Đáp án:

Giải thích các bước giải:

 1) (x+4)(x-7) = x²-7x+4x-28 = x²-3x-28

 2) (a-3)(5-a) = 5a-a²-15+3a = -a²+8a-15 = -(a²-8a+15)

 3) (3x+y)(x-2y) = 3x²-6xy+xy-2y² = 3x²-5xy-2y²

 4) (x-2y)(x+y) = x²+xy-2xy-2y² = x²-xy-2y²

 5) (x²-2x)(x+1) = x³+x²-2x²-2x = x³-x²-2x

 6) (x-3)(x²+x+1) = x³+x²+x-3x²-3x-1 = x³-2x²-2x-1

 7) (x+y)(x-y+5) = x²-xy+5x+xy-y²+5y = x²+5x+5y-y²

 8) (x-6)(x+y-z) = x²+xy-xz-6x-6x+6z 

 9) -(x+y)(x-2y) = -(x²-2xy+xy-2y²) = -(x²-xy-2y²)

10) -(x+2)(x²+2x+2) = -(x³+2x²+2x+2x²+4x+4) = -(x³+4x²+6x+4)

11) 5x+(x-3)(x-4) = 5x+x²-4x-3x+12 = x²-2x+12

12) x²-2x-(x+7)(x+2) = x²-2x-x²-2x-7x-14 = -(11x+14)

13) (3x-2)(x+9)-3x = 3x²+27x-2x-18-3x = 3x²+22x-18

14) (x-4)(x-1)+(x-2)(3-x) = x²-x-4x+4+3x-x²-6+2x = -6

15) (x+7)(x-5)+(x-1)(2x+3) = x²-5x+7x-35+2x²+3x-2x-3 = 3x²-x-38

16) (x-2)(x+3)-(x-1)(2x-1) = x²+3x-2x-6-2x²+x+2x-1 = -(x²-4x+7)

17) 3x²-3(x+1)(x+2) = 3x²-3(x²+3x+2) = 3x²-3x²-9x-6 = -(9x+6)

18) 5x-3+(x-5)(x+4) = 5x-3+x²+4x-5x-20 = x²+4x-23

Đáp án:

$\begin{array}{l}
2)\left( {a - 3} \right).\left( {5 - a} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
a - 3 = 0\\
5 - a = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
a = 3\\
a = 5
\end{array} \right.\\
Vậy\,a = 3;a = 5\\
4)\left( {x - 2y} \right)\left( {x + y} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 2y = 0\\
x + y = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 2y\\
x =  - y
\end{array} \right.\\
Vậy\,x = 2y;x =  - y\\
6)\left( {x - 3} \right)\left( {{x^2} + x + 1} \right) = 0\\
 \Leftrightarrow x - 3 = 0\left( {do:{x^2} + x + 1 > 0} \right)\\
 \Leftrightarrow x = 3\\
Vậy\,x = 3\\
8)\left( {x - 6} \right)\left( {x + y - z} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 6\\
x = z - y
\end{array} \right.\\
Vậy\,x = 6;x = z - y\\
10) - \left( {x + 2} \right)\left( {{x^2} + 2x + 2} \right) = 0\\
 \Leftrightarrow x + 2 = 0\\
 \Leftrightarrow x =  - 2\\
Vậy\,x =  - 2\\
12){x^2} - 2x - \left( {x + 7} \right)\left( {x + 2} \right) = 0\\
 \Leftrightarrow {x^2} - 2x - \left( {{x^2} + 7x + 2x + 14} \right) = 0\\
 \Leftrightarrow {x^2} - 2x - {x^2} - 9x - 14 = 0\\
 \Leftrightarrow 11x =  - 14\\
 \Leftrightarrow x =  - \dfrac{{14}}{{11}}\\
Vậy\,x =  - \dfrac{{14}}{{11}}\\
14)\left( {x - 4} \right)\left( {x - 1} \right) + \left( {x - 2} \right)\left( {3 - x} \right) = 0\\
 \Leftrightarrow {x^2} - 5x + 4 - {x^2} + 5x - 6 = 0\\
 \Leftrightarrow  - 2 = 0\left( {ktm} \right)\\
Vậy\,x \in \emptyset \\
16)\left( {x - 2} \right)\left( {x + 3} \right) - \left( {x + 1} \right)\left( {2x - 1} \right) = 0\\
 \Leftrightarrow {x^2} + x - 6 - \left( {2{x^2} + x - 1} \right) = 0\\
 \Leftrightarrow {x^2} + x - 6 - 2{x^2} - x + 1 = 0\\
 \Leftrightarrow {x^2} =  - 5\left( {ktm} \right)\\
Vậy\,x \in \emptyset \\
18)5x - 3 + \left( {x - 5} \right)\left( {x + 4} \right) = 0\\
 \Leftrightarrow 5x - 3 + {x^2} - x - 20 = 0\\
 \Leftrightarrow {x^2} + 4x - 23 = 0\\
 \Leftrightarrow {x^2} + 4x + 4 - 27 = 0\\
 \Leftrightarrow {\left( {x + 2} \right)^2} = 27 = {\left( {3\sqrt 3 } \right)^2}\\
 \Leftrightarrow x =  - 2 \pm 3\sqrt 3 \\
Vậy\,x =  - 2 \pm 3\sqrt 3 
\end{array}$