Đáp án`+`Giải thích các bước giải:

 Câu `1:`

`a)` `3xy(x^2-2xy+5)`

`=3xy.x^2-3xy.2xy+3xy.5`

`=3x^3y-6x^2y^2+15xy`

`**` Nhân đơn thức với đa thức

`b)` `x^2+2xy-25+y^2`

`=(x^2+2xy+y^2)-25`

`=(x+y)^2-5^2`

`=(x+y+5)(x+y-5)`

`**` Nhóm các hạng tử

`**` HĐT `3:A^2-B^2=(A+B)(A-B)`

$\\$

 Câu `2:`

`a)` `4x^2+20x=0`

`4x(x+5)=0`

`=>[(4x=0),(x+5=0):}`

`=>[(x=0),(x=-5):}`

Vậy `x∈{0;-5}`

`**` Đặt nhân tử chung: `A.B=0=>[(A=0),(B=0):}`

`b)` `x(x+3)-3x-9=0`

`x(x+3)-(3x+9)=0`

`x(x+3)-3(x+3)=0`

`(x+3)(x-3)=0`

`=>[(x+3=0),(x-3=0):}`

`=>[(x=-3),(x=3):}`

Vậy `x∈{3;-3}`

`**` Đặt nhân tử chung: `A.B=0=>[(A=0),(B=0):}`

$\\$

 Câu `3:`

`a)` `M=18/(x^2-9)+5/(x-3)+3/(x+3)`     `(x\ne±3)`

`=18/((x-3)(x+3))+5/(x-3)+3/(x+3)`    

`=18/((x-3)(x+3))+(5(x+3))/((x-3)(x+3))+(3(x-3))/((x-3)(x+3))`    

`=(18+5x+15+3x-9)/((x-3)(x+3))`

`=(8x+24)/((x-3)(x+3))`

`=(8(x+3))/((x-3)(x+3))`

`=8/(x-3)`

Vậy `M=8/(x-3)`

`b)` Thay `x=11` vào `M` được:

`M=8/(11-3)=8/8=1`

Vậy `M=1` khi `x=11`

#andy

Bài 1: $\href{https://hoidap247.com/cau-hoi/3565100}{\color{black}{\text{ tại đây }}}$

\[\begin{array}{l}
2)\\
a)4{x^2} + 20x = 0\\
 \Leftrightarrow 4x\left( {x + 5} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
4x = 0\\
x + 5 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x =  - 5
\end{array} \right.\\
 \Rightarrow S \in \left\{ {0; - 5} \right\}\\
b)x\left( {x + 3} \right) - 3x - 9 = 0\\
 \Leftrightarrow x\left( {x + 3} \right) - \left( {3x + 9} \right) = 0\\
 \Leftrightarrow x\left( {x + 3} \right) - 3\left( {x + 3} \right) = 0\\
 \Leftrightarrow \left( {x - 3} \right)\left( {x + 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 3 = 0\\
x + 3 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x =  - 3
\end{array} \right.\\
 \Leftrightarrow S \in \left\{ {3; - 3} \right\}\\
3)M = \dfrac{{18}}{{{x^2} - 9}} + \dfrac{5}{{x - 3}} + \dfrac{3}{{x + 3}}\\
DKXD:\left\{ \begin{array}{l}
x - 3 \ne 0\\
x + 3 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne 3\\
x \ne  - 3
\end{array} \right.\\
 = \dfrac{{18}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} + \dfrac{5}{{x - 3}} + \dfrac{3}{{x + 3}}\\
 = \dfrac{{18}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} + \dfrac{{5\left( {x + 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} + \dfrac{{3\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
 = \dfrac{{18}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} + \dfrac{{5x + 15}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} + \dfrac{{3x - 9}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
 = \dfrac{{18 + 5x + 15 + 3x - 9}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
 = \dfrac{{8x + 24}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
 = \dfrac{{8\left( {x + 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
 = \dfrac{8}{{x - 3}}\\
thay\,\,x = 11\\
 \Rightarrow M = \dfrac{8}{{11 - 3}}\\
 = \dfrac{8}{8} = 1
\end{array}\]