Cái này nếu lớp 6 thì chắc phải thêm ĐK a là số tự nhiên

Giải:

Gọi số cần tìm là \(a\)

Theo đề bài ta có:

\(a\div120\) dư \(58\Rightarrow a=120k_1+58\)

\(a\div135\) dư \(88\Rightarrow a=135k_2+88\)

\(\Rightarrow120k_1+58=135k_2+88\)

\(\Rightarrow120k_1-120k_2=15k_2+88-58\)

\(\Rightarrow120\left(k_1-k_2\right)=15k_2+30\)

Mà \(a\) nhỏ nhất \(\Rightarrow k_1-k_2\) cũng phải nhỏ nhất

\(\Rightarrow k_1-k_2=1\). Lúc đó ta có:

\(15k_2+30=120\Rightarrow15k_2=90\Rightarrow k_2=6\)

\(\Rightarrow a=135.6+88=898\)

Vậy số cần tìm là \(898\)

a) Ta có:

\(\left(2x+1\right)^4=\left(2x+1\right)^6\)

\(\Leftrightarrow\left(2x+1\right)^4-\left(2x+1\right)^6=0\)

\(\Leftrightarrow\left(2x+1\right)^4\left[\left(2x+1\right)^2-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(2x+1\right)^4=0\\\left(2x+1\right)^2-1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-0,5\\\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=-0,5\\x=0\\x=-1\end{matrix}\right.\)

b) Ta có:

\(|\left|x+3\right|-8|=20\)

\(\Rightarrow\left[{}\begin{matrix}\left|x+3\right|-8=20\\\left|x+3\right|-8=-20\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\left|x+3\right|=28\\\left|x+3\right|=-12\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x+3=28\\x+3=-28\end{matrix}\right.\\x+3\in\varnothing\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=25\\x=-31\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=25\\x=-31\end{matrix}\right.\)

c) \(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)

\(\Leftrightarrow\dfrac{x-1}{2009}-1+\dfrac{x-2}{2008}-1=\dfrac{x-3}{2007}-1+\dfrac{x-4}{2006}-1\)

\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}=\dfrac{x-2010}{2007}+\dfrac{x-2010}{2006}\)

\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)

\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\)

Mà \(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\)

\(\Rightarrow x-2010=0\Leftrightarrow x=2010\)

Vậy \(x=2010\)

Giải:

Giả sử số có 3 chữ số đó là \(\overline{aaa}=111a\left(a\ne0;a< 10\right)\)

Gọi số số hạng của tổng là \(n\) ta có:

\(\dfrac{n\left(n+1\right)}{2}=111a=3.37.a\) Hay \(n\left(n+1\right)=2.3.37.a\)

\(\Rightarrow n\left(n+1\right)⋮37\)

Mà \(37\) là số nguyên tố và \(n+1< 74\) (Nếu \(n=74\) thì không thỏa mãn)

Do đó \(\left[{}\begin{matrix}n=37\\n+1=37\end{matrix}\right.\)

Nếu \(n=37\Rightarrow n+1=38\Rightarrow\dfrac{n\left(n+1\right)}{2}=703\) (loại)

Nếu \(n+1=37\Rightarrow n=36\Rightarrow\dfrac{n\left(n+1\right)}{2}=666\) (chọn)

Vậy số số hạng của tổng là \(36\)

Ta có:

\(A=\dfrac{1}{1.1981}+\dfrac{1}{2.1982}+...+\dfrac{1}{n\left(1980+n\right)}+...+\dfrac{1}{25.2005}\)

\(=\dfrac{1}{1980}\left(\dfrac{1981-1}{1.1981}+\dfrac{1982-2}{2.1982}+...+\dfrac{1980+n-n}{n\left(1980+n\right)}+...+\dfrac{2005-25}{25.2005}\right)\)

\(=\dfrac{1}{1980}\left(1-\dfrac{1}{1981}+\dfrac{1}{2}-\dfrac{1}{1982}+...+\dfrac{1}{n}-\dfrac{1}{1980+n}+...+\dfrac{1}{25}-\dfrac{1}{2005}\right)\)

\(=\dfrac{1}{1980}\left[\left(1+\dfrac{1}{2}+...+\dfrac{1}{25}\right)-\left(\dfrac{1}{1981}+\dfrac{1}{1982}+...+\dfrac{1}{2005}\right)\right]\)

Lại có:

\(B=\dfrac{1}{1.26}+\dfrac{1}{2.27}+...+\dfrac{1}{m\left(m+25\right)}+...+\dfrac{1}{1980.2005}\)

\(=\dfrac{1}{25}\left(\dfrac{26-1}{1.26}+\dfrac{27-2}{2.27}+...+\dfrac{25+m-m}{m\left(25+m\right)}+...+\dfrac{2005-1980}{1980.2005}\right)\)

\(=\dfrac{1}{25}\left(\dfrac{1}{1}-\dfrac{1}{26}+\dfrac{1}{2}-\dfrac{1}{27}+...+\dfrac{1}{m}-\dfrac{1}{25+m}+...+\dfrac{1}{1980}-\dfrac{1}{2005}\right)\)

\(=\dfrac{1}{25}\left[\left(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{1980}\right)-\left(\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{2005}\right)\right]\)

\(=\dfrac{1}{25}\left[\left(1+\dfrac{1}{2}+...+\dfrac{1}{25}\right)-\left(\dfrac{1}{1981}+\dfrac{1}{1982}+...+\dfrac{1}{2005}\right)\right]\)

\(\Rightarrow\dfrac{A}{B}=\dfrac{\dfrac{1}{1980}}{\dfrac{1}{25}}=\dfrac{5}{396}\)

Vậy tỉ số của \(A\) và \(B\) là \(\dfrac{5}{396}\)