Bài 15:

a: 2x+x=45

=>3x=45

=>\(x=\frac{45}{3}=15\)

b: 2x+7x=918

=>\(x\cdot\left(7+2\right)=918\)

=>9x=918

=>\(x=\frac{918}{9}=102\)

c: 2x+3x=60+5

=>5x=65

=>\(x=\frac{65}{5}=13\)

d: \(11x+22x=33\cdot2\)

=>33x=66

=>\(x=\frac{66}{33}=2\)

Bài 14:

a: (12-x)(2-x)=0

=>\(\left[\begin{array}{l}12-x=0\\ 2-x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=12\\ x=2\end{array}\right.\)

b: (x-33)(11-x)=0

=>\(\left[\begin{array}{l}x-33=0\\ 11-x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=33\\ x=11\end{array}\right.\)

c: (21-x)(12-x)=0

=>\(\left[\begin{array}{l}21-x=0\\ 12-x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=21\\ x=12\end{array}\right.\)

d: (50-x)(x-150)=0

=>\(\left[\begin{array}{l}50-x=0\\ x-150=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=50\\ x=150\end{array}\right.\)

Bài 13:

a: (x-2)(x-3)=0

=>\(\left[\begin{array}{l}x-2=0\\ x-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=3\end{array}\right.\)

b: (x-3)(x-4)=0

=>\(\left[\begin{array}{l}x-3=0\\ x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=4\end{array}\right.\)

c: (x-7)(6-x)=0

=>\(\left[\begin{array}{l}x-7=0\\ 6-x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=7\\ x=6\end{array}\right.\)

d: (x-3)(x-13)=0

=>\(\left[\begin{array}{l}x-3=0\\ x-13=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=13\end{array}\right.\)

a: \(x\in\left\{0;25\right\}\)

c: \(x\in\left\{0;5\right\}\)

a. x( x+ 3)= 0 

⇔ x= 0 hoặc x+ 3= 0

⇔ x= 0          x = -3

b. x( 2x− 1)+ 2( 2x− 1) =0 

⇔ ( 2x− 1)(x+ 2) =0

⇔ 2x− 1 =0 hoặc  x+ 2 =0

⇔ 2x       =1          x      = -2

⇔   x       =\(\dfrac{1}{2}\)         x      = -2

a

\(x^2\left(2x+15\right)+4\left(2x+15\right)=0\\ \Leftrightarrow\left(2x+15\right)\left(x^2+4\right)=0\\ \Leftrightarrow2x+15=0\left(x^2+4>0\forall x\right)\\ \Leftrightarrow2x=-15\\ \Leftrightarrow x=-\dfrac{15}{2}\)

b

\(5x\left(x-2\right)-3\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(5x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\5x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0+2=2\\x=\dfrac{0+3}{5}=\dfrac{3}{5}\end{matrix}\right.\)

c

\(2\left(x+3\right)-x^2-3x=0\\ \Leftrightarrow2\left(x+3\right)-\left(x^2+3x\right)=0\\ \Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\2-x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0-3=-3\\x=2-0=2\end{matrix}\right.\)

a: =>(2x+15)(x^2+4)=0

=>2x+15=0

=>2x=-15

=>x=-15/2

b; =>(x-2)(5x-3)=0

=>x=2 hoặc x=3/5

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

=>x=2 hoặc x=-3

a: \(\Leftrightarrow\left(x+2\right)\left(x+2-2x+10\right)=0\)

\(\Leftrightarrow x\in\left\{-2;12\right\}\)

3) \(x\left(x-4\right)+\left(x-4\right)^2=0\Leftrightarrow\left(x-4\right)\left(x+x-4\right)=0\Leftrightarrow2\left(x-4\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

4x.(x+1)-8(x+1)=0

(4x-8)(x+1)=0

suy ra x=2 hoặc x=-1

`x(2x-4)-(x-2)(2x+3)=0`

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

`<=>(x-2)(2x-2x-3)=0`

`<=>(x-2)*(-3)=0`

`<=>x-2=0`

`<=>x=2`

a) x = 1; x = - 1 3                 b) x = 2.

c) x = 3; x = -2.                 d) x = -3; x = 0; x = 2.