\(\frac{1}{2}x+150\%x=2014\)

\(\frac{1}{2}x+\frac{3}{2}x=2014\)

\(\left(\frac{1}{2}+\frac{3}{2}\right).x=2014\)

\(2.x=2014\)

\(\Rightarrow x=1007\)

\(\frac{1}{2}x+150\%x=2014\)

=>\(\frac{1}{2}x+\frac{3}{2}x=2014\)

=>x(\(\frac{1}{2}+\frac{3}{2}\))=2014

=>x.2=2014

=>x=1007

vậy x=1007

\(2x+\left(x-3\right)=\frac{1}{2}\)

\(2x+x-3=\frac{1}{2}\)

\(3x-3=\frac{1}{2}\)

\(3x=\frac{1}{2}+3\)

\(3x=\frac{7}{2}\)

\(x=\frac{7}{2}:3\)

\(\Rightarrow x=\frac{7}{6}\)

\(2x+\left(x-3\right)=\frac{1}{2}\)

\(2x+x-3=\frac{1}{2}\)

\(3x=\frac{7}{2}\)

\(\Leftrightarrow x=\frac{7}{6}\)

số nào x0 cx bằng 0

Ai tl tk hết nhé

\(A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)

\(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)

\(2A=2+\frac{2}{2}+\frac{2}{2^2}+...+\frac{2}{2^9}\)

\(2A=2+1+\frac{1}{2}+...+\frac{2}{2^8}\)

=>\(A=2+\frac{-1}{29}\)

=>\(A=\frac{57}{29}\)

\(A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)

=>  \(2A=2+1+\frac{1}{2}+...+\frac{1}{2^8}\)

=> \(A=\left(2+1+\frac{1}{2}+...+\frac{1}{2^8}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)

=> \(A=2+1+\frac{1}{2}+...+\frac{1}{2^8}-1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^9}\)

=> \(A=2-\frac{1}{2^9}\)

Hình (tự vẽ)

GT: - ΔABC đều (AB = AC = BC ;  = B = C = 60^o)

       - BM = CM 

KL: BÂM = 30^o  

Xét ΔABM và ΔACM có:

AB = AC (gt)

B = C = 60^o

BM = CM (gt)

Do đó:  ΔABM = ΔACM (c-g-c)

⇒ BÂM = MÂC (hai góc tương ứng)

Mà BÂM + MÂC = Â = 60^o

⇒ BÂM = MÂC = 60^o : 2 = 30^o

Tìm x

a) | 2x + 1 | = | x - 3 |

TH1:  2x + 1 = x - 3 ⇒ 2x - x = -3 - 1 ⇒ x = -4

TH2: 2x + 1 = -x + 3 ⇒ 2x + x = 3 - 1 ⇒ 3x = 2 ⇒ x = 2/3 

Vậy x = -4 hoặc x = 2/3

b) (2x + 1/3 ) ² = 1/100         (sửa đề) 

TH1: 2x + 1/3 =  1/10 ⇒ 2x = 1/10 - 1/3 = -7/30 ⇒ x = -7/30 : 2 = -7/60

TH2: 2x + 1/3 = -1/10 ⇒ 2x = -1/10 - 1/3 = -13/30 ⇒ x = -13/30 : 2 = -13/60

Vậy x = -7/60 hoặc x = -13/60

\(A=\frac{3n+2}{n-2}=\frac{3\left(n-2\right)+8}{n-2}=3+\frac{8}{n-2}\)

A nguyên khi  \(\frac{8}{n-2}\) nguyên

=> \(8⋮n-2\)      => \(n-2\inƯ\left(8\right)\)

=>  \(n-2\in\left\{1;2;4;8;-1;-2;-4;-8\right\}\)

=> \(n\in\left\{3;4;6;10;1;0;-2;-4;-6\right\}\)

x=\(\left(1-\frac{1}{2}\right)\).\(\left(1-\frac{1}{3}\right)\).\(\left(1-\frac{1}{4}\right)\)...\(\left(1-\frac{1}{10}\right)\)

x=\(\frac{1}{2}\).\(\frac{2}{3}\).\(\frac{3}{4}\)...\(\frac{9}{10}\)

x=\(\frac{1}{10}\)

vậy x=\(\frac{1}{10}\)

Cảm ơn bạn nhiều .

Mình k cho bạn 3k rồi .

Đó là món quà dành cho bạn khi giúp đỡ mình .

\(B=\frac{3^2}{8.11}+\frac{3^2}{11.14}+...+\frac{3^2}{197.200}\)

\(B=3.\left(\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{197.200}\right)\)

\(B=3.\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)

\(B=3.\left(\frac{1}{8}-\frac{1}{200}\right)\)

\(B=3.\frac{3}{25}\)

\(\Rightarrow B=\frac{9}{25}\)

\(B=\frac{3^2}{8.11}+\frac{3^2}{11.14}+...+\frac{3^2}{197.200}.\)

\(=3\left(\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{197.200}\right)\)

\(=3\left(\frac{11-8}{8.11}+\frac{14-11}{11.14}+...+\frac{200-197}{197.200}\right)\)

\(=3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)

\(=3\left(\frac{1}{8}-\frac{1}{200}\right)\)

\(=3\cdot\frac{3}{25}\)

\(=\frac{9}{25}\)