a, \(2\left|x\right|=1,8\)

\(\left|x\right|=1,8:2\)

\(\left|x\right|=0,9\)

\(\Rightarrow x=0,9\) hoặc \(x=-0,9\)

b, \(\left|2x-\dfrac{1}{2}\right|-0.4=\dfrac{3}{5}\)

\(\left|2x-0,5\right|-0,4=0.6\)

\(\left|2x-0,5\right|=0,6+0,4\)

\(\left|2x-0,5\right|=1\)

\(\Rightarrow\left[{}\begin{matrix}2x-0,5=1\\2x-0,5=-1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}2x=1,5\\2x=-0,5\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0,75\\x=-0,25\end{matrix}\right.\)

Vậy \(x=0,75\) hoặc \(x=-0,25\)

1. Mk sửa lại đề bài như sau:

\(A=\dfrac{1}{1.5}+\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{89.93}\)

\(\Rightarrow4A=\dfrac{4}{1.5}+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{89.93}\)

\(4A=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{89}-\dfrac{1}{93}\)

\(4A=1-\dfrac{1}{93}\)

\(4A=\dfrac{92}{93}\)

\(A=\dfrac{92}{93}:4\)

\(A=\dfrac{23}{93}\)

2. Mk cux sửa lại đề bài:

\(A=3+3^2+3^3+3^4+3^5+...+3^{100}\)

\(=\left(3+3^2+3^3+3^4\right)+...+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)

\(=3\left(1+3+9+27\right)+...+3^{97}\left(1+3+9+27\right)\)

\(=3.40+...+3^{97}.40\)

\(=\left(3+3^{97}\right)⋮4.10\)

\(\Rightarrow A⋮4;10\)

b, Ta có:

\(14A=\dfrac{7^{2013}+14}{7^{2013}+1}=\dfrac{7^{2013}+1+13}{7^{2013}+1}=\dfrac{7^{2013}+1}{7^{2013}+1}+\dfrac{13}{7^{2013}+1}=1+\dfrac{13}{7^{2013}+1}\)

\(14B=\dfrac{7^{2015}+14}{7^{2015}+1}=\dfrac{7^{2015}+1+13}{7^{2015}+1}=\dfrac{7^{2015}+1}{7^{2015}+1}+\dfrac{13}{7^{2015}+1}=1+\dfrac{13}{7^{2015}+1}\)

\(\)Vì \(7^{2013}+1< 7^{2015}+1\)

\(\dfrac{\Rightarrow13}{7^{2013}+1}>\dfrac{13}{7^{2015}+1}\)

\(\Rightarrow1+\dfrac{13}{7^{2013}+1}>1+\dfrac{13}{7^{2015+1}}\)

\(\Leftrightarrow14A>14B\)

\(\Rightarrow A>B\)

Gọi ƯCNL \(\left(5m+3;3m+2\right)=d\left(d\in Z\right)\)

\(\Leftrightarrow5m+3⋮d\)

\(3m+2⋮d\)

\(\Rightarrow\left[\left(3m+2\right)-\left(5m+3\right)\right]⋮d\)

\(\Rightarrow5\left(3m+2\right)-3\left(5m+3\right)⋮d\)

\(\Rightarrow15m+10-15m+9⋮d\)

\(\Rightarrow1⋮d\Leftrightarrow d=\pm1\)

Vậy \(5m+3\) và \(3m+2\) là hai số nguyên tố cùng nhau.

\(\)

a, \(\left|2x+1\right|< 2\)

\(\left[{}\begin{matrix}2x+1=0\\2x+1=1\\2x+1=\left(-1\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\left(l\right)\\x=0\\x=-1\end{matrix}\right.\)

\(25\le5^x< 3125\)

\(\Leftrightarrow5^2\le5^x< 5^5\)

\(\Rightarrow2\le x< 5\)

\(\Rightarrow x=\left\{2;3;4\right\}\)

a, Ta có:

\(2009^{2008}=\left(2009^2\right)^{1004}=\left(\overline{...1}\right)^{1004}=\overline{...1}\)

Vậy chữ số tận cùng của \(2009^{2008}\) là \(1.\)

b, Ta có:

\(72^{1982}=\left(72^2\right)^{991}=\left(\overline{..76}\right)^{991}=\left(\overline{..76}\right)\)

Vậy chữ số tận cùng của \(72^{1982}\) là \(6.\)

\(a,15-2\left|x+\dfrac{1}{3}\right|=\dfrac{4}{7}\)

\(2\left|x+\dfrac{1}{3}\right|=15-\dfrac{4}{7}\)

\(2\left|x+\dfrac{1}{3}\right|=\dfrac{101}{7}\)

\(\left|x+\dfrac{1}{3}\right|=\dfrac{101}{7}:2\)

\(\left|x+\dfrac{1}{3}\right|=\dfrac{101}{14}\)

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{101}{14}\) hoặc \(x+\dfrac{1}{3}=-\dfrac{101}{14}\)

\(x=\dfrac{101}{14}-\dfrac{1}{3}\) \(x=-\dfrac{101}{14}-\dfrac{1}{3}\)

\(x=\dfrac{302}{3}\) \(x=-\dfrac{317}{42}\)

a, \(3^x-3^{x+3}=-234\)

\(3^x\left(1-3^3\right)=-234\)

\(3^x.\left(-26\right)=-234\)

\(3^x=-234:\left(-26\right)\)

\(3^x=9\)

\(3^x=3^2\Rightarrow x=2\)

b, Mk sửa lại đề:

\(3.3^x+5.3^{x+1}=162\) \(\)

\(3^{x+1}+5.3^{x+1}=162\)

\(3^{x+1}\left(1+5\right)=162\)

\(3^{x+1}.6=162\)

\(3^{x+1}=162:6\)

\(3^{x+1}=27\)

\(3^{x+1}=3^3\)

\(\Rightarrow x+1=3\)

\(x=3-1\)

\(\Rightarrow x=2\)