\(a,\left|\frac{1}{2}-2x\right|+\frac{2}{3}=\frac{7}{3}\)

\(\Leftrightarrow\left|\frac{1}{2}-2x\right|=\frac{7}{3}-\frac{2}{3}\)

\(\Leftrightarrow\left|\frac{1}{2}-2x\right|=\frac{5}{3}\)

\(\Rightarrow\orbr{\begin{cases}\frac{1}{2}-2x=\frac{5}{3}\\\frac{1}{2}-2x=-\frac{5}{3}\end{cases}}\Rightarrow\orbr{\begin{cases}2x=-\frac{7}{6}\\2x=\frac{13}{6}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=-\frac{7}{6}:2\\x=\frac{13}{6}:2\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{7}{12}\\x=\frac{13}{12}\end{cases}}\)

Vậy \(x=-\frac{7}{12}\)hoặc \(x=\frac{13}{12}\)

\(b,\left(2x+1\right)+100^0=2^4:50^0\)

\(\Leftrightarrow\left(2x+1\right)+1=2^4:1\)

\(\Leftrightarrow\left(2x+1\right)+1=16\)

\(\Leftrightarrow2x+1=16-1\)

\(\Leftrightarrow2x+1=15\)

\(\Leftrightarrow2x=14\)

\(\Rightarrow x=14:2\)

\(\Rightarrow x=7\)

các bạn ơi 12n+5 nhé mình viết thiếu mất :)

Cho \(A=\frac{12n+5}{2n+3}=\frac{6\left(2n+3\right)-13}{2n+3}=\frac{6\left(2n+3\right)}{2n+3}-\frac{13}{2n+3}\in Z\)

Để \(A\in Z\Rightarrow13⋮\left(2n+3\right)\)hay \(2n+3\inƯ\left(13\right)\)

Ta có :

\(Ư\left(13\right)\in\left\{\pm1;\pm13\right\}\Rightarrow2n+3\in\left\{\pm1;\pm13\right\}\)

Vậy để A nguyên \(\Rightarrow n\in\left\{-1;-2;5;-8\right\}\)

a, Gọi ƯCLN 2n + 5 ; n + 3 = d \(\left(d\inℕ^∗\right)\)

Ta có : \(2n+5⋮d\)(1) 

\(n+3⋮d\Rightarrow2n+6⋮d\)(2) 

Lấy (2) - (1) ta được : \(2n+6-2n-5⋮d\Rightarrow1⋮d\Rightarrow d=1\)

b, Để  \(B=\frac{2n}{n+3}+\frac{5}{n+3}=\frac{2n+5}{n+3}\)nhận giá trị nguyên khi 

\(2n+5⋮n+3\Leftrightarrow2\left(n+3\right)-1⋮n+3\)

\(\Rightarrow n+3\inƯ\left(1\right)=\left\{\pm1\right\}\)

1/5:x+1/2=1/3+2/5

1/5:x+1/2=11/15

1/5:x=11/15-1/2

1/5:x=7/30

x=7/30*1/5

x=7/150