\(\frac{1}{3}-\frac{3}{4}-\left(\frac{-3}{5}\right)+\frac{1}{72}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}\)

\(=\left(\frac{24}{72}-\frac{54}{72}+\frac{1}{72}-\frac{16}{72}-\frac{2}{72}\right)+\left(\frac{1}{15}+\frac{9}{15}\right)\)

\(=\frac{-47}{72}+\frac{2}{3}\)

\(=\frac{1}{72}\)

\(\frac{x+7}{x-3}=\frac{4}{15}\)

\(\left(x+7\right)\cdot15=\left(x-3\right)\cdot4\)

\(15x+105=4x-12\)

\(11x=-117\)

\(x=-\frac{117}{11}\)

\(-\frac{1}{21}-\frac{1}{28}=-\frac{4}{84}-\frac{3}{84}=-\frac{7}{84}=-\frac{1}{12}\)

\(n^3-n^2+2n+7=n^3+n-n^2-1+n+8\)chia hết cho \(n^2+1\)

tương đương \(n+8\)chia hết cho \(n^2+1\).

Với \(n=8\)thấy không thỏa. 

Với \(n\ne8\Rightarrow\left(n+8\right)\left(n-8\right)=n^2-64=n^2+1-65⋮\left(n^2+1\right)\)

\(\Leftrightarrow65⋮\left(n^2+1\right)\)

mà \(n\)là số nguyên nên \(n^2+1\)là ước của \(65\).

Do đó \(n^2+1\in\left\{1;5;13;65\right\}\)

\(\Rightarrow n\in\left\{\pm1,\pm2,\pm8\right\}\).

Thử lại \(n\in\left\{-8,2\right\}\)thỏa mãn. 

Đây nha học tốt

2 mũ 13 -2 nhé

cả lời giải bạn ơi

\(a.\frac{2}{5}-x=-\frac{3}{10}\Leftrightarrow x=\frac{2}{5}+\frac{3}{10}=\frac{7}{10}\)

\(b.x-\frac{1}{15}=\frac{3}{2}\Leftrightarrow x=\frac{1}{15}+\frac{3}{2}=\frac{47}{30}\)

\(c.\frac{3}{8}-x=-\frac{5}{12}\Leftrightarrow x=\frac{3}{8}+\frac{5}{12}=\frac{19}{24}\)

\(d.\frac{11}{13}-\left(\frac{5}{42}-x\right)=0\Leftrightarrow x=\frac{5}{42}-\frac{11}{13}=-\frac{397}{546}\)

\(e.\frac{2}{3}x+\frac{5}{7}=\frac{3}{10}\Leftrightarrow\frac{2}{3}x=\frac{3}{10}-\frac{5}{7}=-\frac{29}{70}\Leftrightarrow x=-\frac{29}{70}:\frac{2}{3}=-\frac{87}{140}\)

\(g.-5x\left(x-\frac{5}{4}\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x-\frac{5}{4}=0\Leftrightarrow x=\frac{5}{4}\end{cases}}\)

\(h.\left|x+\frac{3}{4}\right|=\frac{1}{2}\Leftrightarrow\orbr{\begin{cases}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{5}{4}\end{cases}}}\)

\(k.-\frac{21}{13}x+\frac{1}{3}=-\frac{2}{3}\Leftrightarrow-\frac{21}{13}x=-\frac{1}{3}-\frac{2}{3}=-1\Leftrightarrow x=\frac{13}{21}\)

2x + 3 = 5 

=>  2x =  2

=>  x  =  1

giải 

th1

2x+3=5

2x     =2

 x      = 1

th2

2x+3=-5

2x   =-5 -3

2x    =  -8

x    = -4

Bài 1:

a)  \(|2x-5|=-4\)

=> ko có giá trị x nào thoả mãn

b)   \(\frac{1}{3}-|\frac{5}{4}-2x|=\frac{1}{4}\)

=>  \(|\frac{5}{4}-2x|=\frac{1}{12}\)

=>  \(\orbr{\begin{cases}\frac{5}{4}-2x=\frac{1}{12}\\\frac{5}{4}-2x=-\frac{1}{12}\end{cases}}\)

=>  \(\orbr{\begin{cases}2x=\frac{7}{6}\\2x=\frac{4}{3}\end{cases}}\)

=>  \(\orbr{\begin{cases}x=\frac{7}{12}\\x=\frac{2}{3}\end{cases}}\)

Bài 2 :

a)  \(2|2x-3|=\frac{1}{2}\)

=>  \(|2x-3|=\frac{1}{4}\)

=>  \(\orbr{\begin{cases}2x-3=\frac{1}{4}\\2x-3=-\frac{1}{4}\end{cases}}\)

=>  \(\orbr{\begin{cases}2x=\frac{13}{4}\\2x=\frac{11}{4}\end{cases}}\)

=>   \(\orbr{\begin{cases}x=\frac{13}{8}\\x=\frac{11}{8}\end{cases}}\)

b)   \(7,5-3|5-2x|=-4,5\)

=>  \(3|5-2x|=12\)

=>\(|5-2x|=4\)

=>  \(\orbr{\begin{cases}5-2x=4\\5-2x=-4\end{cases}}\)

=>   \(\orbr{\begin{cases}2x=1\\2x=9\end{cases}}\)

=>  \(\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{9}{2}\end{cases}}\)

Bài 3 :

a)  \(2|3x-1|+1=5\)

=>  \(2|3x-1|=4\)

=>   \(|3x-1|=2\)

=>   \(\orbr{\begin{cases}3x-1=2\\3x-1=-2\end{cases}}\)

=>   \(\orbr{\begin{cases}3x=3\\3x=-1\end{cases}}\)

=>   \(\orbr{\begin{cases}x=1\\x=-\frac{1}{3}\end{cases}}\)

b)  \(|\frac{x}{2}-1|=3\)

=>  \(\orbr{\begin{cases}\frac{x}{2}-1=3\\\frac{x}{2}-1=-3\end{cases}}\)

=>   \(\orbr{\begin{cases}\frac{x}{2}=4\\\frac{x}{2}=-2\end{cases}}\)

=>   \(\orbr{\begin{cases}x=8\\x=-4\end{cases}}\)