Ta có:

\(\dfrac{1}{a}-\dfrac{1}{a+1}=\dfrac{a+1}{a\left(a+1\right)}-\dfrac{a}{a\left(a+1\right)}\)

\(=\dfrac{a+1-a}{a\left(a+1\right)}=\dfrac{1}{a\left(a+1\right)}\)

\(\Leftrightarrow\dfrac{1}{a\left(a+1\right)}=\dfrac{1}{a}-\dfrac{1}{a+1}\)

\(\Leftrightarrow\dfrac{1}{a}-\dfrac{1}{a+1}=\dfrac{1}{a\left(a+1\right)}\)

\(\Leftrightarrow\dfrac{1}{a}=\dfrac{1}{a\left(a+1\right)}+\dfrac{1}{a+1}\)

\(\Leftrightarrow\dfrac{1}{a}=\dfrac{1}{a+1}+\dfrac{1}{a\left(a+1\right)}\) (Đpcm)

Đặt \(A=1.3.5.7...99\)

\(B=\dfrac{51}{2}.\dfrac{52}{2}.\dfrac{53}{2}...\dfrac{100}{2}\)

Ta có:

\(A=1.3.5.7...99\)

\(\Rightarrow A=\dfrac{\left(1.3.5.7...99\right)\left(2.4.6.8...100\right)}{2.4.6.8...100}\)

\(\Rightarrow A=\dfrac{1.2.3.4...100}{2.4.6.8...100}\)

\(\Rightarrow A=\dfrac{1.2.3.4...100}{\left(2.1\right)\left(2.2\right)\left(2.3\right)...\left(2.50\right)}\)

\(\Rightarrow A=\dfrac{\left(1.2.3.4...50\right)\left(51.52.53...100\right)}{\left(1.2.3.4...50\right)\left(2.2.2.2...2\right)}\)

\(\Rightarrow A=\dfrac{51.52.53.54...100}{2.2.2.2...2}\)

\(\Rightarrow A=\dfrac{51}{2}.\dfrac{52}{2}.\dfrac{53}{2}....\dfrac{100}{2}\)

\(\Rightarrow A=B\)

Vậy \(1.3.5.7...99=\dfrac{51}{2}.\dfrac{52}{2}.\dfrac{53}{2}...\dfrac{100}{2}\) (Đpcm)

Ta có:

\(4^{30}=2^{30}.2^{30}=2^{30}.\left(2^2\right)^{15}=2^{30}.4^{15}\)

Lại có:

\(3.24^{10}=3.3^{10}.8^{10}=3^{11}.\left(2^3\right)^{10}=3^{11}.2^{30}\)

Mà \(4^{15}>3^{11}\Rightarrow4^{30}>3^{11}\)

\(\Rightarrow2^{30}+3^{30}+4^{30}>3.24^{10}\)

Vậy \(2^{30}+3^{30}+4^{30}>3.24^{10}\)

Giải:

Gọi số tự nhiên cần tìm là \(a\)

Ta có:

\(a\div29\) dư \(5\)

\(\Rightarrow a=29k+5\left(k\in N\right)\)

\(a\div31\) dư \(28\)

\(\Rightarrow a=31q+28\left(q\in N\right)\)

\(\Leftrightarrow29k+5=31q+28\Rightarrow29\left(k-q\right)=2q+23\)

Lại có:

\(2q+23\) là số lẻ \(\Rightarrow29\left(k-q\right)\) là số lẻ \(\Rightarrow k-q\ge1\)

Vì \(a\) nhỏ nhất \(\Rightarrow q\) cũng phải nhỏ nhất \(\left(a=31q+28\right)\)

\(\Rightarrow2q=29\left(k-q\right)-23\) nhỏ nhất

\(\Rightarrow k-q\) nhỏ nhất

Do đó: \(k-q=1\Rightarrow2q=29-23=6\Leftrightarrow q=3\)

\(\Rightarrow a=31q+28=31.3+28=121\)

Vậy số cần tìm là \(121\)

tap sau dau Nguyễn Ngọc Huyền

Áp dụng tính chất \(\dfrac{a}{b}>1\Rightarrow\dfrac{a}{b}< \dfrac{a+m}{b+m}\) ta có:

\(B=\dfrac{10^{2014}+1}{10^{2013}+1}< \dfrac{10^{2014}+1+9}{10^{2013}+1+9}\)

\(=\dfrac{10^{2014}+10}{10^{2013}+10}=\dfrac{10\left(10^{2013}+1\right)}{10\left(10^{2012}+1\right)}=\dfrac{10^{2013}+1}{10^{2012}+1}\)

\(\Rightarrow\dfrac{10^{2014}+1}{10^{2013}+1}< \dfrac{10^{2013}+1}{10^{2012}+1}\)

Hay \(A< B\)

Giải:

Ta có:

\(ab-ac+bc-c^2=-1\)

\(\Leftrightarrow a\left(b-c\right)+c\left(b-c\right)=-1\)

\(\Leftrightarrow\left(b-c\right)\left(a+c\right)=-1\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}b-c=1\\a+c=-1\end{matrix}\right.\\\left\{{}\begin{matrix}b-c=-1\\a+c=1\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(b-c\right)+\left(a+c\right)=1+\left(-1\right)\\\left(b-c\right)+\left(a+c\right)=\left(-1\right)+1\end{matrix}\right.\)

\(\Leftrightarrow b+a=0\)

\(\Leftrightarrow a;b\) là hai số nguyên tố cùng nhau

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{a}{b}=\dfrac{-a}{a}=-1\\\dfrac{a}{b}=\dfrac{a}{-a}=-1\end{matrix}\right.\)

Vậy \(\dfrac{a}{b}=-1\)