Bài 2

 x chia hết cho 12; 21; 28 => x ∈ BC(12;21;28) 

12 = 2².3 ; 21 = 3.7; 28 = 2².7 => BCNN (12;21;28) = 2².3,7 = 84

=> x ∈ {0;84; 168; 252; 336;...} 

Vì 150 < x < 300 nên x = 168 hoặc x = 252

ta có : 144=2⁴.3²

Bài 1 : ta có : 192=2⁶.3 và 144=2⁴.3²

Vậy ƯCLN_(144;192)=2⁴.3=48

Vậy ƯC_(144;192)={1;2;3;4;6;8;12;16;24;48}

Vậy các số cần tìm là : 24;48

\(1,\) Ta có \(144=3^2\cdot2^4;192=3\cdot2^6\)

\(\RightarrowƯCLN\left(144;192\right)=3\cdot2^4=48\)

\(\Rightarrow a\inƯ\left(48\right)=\left\{1;2;34;6;8;12;16;24;48\right\}\)

Mà \(a>20\)

\(\Rightarrow a\in\left\{24;48\right\}\)

TL :

\(12\left(x-81\right):3=4^3+2^3\)

\(12\left(x-81\right):3=64+8\)

\(12\left(x-81\right):3=72\)

\(12\left(x-81\right)=72.3\)

\(12\left(x-81\right)=216\)

\(x-81=216:12\)

\(x-81=18\)

\(\Rightarrow x=109\)

Câu b đang suy nghĩ. ( mik ngu vl )

\(3x-6-x-14=0\)

\(3x-6-1x=0+14\)

\(3x-6-x=14\)

\(\left(3-1\right)x-6=14\)

\(2x=14+6\)

\(2x=20\)

\(\Rightarrow x=10\)

Tks bạn nha

a: \(35=5\cdot7;105=3\cdot5\cdot7\)

=>\(ƯCLN\left(35;105\right)=5\cdot7=35\)

\(35⋮x;105⋮x\)

=>\(x\inƯC\left(105;35\right)\)

=>\(x\inƯ\left(35\right)\)

=>\(x\in\left\{1;5;7;35\right\}\)

mà x>5

nên \(x\in\left\{7;35\right\}\)

b: \(144=2^4\cdot3^2;192=2^6\cdot3;240=2^4\cdot3\cdot5\)

=>\(ƯCLN\left(144;192;240\right)=2^4\cdot3=48\)

\(144⋮x;192⋮x;240⋮x\)

=>\(x\inƯC\left(192;144;240\right)\)

=>\(x\inƯ\left(48\right)\)

=>\(x\in\left\{1;2;3;4;6;8;12;16;24;48\right\}\)

mà 10<=x<=99

nên \(x\in\left\{12;16;24;48\right\}\)

\(\left(x+2\right)-2=0\)

\(\Rightarrow x+2-2=0\)

\(\Rightarrow x=0\)

\(\left(x+3\right)+1=7\)

\(\Rightarrow x+3+1=7\)

\(\Rightarrow x+4=7\)

\(\Rightarrow x=3\)

\(\left(3x-4\right)+4=12\)
\(\Rightarrow3x-4+4=12\)

\(\Rightarrow3x=12\)

\(\Rightarrow x=4\)

\(\left(5x+4\right)-1=13\)

\(\Rightarrow5x+4-1=13\)

\(\Rightarrow5x+3=13\)

\(\Rightarrow5x=10\)

\(\Rightarrow x=2\)

\(\left(4x-8\right)-3=5\)

\(\Rightarrow4x-8-3=5\)

\(\Rightarrow4x-11=5\)

\(\Rightarrow4x=16\)

\(\Rightarrow x=4\)

\(8-\left(2x+4\right)=2\)

\(\Rightarrow8-2x-4=2\)

\(\Rightarrow4-2x=2\)

\(\Rightarrow2x=2\)

\(\Rightarrow x=1\)

\(7+\left(5x+2\right)=14\)

\(\Rightarrow7+5x+2=14\)

\(\Rightarrow9+5x=14\)

\(\Rightarrow5x=5\)

\(\Rightarrow x=1\)

\(5-\left(3x-11\right)=1\)

\(\Rightarrow5-3x+11=1\)

\(\Rightarrow16-3x=1\)

\(\Rightarrow3x=15\)

\(\Rightarrow x=5\)

tích tao đi

1.

(12x-5)(4x-1)+(3x-7)(1-16x)=81

48x²-12x-20x+5+3x-48x²-7+112x=81

48x²-48x²-12x-20x+3x+112x+5-7=81

83x-2=81

83x=81+2

83x=83

x=83:83

x=1

         4x(12x-5)-1(12x-5)+3x(1-16x)-7(1-16x)=81

         48x²-20x-12x+5+3x-48x²-7+112x=81

             83x-2=81

             83x=83

             x=1

2.Gọi 3 số đó là 2k-2;2k;2k+2

Theo đầu bài ta có 

2k(2k+2)-2k(2k-2)=192

4k²+4k-4k²+4k=192

8k=192

k=24

=>Số thứ nhất là:2.24-2=46

    Số thứ hai là :2.24=48

    Số thứ ba là:2.24+2=50 

Bài 1:

( 12x - 5 )( 4x - 1 ) + ( 3x - 7 )( 1 - 16x ) = 81

<=>48x²-12x-20x+5+3x-48x²-7+112x=81

<=>(48x²-48x²)-(12x+20x-3x-112x)+(5-7)=81

<=>-(-83x)-2=81

<=>83x=81+2

<=>83x=83

<=>x=1

Bài 2:

Gọi 3 số chẵn liên tiếp là a;a+2;a+4

Ta có: 

(a+2)(a+4)-a(a+2)=192

a²+4a+2a+8-a²-2a=192

(a²-a²)+(2a-2a)+4a+8=192

4a+8=192

4a=184

a=46

Vậy 3 số đó là 46;48;50

  (12+3x)^2=1a96

=>(12+3x)^2=1296=36^2

=>12+3x=36=-36

=>3x=24=-48

=>x=8=-16

nhận đúng nha

không có câu trả lời nhưng cứ cho

(12+3x)^2=1a96

=> (12+3x)^2=1a96

    (12+3x)^2=36^2

=>(12+3x)=36

=>12+3x=36 => 3x=24 => x=8

=>12+3x=-36 => 3x=-48 =>x=-16

Nhớ cho mk nhé, các bạn!!!

a) \(18-\left(2x+5\right)=9\)

\(2x+5=18-9\)

\(2x+5=9\)

\(2x=9-5\)

\(2x=4\)

\(x=2\)

a) \(18-\left(2x+5\right)=9\)

\(\Rightarrow2x+5=18-9=9\)

\(\Rightarrow2x=9-5=4\Rightarrow x=4:2=2\)

b) \(23x-4=32\Rightarrow23x=32+4=36\Rightarrow x=\dfrac{36}{23}\)

c) \(\left(3x+2\right)^2=64\)

\(\Rightarrow\left[{}\begin{matrix}3x+2=8\\3x+2=-8\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{10}{3}\end{matrix}\right.\)

d) \(x\left(2x-12\right)=0\Rightarrow6x\left(x-2\right)=0\)

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

2x+5=18-9

2x+5=9

2x=9-5

2x=4

  x=4:2

  x=2

c)3x+2=64:2

   3x+2=32

    3x=32-2

    3x=30

     x=30:3

     x=10

a: \(50=2\cdot5^2;75=5^2\cdot3\)

=>\(ƯCLN\left(50;75\right)=5^2=25\)

=>\(ƯC\left(50;75\right)=Ư\left(25\right)=\left\{1;5;25\right\}\)

\(x\inƯC\left(50;75\right)\)

=>\(x\in\left\{1;5;25\right\}\)

mà x<=20

nên \(x\in\left\{1;5\right\}\)

b: \(35=5\cdot7;105=3\cdot5\cdot7\)

=>\(ƯCLN\left(35;105\right)=5\cdot7=35\)

\(35⋮x;105⋮x\)

=>\(x\inƯC\left(35;105\right)\)

=>\(x\inƯ\left(35\right)\)

=>\(x\in\left\{1;5;7;35\right\}\)

mà x>5

nên \(x\in\left\{7;35\right\}\)

c: \(144=2^4\cdot3^2;192=2^6\cdot3;240=2^4\cdot3\cdot5\)

=>\(ƯCLN\left(144;192;240\right)=2^4\cdot3=48\)

\(144⋮x;192⋮x;240⋮x\)

=>\(x\inƯC\left(144;192;240\right)\)

mà x lớn nhất

nên x=ƯCLN(144;192;240)=48

d: \(54=3^3\cdot2;14=2\cdot7\)

=>\(ƯCLN\left(54;14\right)=2\)

\(x\inƯC\left(54;14\right)\)

mà x lớn nhất

nên x=ƯCLN(54;14)

=>x=2