bn hỏi 1 lần thôi đừng đăng nhiều chật lắm

12x = 1800 - 264

12x = 1536

x = 1536 : 12

x = 128

( x + 73 ) : 19 = 321

( x + 73 ) = 321 x 19

 x + 73 = 6099

x = 6099 - 73

x = 6026

2x + 3x + 4x = 18 

x . ( 2 + 3 + 4 ) = 18

x .          9          = 18

x = 18 : 9

x = 2

10 < 2x < 20

x = 6 ; 7 ; 8 ; 9

264 + 2x = 1800

2x = 1800 - 246 

2x = 1554

x = 1554 : 2

x = 777

( x + 73 ) : 19 = 321

x + 73 = 321 x 19

x  + 73 = 6099

x = 6099 - 73

x = 6026

2x + 3x + 4x = 18

x = 2

Câu cuối của bài 1 và bài 2 mình ko biết ~

1 /

a )         264 + 2x = 1800

              2 x = 1800 - 264 = 1536

              x = 1536 : 2 = 768

b )          ( x + 73 ) : 19 = 321

                x + 73 = 321 x 19 = 6099

               x = 6099 - 73 = 6026

c )           2 x + 3 x + 4 x = 18

               9 x = 18

               x = 18 : 9 = 2

d )          x = 5 , 6 , 7 , 8 , 9 ,  10 

nhé !

ước của 300 là : 1 , 2 , 3 , 4 , 5 , 6 , 10 , 12 , 15 , 20 , 30 , ........ , 300

bội của 20  : 20 , 40 , 60 , 80 , 100 , 120 , 140 , 160 , 180 , 200 , ........... , 300 

có 5 số 

nhé !

a) \(\text{264 + 12x = 1800}\)

\(\Rightarrow12x=1800-264\)

\(\Rightarrow12x=1536\)

\(\Rightarrow x=1536:12\)

\(\Rightarrow x=128\)

b) \(\text{( x + 73 ) : 19 = 321}\Rightarrow x+73=6099\Rightarrow x=6026\)

c) 2x + 3x + 4x = 18

=> x.(2+3+4)=18

=> x.9 = 18

=> x=18 : 2

=> x= 2

d) 10 < 2x <  20   ( x thuộc N )  

Ta có : số lớn hơn hoặc bằng 10 và bé hơn hoặc bằng 20 là :

10;11;12;13;14;15;16;17;18;19;20

=>  ... tiếp đi nha bạn tự làm đc mà

ko làm đc hỏi mk

Ai giúp mk với 

\(1,\Leftrightarrow7x=42\Leftrightarrow x=6\\ 2,\Leftrightarrow36-4x=4\Leftrightarrow4x=32\Leftrightarrow x=8\\ 3,\Leftrightarrow3x=35\Leftrightarrow x=\dfrac{35}{3}\\ 4,\Leftrightarrow x-12=144\Leftrightarrow x=156\\ 5,\Leftrightarrow x-14=16\Leftrightarrow x=30\\ 6,\Leftrightarrow3x-24=\dfrac{148}{73}\Leftrightarrow3x=\dfrac{1900}{73}\Leftrightarrow x=\dfrac{1900}{219}\\ 7,\Leftrightarrow33+x=45\Leftrightarrow x=12\\ 8,Sai.đề\\ 9,\Leftrightarrow\left(x+9\right):2=39\Leftrightarrow x+9=78\Leftrightarrow x=69\\ 11,\Leftrightarrow2\left(x+7\right)=38\Leftrightarrow x+7=19\Leftrightarrow x=12\\ 13,\Leftrightarrow2\left(x-51\right)=66\Leftrightarrow x-51=33\Leftrightarrow x=84\)

\(15,\Leftrightarrow x-19=9\Leftrightarrow x=28\\ 17,\Leftrightarrow\left(x-3\right):2=48\Leftrightarrow x-3=96\Leftrightarrow x=99\\ 19,\Leftrightarrow0x=46\Leftrightarrow x\in\varnothing\\ 8,Sai.đề\\ 14,\Leftrightarrow2x=209\Leftrightarrow x=\dfrac{209}{2}\\ 16,\Leftrightarrow2x+6=157\Leftrightarrow2x=151\Leftrightarrow x=\dfrac{151}{2}\\ 18,\Leftrightarrow5\left(x+4\right)=100\Leftrightarrow x+4=20\Leftrightarrow x=16\\ 20,\Leftrightarrow\left(3x-5\right)^3=27=3^3\Leftrightarrow3x-5=3\Leftrightarrow x=\dfrac{8}{3}\)

1

a, -114

b, -1000

c, 87

d, 164

e, 981

2

a, x=14

b, x E {-32,2}

c, đề sai

d, 262

e, -3

g, đề sai

h, -3

Ta có : 3x + 8 = 3(x - 1) + 11

Do x - 1 \(⋮\)x - 1 => 3(x - 1) \(⋮\)x - 1

Để 3x  + 8 \(⋮\)x - 1 thì 11 \(⋮\)x - 1 => x - 1 \(\in\)Ư(11) = {1; 11; -1; -11}

Lập bảng : 

Vậy ...

Cảm ơn nha!

a) \(58+7x=100\)

\(=>7x=100-58\)

\(=>7x=42\)

\(=>x=42:7\)

\(=>x=6\)

b) \(3x-7=28\)

\(=>3x=28+7\)

\(=>3x=35\)

\(=>x=35:3\)

\(=>x=\dfrac{35}{3}\)

c) \(x-56:4=16\)

\(=>x-14=16\)

\(=>x=16+14\)

\(=>x=30\)

d) \(101+\left(36-4x\right)=105\)

\(=>36-4x=105-101\)

\(=>36-4x=4\)

\(=>4x=36-4\)

\(=>4x=32\)

\(=>x=32:4\)

\(=>x=8\)

e) \(\left(x-12\right):12=12\)

\(=>x-12=12.12\)

\(=>x-12=144\)

\(=>x=144-12\)

\(=>x=132\)

f) \(\left(3x-2^4\right).7^3=2.7^4\)

\(=>3x-2^4=2.7^4:7^3\)

\(=>3x-16=2.7=14\)

\(=>3x=14+16\)

\(=>3x=30\)

\(=>x=30:3\)

\(=>x=10\)

i) \(\left(10+2x\right).4^{2011}=4^{2013}\)

\(=>10+2x=4^{2013}:4^{2011}\)

\(=>10+2x=4^2=16\)

\(=>2x=16-10\)

\(=>2x=6\)

\(=>x=6:2\)

\(=>x=3\)

\(#WendyDang\)

a: \(x^3-4x^2-x+4=0\)

=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)

=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)

=>\(\left(x-4\right)\left(x^2-1\right)=0\)

=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)

b: Sửa đề: \(x^3+3x^2+3x+1=0\)

=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)

=>\(\left(x+1\right)^3=0\)

=>x+1=0

=>x=-1

c: \(x^3+3x^2-4x-12=0\)

=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)

=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)

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

=>\(\left(x+3\right)\left(x-2\right)\left(x+2\right)=0\)

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

d: \(\left(x-2\right)^2-4x+8=0\)

=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)

=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)

=>\(\left(x-2\right)\left(x-2-4\right)=0\)

=>(x-2)(x-6)=0

=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)