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lấy máy tính tính 2 vế
xong thay x vào để thỏa mãn điều kiện
hok tốt
Ta có :
\(\frac{1}{5}+\frac{2}{30}+\frac{121}{165}\le x\le\frac{1}{2}+\frac{156}{72}+\frac{1}{3}\)
\(\Leftrightarrow\)\(\frac{3}{15}+\frac{1}{15}+\frac{11}{15}\le x\le\frac{3}{6}+\frac{13}{6}+\frac{2}{6}\)
\(\Leftrightarrow\)\(\frac{15}{15}\le x\le\frac{18}{6}\)
\(\Leftrightarrow\)\(1\le x\le3\)
\(\Rightarrow\)\(x\in\left\{1;2;3\right\}\)
Vậy \(x\in\left\{1;2;3\right\}\)
Chúc bạn học tốt ~
\(VT=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{101}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{102}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{101}+\frac{1}{102}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{102}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}+\frac{1}{102}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{51}\)
\(=\frac{1}{52}+\frac{1}{53}+\frac{1}{54}+...+\frac{1}{102}\)
\(=VP\)
Đặt \(A=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}\)
\(\Rightarrow A< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)
Vậy \(A< \frac{1}{2}\left(đpcm\right)\)
Ta có: \(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)
1)
Dễ thấy \(B=\dfrac{10^{19}}{10^{19}-3}>1\)
\(\Rightarrow B=\dfrac{10^{19}}{10^{19}-3}>\dfrac{10^{19}+2}{10^{19}-3+2}=\dfrac{10^{19}+2}{10^{19}-1}=A\)
\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+....+\frac{1}{\left(2n\right)^2}\)= \(\frac{1}{2^2.2^2}+\frac{1}{2^2.3^2}+\frac{1}{2^2.4^2}+...+\frac{1}{2^2+n^2}\)
= \(\frac{1}{2^2}\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)\)
= \(\frac{1}{4}\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)\)
Coi A=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\)
A= \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{n.n}\)
Vì \(\frac{1}{2.2}< \frac{1}{1.2}\)
\(\frac{1}{3.3}< \frac{1}{2.3}\)
....
\(\frac{1}{n.n}< \frac{1}{\left(n-1\right)n}\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=B\)
=> B= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)
=> B= \(1-\frac{1}{n}\)
=> B<1 <=> A<B<1
=> A<1
=> \(\frac{1}{4}\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)< \frac{1}{4}\)
Vậy \(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{4}\)
(1/1×2 + 1/2×3 + ... + 1/9×10) × x < 2/1×3 + 2/3×5 + ... + 2/9×11
(1 - 1/2 + 1/2 - 1/3 + ... + 1/9 - 1/10) × x < 1 - 1/3 + 1/3 - 1/5 + ... + 1/9 - 1/11
(1 - 1/10) × x < 1 - 1/11
9/10 × x < 10/11
x < 10/11 : 9/10
x < 10/11 × 10/9
x < 100/99
Mà x là số tự nhiên => x = 0 hoặc 1
Ta có :
\(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\)
\(2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\)
\(2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)\)
\(B=1-\frac{1}{2^{2016}}\)
\(B=\frac{2^{2016}-1}{2^{2016}}< 1\)
Vậy \(B< 1\)
Chúc bạn học tốt ~
Ta có: 2B=1+1/2+1/2^2+...+1/2^2015
2B-B=(1+1/2+1/2^2+...+1/2^2015)-(1/2+1/2^2+1/2^3+...+1/2^2016)
B=1-1/2^2015<1
Vậy B<1