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\(S=\left(1-\dfrac{1}{4}\right)+\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{1}{16}\right)+...+\left(1-\dfrac{1}{n^2}\right)\\ S=\left(1+1+...+1\right)-\left(\dfrac{1}{4}+\dfrac{1}{9}+...+\dfrac{1}{n^2}\right)\\ S=n-1-\left(\dfrac{1}{4}+\dfrac{1}{9}+...+\dfrac{1}{n^2}\right)< n-1\)
Lại có \(\dfrac{1}{4}+\dfrac{1}{9}+..+\dfrac{1}{n^2}=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\)
\(\Rightarrow\dfrac{1}{4}+\dfrac{1}{9}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{n\left(n-1\right)}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}=1-\dfrac{1}{n}< 1\)
\(\Rightarrow S>n-1-1=n-2\\ \Rightarrow n-2< S< n-1\\ \Rightarrow S\notin N\)
a) \(\dfrac{12}{\left(-2\right)^n}=\dfrac{-12}{8}\)
\(\Rightarrow12.8=\left(-2\right)^n.\left(-12\right)\)
\(\Rightarrow96=\left(-2\right)^n.\left(-12\right)\)
\(\Rightarrow\left(-2\right)^n=\dfrac{96}{-12}\)
\(\Rightarrow\left(-2\right)^n=-8\)
\(\Rightarrow\left(-2\right)^n=\left(-2\right)^3\)
\(\Rightarrow n=3\)
Vậy \(n=3\)
2)
a) \(\dfrac{4}{9}\) và \(\dfrac{5}{8}\) Mẫu chung: 72
\(\dfrac{4}{9}=\dfrac{4.8}{72}=\dfrac{32}{72}\)
\(\dfrac{5}{8}=\dfrac{5.9}{72}=\dfrac{45}{72}\)
Vì \(\dfrac{32}{72}< \dfrac{45}{72}\)
Vậy \(\dfrac{4}{9}< \dfrac{5}{8}\)
b) \(-\sqrt{\dfrac{4}{9}}\) và \(\dfrac{-3}{4}\) MTC: 12
\(-\sqrt{\dfrac{4}{9}}=-\sqrt{\left(\dfrac{2}{3}\right)^2}=-\dfrac{2}{3}=\dfrac{-2.4}{12}=\dfrac{-8}{12}\)
\(-\dfrac{3}{4}=\dfrac{-3.3}{12}=\dfrac{-9}{12}\)
Vì \(\dfrac{-8}{12}>\dfrac{-9}{12}\)
Vậy \(-\sqrt{\dfrac{4}{9}}>\dfrac{-3}{4}\)
Bài 1: Vì: 2x^3 - 1 = 15
=> 2x^3 = 16
=> x^3 = 8
=> x = 2 (1)
Ta có:
* (x + 16)/9 = (y - 25)/16
<=> (2 + 16)/9 = (y - 25)/16
<=> 18/9 = (y - 25)/16
<=> 2 = (y - 25)/16
<=> y - 25 = 16.2 = 32
=> y = 32+25 = 57 (2)
* (x + 16)/9 = (z + 9)/25
<=> (2 + 16)/9 = (z + 9)/25
<=> 2 = (z + 9)/25
<=> z + 9 = 25.2 = 50
=> z = 50 - 9 = 41 (3)
Từ (1), (2) và (3) => x + y + z = 2 + 57 + 41 = 100
Bài 2:
c) vì a,b,c là độ dài các cạnh của tam giác:
\(\Rightarrow\left\{{}\begin{matrix}a< b+c\\b< a+c\\c< a+b\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{b+c}< 1\\\dfrac{b}{a+c}< 1\\\dfrac{c}{a+b}< 1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{b+c}< \dfrac{2a}{a+b+c}\\\dfrac{b}{a+c}< \dfrac{2b}{a+b+c}\\\dfrac{c}{a+b}< \dfrac{2c}{a+b+c}\end{matrix}\right.\)
\(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}< \dfrac{2a}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2c}{a+b+c}\)
\(\Rightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}< 2\) (đpcm)
= \(49-\left(\dfrac{1}{2}-\dfrac{1}{51}\right)=\dfrac{4949}{102}\notin N\)
Vậy \(S\notin N\)
\(\Rightarrow S=49-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)>49-1\)\(S=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+...+\dfrac{2499}{2500}\)
\(\Rightarrow S=1-\dfrac{1}{4}+1-\dfrac{1}{9}+1-\dfrac{1}{16}+...+1-\dfrac{1}{2500}\)
\(\Rightarrow S=1-\dfrac{1}{2^2}+1-\dfrac{1}{3^2}+1-\dfrac{1}{4^2}+...+1-\dfrac{1}{50^2}\)
\(\Rightarrow S=\left(1+1+...+1\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)\)
Từ 2-50 có 49 số nên có 49 số 1
\(\Rightarrow S=49-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)< 49\)
Nhận xét: \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{50^2}< \dfrac{1}{49.50}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...-\dfrac{1}{50}=1-\dfrac{1}{50}< 1\)
\(\Rightarrow-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)>-1\)
\(\Rightarrow S=49-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)>49-1\)
\(\Rightarrow S=49-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)>48\) (2)
Từ (1) và (2) \(\Rightarrow48< S< 49\)
Vậy \(S\notin N\)
2: \(A=9^n\cdot81-9^n+3^n\cdot9+3^n\)
\(=9^n\cdot80+3^n\cdot10\)
\(=10\left(9^n\cdot8+3^n\right)⋮10\)
\(S_n=1-\dfrac{1}{n^2}\) xét tổng \(U_n=\dfrac{1}{n^2}\) với n >=2
cơ bản có \(\dfrac{1}{n^2}< \dfrac{1}{n\left(n-1\right)}=\dfrac{1}{n-1}-\dfrac{1}{n}\)
<=>\(U< 1-\dfrac{1}{n-1}\)
cơ bản có \(\dfrac{1}{n^2}>\dfrac{1}{n\left(n+1\right)}=\dfrac{1}{n}-\dfrac{1}{n+1}\)
<=>\(U>1-\dfrac{1}{n+1}\)
<=>\(1-\dfrac{1}{n-1}< U< 1-\dfrac{1}{n+1}\)
với n >2 => 1/(n-1) ; 1/(n+1) là hai phân số <1
=> U không phải là số nguyên
=> S không là số nguyên => dpcm
vế phải đâu