Đáp án:

$\begin{array}{l}
a){x^2} - 5x + 3\left( {x - 5} \right) = 0\\
 \Leftrightarrow x\left( {x - 5} \right) + 3\left( {x - 5} \right) = 0\\
 \Leftrightarrow \left( {x - 5} \right)\left( {x + 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 5 = 0\\
x + 3 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 5\\
x =  - 3
\end{array} \right.\\
\text{Vậy}\,x = 5;x =  - 3\\
b)2{x^2} + 9x - 5 = 0\\
 \Leftrightarrow 2{x^2} - x + 10x - 5 = 0\\
 \Leftrightarrow x\left( {2x - 1} \right) + 5\left( {2x - 1} \right) = 0\\
 \Leftrightarrow \left( {2x - 1} \right)\left( {x + 5} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - 1 = 0\\
x + 5 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \frac{1}{2}\\
x =  - 5
\end{array} \right.\\
\text{Vậy}\,x = \frac{1}{2};x =  - 5\\
c){\left( {3x - 5} \right)^2} - {\left( {x + 1} \right)^2} = 0\\
 \Leftrightarrow \left( {3x - 5 - x - 1} \right)\left( {3x - 5 + x + 1} \right) = 0\\
 \Leftrightarrow \left( {2x - 6} \right)\left( {4x - 4} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = 1
\end{array} \right.\\
\text{Vậy}\,x = 1;x = 3\\
d)3x\left( {x - 4} \right) - x\left( {5 + 3x} \right) =  - 34\\
 \Leftrightarrow x.\left( {3x - 12} \right) - x.\left( {5 + 3x} \right) =  - 34\\
 \Leftrightarrow x\left( {3x - 12 - 5 - 3x} \right) =  - 34\\
 \Leftrightarrow x.\left( { - 17} \right) =  - 34\\
 \Leftrightarrow x = 2\\
\text{Vậy}\,x = 2\\
e){x^3} - 25x = 0\\
 \Leftrightarrow x\left( {{x^2} - 25} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
{x^2} = 25
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 5\\
x =  - 5
\end{array} \right.\\
\text{Vậy}\,x = 0;x = 5;x =  - 5\\
f){\left( {x + 1} \right)^3} - {x^2}\left( {x + 3} \right) = 2\\
 \Leftrightarrow {x^3} + 3{x^2} + 3x + 1 - {x^3} - 3{x^2} - 2 = 0\\
 \Leftrightarrow 3x - 1 = 0\\
 \Leftrightarrow x = \frac{1}{3}\\
\text{Vậy}\,x = \frac{1}{3}\\
g){x^2}\left( {x - 5} \right) + 20 - 4x = 0\\
 \Leftrightarrow {x^2}\left( {x - 5} \right) - 4\left( {x - 5} \right) = 0\\
 \Leftrightarrow \left( {x - 5} \right)\left( {{x^2} - 4} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 5\\
x = 2\\
x =  - 2
\end{array} \right.\\
\text{Vậy}\,x = 5;x = 2;x =  - 2\\
h){x^2} - x - 6 = 0\\
 \Leftrightarrow {x^2} - 3x + 2x - 6 = 0\\
 \Leftrightarrow \left( {x - 3} \right)\left( {x + 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x =  - 2
\end{array} \right.\\
\text{Vậy}\,x = 3;x =  - 2
\end{array}$

Đáp án:

a) x = 5 hoặc x = -3

b) x = -5 hoặc x = 1/2

c) x = 1 hoặc x = 3

d) x = 2

e) x = 0 , x=  5 , x = -5

f) x = 1/3

g) x = 5 , x = -2 , x = 2

h) x = -2 hoặc x = 3

Giải thích các bước giải:

a) x^2 - 5x + 3(x - 5) = 0

=> x(x - 5) + 3(x - 5) = 0

=> (x - 5)(x + 3) = 0

=> x = 5 hoặc x = -3

b) 2x^2 + 9x - 5 = 0

=> 2x^2 + 10x - x - 5 = 0

=> 2x(x + 5) - (x + 5) = 0

=> (x + 5)(2x - 1) = 0

=> x = -5 hoặc x = 1/2

c) (3x - 5)^2 - (x + 1)^2 = 0

=> [3x - 5 + (x + 1)][3x - 5 - (x + 1)] = 0

=> (3x - 5 + x + 1)(3x - 5 - x - 1) = 0

=> (4x - 4)(2x - 6) = 0

=> 4x - 4 = 0 hoặc 2x - 6 = 0

=> 4x = 4 hoặc 2x = 6

=> x = 1 hoặc x = 3

d) 3x(x - 4) - x(5 + 3x) = -34

=> 3x^2 - 12x - 5x - 3x^2 = -34

=> -17x = -34

=> x = 2

e) x^3 - 25x = 0

=> x(x^2 - 25) = 0

=> x(x + 5)(x - 5) = 0

=> x = 0 , x=  -5 , x = 5

f) (x + 1)^3 - x^2(x + 3) = 2

=> x^3  +3x^2 + 3x + 1 - x^3 - 3x^2 - 2 = 0

=> 3x - 1 = 0

=> 3x = 1 => x = 1/3

g) x^2(x - 5) + 20 - 4x = 0

=> x^2(x - 5) + (-4x + 20) = 0

=> x^2(x - 5) - 4x+  20 = 0

=> x^2(x - 5) - 4 (x - 5) = 0

=> (x - 5)(x^2 - 4) = 0

=> (x - 5)(x + 2)(x - 2) = 0

=> x = 5 , x = -2 , x = 2

h) x^2 - x - 6 = 0

=> x^2 + 2x - 3x - 6 = 0

=> x(x + 2) - 3(x + 2) = 0

=> (x + 2)(x - 3) = 0

=> x = -2 hoặc x = 3