b) ĐKXĐ: \(x\notin\left\{0;3\right\}\)

Ta có: \(\dfrac{2x+1}{x-3}+\dfrac{5-3x}{x}=\dfrac{2x^2-15}{x^2-3x}\)

\(\Leftrightarrow\dfrac{x\left(2x+1\right)}{x\left(x-3\right)}+\dfrac{\left(5-3x\right)\left(x-3\right)}{x\left(x-3\right)}=\dfrac{2x^2-15}{x\left(x-3\right)}\)

Suy ra: \(2x^2+x+5x-15-3x^2+9x-2x^2+15=0\)

\(\Leftrightarrow-3x^2+15x=0\)

\(\Leftrightarrow-3x\left(x-5\right)=0\)

mà -3<0

nên x(x-5)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=5\left(nhận\right)\end{matrix}\right.\)

Vậy: S={5}

A = - 3\(x\).(\(x-5\)) + 3(\(x^2\) - 4\(x\)) - 3\(x\) - 10

A = - 3\(x^2\) + 15\(x\) + 3\(x^2\) - 12\(x\) - 3\(x\) - 10

A = (- 3\(x^2\) + 3\(x^2\)) + (15\(x\) - 12\(x\) - 3\(x\)) - 10

A = 0 + (3\(x-3x\)) - 10

A = 0  - 10

A = - 10 

\(\frac{3}{7}\left(7x+\frac{2}{3}\right)-2\left(\frac{5}{14}-3x\right)=\frac{15}{5}\)

\(\Leftrightarrow3x+\frac{2}{7}-\frac{5}{7}+6x=3\)

\(\Leftrightarrow9x=\frac{24}{7}\)

\(\Leftrightarrow x=\frac{8}{21}\)

\(\frac{3}{7}.\left(7x+\frac{2}{3}\right)-2.\left(\frac{5}{14}-3x\right)=\frac{15}{5}\)

\(\Leftrightarrow3x+\frac{2}{7}.x-\frac{5}{7}+6x=3\)

\(\Leftrightarrow\frac{65}{7}.x-\frac{5}{7}=3\)

\(\Leftrightarrow\frac{65}{7}.x=3+\frac{5}{7}\)

\(\Leftrightarrow\frac{65}{7}.x=\frac{26}{7}\)

\(\Leftrightarrow x=\frac{26}{7}:\frac{65}{7}\)

\(\Leftrightarrow x=\frac{2}{5}\)

 3/7.(7x+2/3)-2.(5/14-3x)=15/5

\(\Leftrightarrow3x+\frac{2}{7}.x-\frac{5}{7}+6x=3\)

\(\Leftrightarrow\frac{65}{7}.x-\frac{5}{7}=3\)

\(\Leftrightarrow\frac{65}{7}.x=\frac{26}{7}\)\(\Leftrightarrow x=\frac{2}{5}\)

Vậy x =\(\frac{2}{5}\)