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\(A=\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+...+\dfrac{1}{\sqrt{n}}>\sqrt{n}\left(1\right)\)
Với \(n=2\), BĐT \(\left(1\right)\) trở thành \(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}>\sqrt{2}\) (đúng)
Giả sử \(\left(1\right)\) đúng với \(n=k\), nghĩa là \(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+...+\dfrac{1}{\sqrt{k}}>\sqrt{k}\left(2\right)\)
Ta chứng minh \(\left(1\right)\) đúng với \(n=k+1\). Thật vậy, từ \(\left(2\right)\) suy ra:
\(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+...+\dfrac{1}{\sqrt{k}}+\dfrac{1}{\sqrt{k+1}}>\sqrt{k}+\dfrac{1}{\sqrt{k+1}}\)
Vì \(\sqrt{k}+\dfrac{1}{\sqrt{k+1}}=\dfrac{\sqrt{k\left(k+1\right)}+1}{\sqrt{k+1}}>\sqrt{k+1}\)
Nên \(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+...+\dfrac{1}{\sqrt{k}}+\dfrac{1}{\sqrt{k+1}}>\sqrt{k+1}\)
Tức là \(\left(1\right)\) đúng với \(n=k+1\).
Theo nguyên lí quy nạp, (1) đúng với mọi số tự nhiên \(n>1\)
\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+....+\dfrac{1}{18.19.20}=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{18.19}-\dfrac{1}{19.20}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{19.20}\right)\\ =\dfrac{1}{4}-\dfrac{1}{2.19.20}< \dfrac{1}{4}\)
Cái B TT nhé
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+....+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{\left(n-1\right)n}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\\ =1-\dfrac{1}{n}< 1\)
D TT
E mk thấy nó ss ớ
a: Gọi số nguyên cần tìm là x
Theo đề, ta có: \(\dfrac{1}{3}+\left(\dfrac{2}{4}-1\dfrac{2}{5}\right)< x< 2\dfrac{1}{7}+\left(\dfrac{-2}{5}-\dfrac{1}{4}\right)\)
\(\Leftrightarrow\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{7}{5}< x< \dfrac{15}{7}-\dfrac{2}{5}-\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{20}{60}+\dfrac{30}{60}-\dfrac{84}{60}< x< \dfrac{15\cdot20-2\cdot28-35}{140}\)
\(\Leftrightarrow-\dfrac{34}{60}< x< \dfrac{209}{140}\)
mà x là số nguyên
nên \(x\in\left\{0;1\right\}\)
b: Gọi số nguyên cần tìm là x
Theo đề, ta có: \(\dfrac{7}{3}+\dfrac{3}{4}-\dfrac{1}{5}>x>\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{7\cdot20+3\cdot15-12}{60}>x>\dfrac{56-21+2\cdot12}{84}\)
\(\Leftrightarrow\dfrac{173}{60}>x>\dfrac{59}{84}\)
mà x là số nguên
nên \(x\in\left\{2;1\right\}\)
\(\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{a\left(b+2017\right)}{b\left(b+2017\right)}\\\dfrac{a+2017}{b+2017}=\dfrac{b\left(a+2017\right)}{b\left(b+2017\right)}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{ab+2017a}{b^2+2017b}\\\dfrac{a+2017}{b+2017}=\dfrac{ab+2017b}{b^2+2017b}\end{matrix}\right.\)
Ta cần so sánh:
\(ab+2017a\) với \(ab+2017b\)
Cần so sánh \(a\) với \(b\)
Nếu \(a>b\Leftrightarrow\dfrac{a}{b}>\dfrac{a+2017}{b+2017}\)
Nếu \(a< b\Leftrightarrow\dfrac{a}{b}< \dfrac{a+2017}{b+2017}\)
Nếu \(a=b\Leftrightarrow\dfrac{a}{b}=\dfrac{a+2017}{b+2017}\)
Mấy câu sau dễ tương tự
a: \(\Leftrightarrow-\dfrac{23}{5}\cdot\dfrac{50}{23}< x< \dfrac{-13}{5}:\dfrac{21}{15}=\dfrac{-13}{5}\cdot\dfrac{5}{7}=\dfrac{-13}{7}\)
=>-10<x<-13/7
hay \(x\in\left\{-9;-8;-7;-6;-5;-4;-3;-2\right\}\)
b: \(\Leftrightarrow-\dfrac{13}{3}\cdot\dfrac{1}{3}< x< \dfrac{-2}{3}\cdot\dfrac{4-3-9}{12}\)
\(\Leftrightarrow-\dfrac{13}{9}< x< \dfrac{4}{9}\)
mà x là số nguyên
nên \(x\in\left\{-1;0\right\}\)
1
a) Vì \(\dfrac{a}{b}< \dfrac{c}{d}\)
\(\Rightarrow\dfrac{ad}{bd}< \dfrac{bc}{bd}\)
\(\Rightarrow ad< bc\)
2
b) Ta có : \(\dfrac{-1}{3}=\dfrac{-16}{48};\dfrac{-1}{4}=\dfrac{-12}{48}\)
Ta có dãy sau : \(\dfrac{-16}{48};\dfrac{-15}{48};\dfrac{-14}{48};\dfrac{-13}{48};\dfrac{-12}{48}\)
Vậy 3 số hữu tỉ xen giữa \(\dfrac{-1}{3}\) và \(\dfrac{-1}{4}\) là :\(\dfrac{-15}{48};\dfrac{-14}{48};\dfrac{-13}{48}\)
1a ) Ta có : \(\dfrac{a}{b}\) < \(\dfrac{c}{d}\)
\(\Leftrightarrow\) \(\dfrac{ad}{bd}\) < \(\dfrac{bc}{bd}\) \(\Rightarrow\) ad < bc
1b ) Như trên
2b) \(\dfrac{-1}{3}\) = \(\dfrac{-16}{48}\) ; \(\dfrac{-1}{4}\) = \(\dfrac{-12}{48}\)
\(\dfrac{-16}{48}\) < \(\dfrac{-15}{48}\) <\(\dfrac{-14}{48}\) < \(\dfrac{-13}{48}\) < \(\dfrac{-12}{48}\)
Vậy 3 số hữu tỉ xen giữa là.................
Ta có:\(\dfrac{1}{2^3}< \dfrac{1}{1.2.3};\dfrac{1}{3^3}< \dfrac{1}{2.3.4};\dfrac{1}{4^3}< \dfrac{1}{3.4.5};...;\dfrac{1}{n^3}< \dfrac{1}{\left(n-1\right).n.\left(n+1\right)}\)Vậy:\(\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+...+\dfrac{1}{n^3}< \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{\left(n-1\right).n.\left(n+1\right)}\)Ta có:\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{\left(n-1\right).n.\left(n+1\right)}\)
=\(\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right).n}-\dfrac{1}{n.\left(n+1\right)}\right)\)=\(\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{n.\left(n+1\right)}\right)\)
=\(\dfrac{1}{4}-\dfrac{1}{2n.\left(n+1\right)}< \dfrac{1}{4}\)
Vì:\(\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}-\dfrac{1}{2n.\left(n+1\right)}< \dfrac{1}{4}\)
\(\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}\) hay \(A< \dfrac{1}{4}\)