Chọn A

\(2x+\left(1+2+3+...+100\right)=15150\)

\(2x+\left[\left(1+100\right)+\left(2+99\right)+...+\left(50+51\right)\right]=15150\)

\(2x+\left[101+101+...+101\right]=15150\)CÓ 50 SỐ 101

\(2x+\left[101\times50\right]=15150\)

\(2x=15150:5050\)

\(2x=3\)

\(x=3:2\)

\(x=1.5\)

a, 2x + (1+2+3+4+...+100) = 15150 

=> 2x + \(\frac{\left(1+100\right).\left[\left(100-1\right)+1\right]}{2}\)= 15150 

=> 2x + \(\frac{101.100}{2}\)= 15150 

=> 2x + 5050 = 15150 

=> 2x             = 15150 - 5050 

=> 2x             = 10100

=> x              =  10100 : 2 

=> x              = 5050 

Vậy x = 5050 

b, .(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)+(x+7)+(x+8)=36 

=> (x + x + x + x +x + x +x +x ) + (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8) = 36 

=> 8x + 36 = 36 

=> 8x         = 0 

=>  x          = 0 

Vậy x = 0 

c, 0+0+4+6+8+...+2x=110 

Sửa đề :0 + 2 + 4 + 6 + 8 + ... + 2x = 110 = 2 + 4 + 6 + 8 + ... + 2x = 110 

SSH  : \(\frac{\left(2\text{x}-2\right)}{2}+1=x-1+1=x\)

Tổng : \(\frac{\left(2\text{x}+2\right).x}{2}=110\Leftrightarrow\frac{2.\left(x+1\right).x}{2}=110\)

                                                    \(\Leftrightarrow\left(x+1\right)x=110\)

                                                     \(\Leftrightarrow\left(10+1\right).10=110\)

 => x = 10 

Vậy x = 10 

#)Giải :

a) 2x + ( 1 + 2 + 3 + 4 + ... + 100 ) = 15150

=> 2x + [ ( 100 + 1 ) x 100 : 2 ] = 15150

=> 2x + 5050 = 15150

=> 2x = 10100

=> x = 5050

b) ( x + 1 ) + ( x + 2 ) + ... + ( x + 8 ) + ( x + 9 ) = 36

=> ( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 9 ) = 36

=> ( x + x + x + ... + x ) + 45 = 36

=> ( x + x + x + ... + x ) = -9

=> x = -1

c) Đề kiểu j v ? xem lại đi bạn ???

1C

2A

1C        2A

\(a,x=ƯCLN\left(24,36\right)=12\\ b,x\in BC\left(4,6\right)=BC\left(12\right)=\left\{0;12;24;36;48;50;...\right\}\\ \text{Mà }0< x< 50\\ \Rightarrow x\in\left\{12;24;36;48\right\}\)

a, (x-35)-120=0

x-35=120+0

x-35=120

x=120+35

x=155

b,7x-8=713

7x=713+8

7x=721

x=721:7

x=103

c,4x:17=0

4x=0.17

4x=0

x=0:4

x=0

d,75+(131-x)=205

131-x=205-75

131-x=130

x=131-130

x=1

e,2x-36=4^6:4^3

2x-36=4^6-3

2x-36=4^3

2x-36=48

2x=48-36

2x=12

x=12:2

x=6

f, 2011^2.2011^x=2011^6

2011^x=2011^6: 2011^2

2011^x=2011^6-2

2011^x=2011^4

x=4

bn

\(\text{a/ ( x-35 ) - 120 = 0}\)

\(\Rightarrow\left(x-35\right)=120\)

\(\Rightarrow x=120+35\)

\(x=155\)

\(\text{b/ 7x - 8 = 713}\)

\(\Rightarrow7x=713+8=721\)

\(x=721:7=103\)

\(\text{c/ 4x : 17 = 0}\)

\(4x=0\)

\(x=0\)

\(\text{d/ 75+(131-x) = 205}\)

\(131-x=205-75=130\)

\(x=131-130=1\)

\(e.2x-36=4^6:4^3\)

\(2x-36=4^3=64\)

\(2x=64+36=100\)

\(x=100:2=50\)

\(f.2011^2.2011^x=2011^6\)

\(\text{Ta có công thức }:x^n.x^m=x^{n+m}\)

\(\Rightarrow x=6-2=4\)

Các bạn ơi, các bạn có biết cách thay đổi tài khoản không?

a)  x = 200   

b) x = 450

c)  x = 21

d)  x = 9 hoặc x = 36

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a) 2 . x + 12 = 36

2 . x = 36 - 12

2 . x = 24

x = 24 : 2

x = 12

Vậy x = 12

b) (x + 21)  : 8 + 12 = 21

( x + 21 ) : 8 = 21 - 12

( x + 21 ) : 8 = 9

x + 21 = 9 . 8

x + 21 = 72

x = 72 - 21

x = 51

Vậy x = 51

c) (3 . x - 18) . (x - 9) = 0

\(\Rightarrow\begin{cases}3.x-18=0\\x-9=0\end{cases}\)

\(\Rightarrow\begin{cases}3.x=18\\x=0+9\end{cases}\)

\(\Rightarrow\begin{cases}x=18:3=6\\x=9\end{cases}\)

Vậy \(x\in\left\{6;9\right\}\)

a)

\(2x+12=36\)

\(\Rightarrow2\left(x+6\right)=36\)

\(\Rightarrow x+6=18\)

=> x = 12

Vậy x = 12

b)

\(\left(x+21\right):8+12=21\)

\(\Rightarrow\left(x+21\right):8=9\)

\(\Rightarrow x+21=72\)

=> x = 51

Vậy x = 51

c)

\(\left(3x-18\right)\left(x-9\right)=0\)

\(\Rightarrow3\left(x-6\right)\left(x-9\right)=0\)

\(\Rightarrow\left(x-6\right)\left(x-9\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}x-6=0\\x-9=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=6\\x=9\end{array}\right.\)

Vậy x = 6 ; x = 9

a) \(2x+12=36\Leftrightarrow2x=24x\Leftrightarrow x=12\)

b) \(x+21:8+12=21\)

\(\Leftrightarrow\frac{x+21}{8}=9\)

\(\Leftrightarrow x+21=72\)

\(\Leftrightarrow x=51\)

c) \(\left(3.x-18\right).\left(x-9\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}3x-18=0\\x-9=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=6\\x=9\end{array}\right.\)

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