CMR: Với mọi số tự nhiên n\(\ge\)2 thì tổng:
\(S=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+...+\dfrac{n^2-1}{n^2}\)
chứng tỏ rằng S = \(\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+...+\dfrac{n^2-1}{n^2}\) không là số tự nhiên với mọi
n\(\in\) N, n>2
\(S=\left(1-\dfrac{1}{4}\right)+\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{1}{16}\right)+...+\left(1-\dfrac{1}{n^2}\right)\\ S=\left(1+1+...+1\right)-\left(\dfrac{1}{4}+\dfrac{1}{9}+...+\dfrac{1}{n^2}\right)\\ S=n-1-\left(\dfrac{1}{4}+\dfrac{1}{9}+...+\dfrac{1}{n^2}\right)< n-1\)
Lại có \(\dfrac{1}{4}+\dfrac{1}{9}+..+\dfrac{1}{n^2}=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\)
\(\Rightarrow\dfrac{1}{4}+\dfrac{1}{9}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{n\left(n-1\right)}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}=1-\dfrac{1}{n}< 1\)
\(\Rightarrow S>n-1-1=n-2\\ \Rightarrow n-2< S< n-1\\ \Rightarrow S\notin N\)
Cho \(S=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+....+\dfrac{n^2-1}{n^2}\)
CMR với mọi số tự nhiên n >=2 thì S không nguyên
Với mọi số tự nhiên n \(\ge\)2 hãy so sánh: P = \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}\)với 1
Ta có:
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
...
\(\dfrac{1}{n^2}< \dfrac{1}{n\left(n-1\right)}\)
\(\Rightarrow P< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n-1\right)}\)
\(\Rightarrow P< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(\Rightarrow P< 1-\dfrac{1}{n}< 1\)
\(\Rightarrow P< 1\)
CMR với mọi số tự nhiên lớn hơn 2 thì :
\(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2^n-1}>\dfrac{n}{2}\)
1)CMR: với mọi số tự nhiên n thì : A=5n+2+26.5n+82n+1
2) Với x \(\ge\) 0. Tìm GTNN của bt
a)P=\(\dfrac{\left(x+2\right)^2}{2x}\)
b)Q=\(\dfrac{\left(x+1\right)^2}{y}+\dfrac{4y}{x}\) với x>0,y>0
\(1,A=5^{n+2}+26\cdot5^n+8^{2n+1}\\ A=5^n\cdot25+26\cdot5^n+8\cdot8^{2n+1}\\ A=51\cdot5^n+8\cdot64^n\)
Ta có \(64:59R5\Rightarrow64^n:59R5\)
Vì vậy \(51\cdot5^n+8\cdot64^n:59R=5^n\cdot51+8\cdot5^n=5^n\left(51+8\right)=5^n\cdot59⋮59\)
Vậy \(A⋮59\)
(\(R\) là dư)
\(2,\\ a,2x\ge0;\left(x+2\right)^2\ge0,\forall x\\ \Leftrightarrow P=\dfrac{\left(x+2\right)^2}{2x}\ge0\\ P_{min}=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\)
\(S+\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+....+\dfrac{n^2-1}{n^2}\) không là số nguyên
1, thực hiện phép tính
\(\dfrac{\left(\dfrac{2}{5}\right)^9.10^9-\left(\dfrac{-9}{4}\right)^5:\left(\dfrac{-3}{16}\right)^{10}}{4^{12}+16^9}\)
2,CMR:\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+.......+\dfrac{1}{n^2}< \dfrac{2}{3}\)với n\(\ge\)4
1.Tìm các số tự nhiên a,b khác 0 sao cho :
\(\dfrac{a}{5}-\dfrac{z}{b}=\dfrac{2}{15}\).
2.Tìm số tự nhiên n, để các biểu thức là số tự nhiên.
a)A=\(\dfrac{4}{n-1}+\dfrac{6}{n-1}-\dfrac{3}{n-1}\).
b)B=\(\dfrac{2n+9}{n+2}-\dfrac{3n}{n+2}+\dfrac{5n+1}{n+2}\).
giúp mình với mai mình nộp rồi
Bài 2:
a) Ta có: \(A=\dfrac{4}{n-1}+\dfrac{6}{n-1}-\dfrac{3}{n-1}\)
\(=\dfrac{4+6-3}{n-1}\)
\(=\dfrac{7}{n-1}\)
Để A là số tự nhiên thì \(7⋮n-1\)
\(\Leftrightarrow n-1\inƯ\left(7\right)\)
\(\Leftrightarrow n-1\in\left\{1;7\right\}\)
hay \(n\in\left\{2;8\right\}\)
Vậy: \(n\in\left\{2;8\right\}\)
ta có B=2n+9/n+2-3n+5n+1/n+2=4n+10/n+2 Để B là STN thì 4n+10⋮n+2 4n+8+2⋮n+2 4n+8⋮n+2 ⇒2⋮n+2 n+2∈Ư(2) Ư(2)={1;2} Vậy n=0
1.Cmr , với mọi số tự nhiên n lớn hơn hoặc bằng 1
a) \(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+....+\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{2}\)
b) \(\dfrac{1}{3^2}+\dfrac{1}{5^2}+\dfrac{1}{7^2}+....+\dfrac{1}{\left(2n+1\right)^2}< \dfrac{1}{4}\)
2.Cmr với mọi số tự nhiên lớn hơn hoặc bằng 2
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{n^2}< \dfrac{2}{3}\)
a) Đặt \(A=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{\left(2n\right)^2}\)
\(A=\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\right)\)
Ta có:
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{\left(n-1\right)n}\)
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< 1-\dfrac{1}{n}\)
\(\Rightarrow1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< 1-\dfrac{1}{n}+1\)
\(\Rightarrow1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< 2-\dfrac{1}{n}\)
\(\Rightarrow\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\right)< \dfrac{1}{2^2}\left(2-\dfrac{1}{2}\right)\)
\(\Rightarrow A< \dfrac{1}{2^2}.2-\dfrac{1}{2^2}.\dfrac{1}{2}\)
\(\Rightarrow A< \dfrac{1}{2}-\dfrac{1}{2^3}< \dfrac{1}{2}\)
Vậy \(A< \dfrac{1}{2}\left(Đpcm\right)\)
b) Đặt \(B=\dfrac{1}{3^2}+\dfrac{1}{5^2}+\dfrac{1}{7^2}+...+\dfrac{1}{\left(2n+1\right)^2}\)
Ta có:
\(B< \dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)
\(B< \dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)
\(B< \dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)
\(B< \dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)\)
\(B< \dfrac{1}{2}\left(\dfrac{2n+1}{2n+1}-\dfrac{1}{2n+1}\right)\)
\(B< \dfrac{1}{2}.\dfrac{2n}{2n+1}\)
\(B< \dfrac{2n}{4n+2}\)
\(B< \dfrac{2n}{2\left(2n+1\right)}\)
\(B< \dfrac{n}{2n+1}\)