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

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

-5 ,ủng hộ mk nha

nếu >0 thì hai nhân tử cùng dấu

<0 thì trái dấu

a) Vì (2x + 4)(x + 2) = 0 => 2x + 4 hoặc x + 2 = 0

*) 2x + 4 = 0 => x = (0 - 4) : 2 = -2

*) x + 2 = 0 => x = -2

Vậy x chỉ có thể  = -2

b) x - (25 - 9) = 11 - (15 + x)

=> x + (15 + x) = 11 + (25 - 9)

=> 2x + 15 = 11 + 16

=> 2x = 27 - 15 = 12

=> x = 12 : 2 = 6

a/ => 2x + 4 = 0 => 2x = -4 => x = -2

hoặc x + 2 = 0 => x = -2

Vậy x = -2

b/ => x - 16 = 11 - 15 - x

=> 2x = 12

=> x = 6

c) (4x - 8)[x + (-3)] = 0

=> 4(x - 2)(x - 3) = 0

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

=> x - 2 = 0 hoặc x - 3 = 0

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

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

Vậy x \(\in\){2;3}

11(x - 6) = 4x + 11

=> 11x - 66 = 4x + 11

=> 11x - 4x = 11 + 66

=> 7x = 77

=> x = 77/7

=> x = 11

a: (2x+3)(y-4)=12

mà 2x+3 lẻ

nên (2x+3;y-4)∈{(1;12);(-1;-12);(3;4);(-3;-4)}

=>(2x;y)∈{(-2;16);(-4;-8);(0;8);(-6;0)}

=>(x;y)∈{(-1;16);(-2;-8);(0;8);(-3;0)}

b; x(2y+1)-4y=3

=>x(2y+1)-4y-2=1

=>x(2y+1)-2(2y+1)=1

=>(x-2)(2y+1)=1

=>(x-2;2y+1)∈{(1;1);(-1;-1)}

=>(x;2y)∈{(3;0);(1;-2)}

=>(x;y)∈{(3;0);(1;-1)}

c: xy+2x+y+11=0

=>x(y+2)+y+2+9=0

=>(x+1)(y+2)=-9

=>(x+1;y+2)∈{(1;-9);(-9;1);(-1;9);(9;-1);(3;-3);(-3;3)}

=>(x;y)∈{(0;-11);(-10;-1);(-2;7);(8;-3);(2;-5);(-4;1)}