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Câu 3:
Ta có: \(1< \dfrac{x+1}{5}-\dfrac{x-2}{3}< \dfrac{7}{5}\)
\(\Leftrightarrow1< \dfrac{3x+3-5x+10}{15}< \dfrac{7}{5}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-2x+13}{15}>1\\\dfrac{-2x+13}{15}< \dfrac{7}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2x+13>15\\-2x+13< 21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-2x>2\\-2x< 8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< -1\\x>-4\end{matrix}\right.\Leftrightarrow-4< x< -1\)
Bài 4:
Sửa đề: \(\left(x+2\right)^2-\left(x-3\right)\left(x+3\right)< =40\)
\(\Leftrightarrow x^2+4x+4-x^2+9< =40\)
=>4x<=27
hay x<=27/4
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:
\(B< \dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)
\(B< \dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)
\(B< \dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)
\(B< \dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)\)
\(B< \dfrac{1}{2}\left(\dfrac{2n+1}{2n+1}-\dfrac{1}{2n+1}\right)\)
\(B< \dfrac{1}{2}.\dfrac{2n}{2n+1}\)
\(B< \dfrac{2n}{4n+2}\)
\(B< \dfrac{2n}{2\left(2n+1\right)}\)
\(B< \dfrac{n}{2n+1}\)
Ta có: \(\dfrac{16}{2x+y+z}\le\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\)
\(\Leftrightarrow\dfrac{1}{2x+y+z}\le\dfrac{1}{16}\left(\dfrac{2}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\left(1\right)\)
Tương tự ta có: \(\left\{{}\begin{matrix}\dfrac{1}{x+2y+z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{1}{z}\right)\left(2\right)\\\dfrac{1}{x+y+2z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{2}{z}\right)\left(3\right)\end{matrix}\right.\)
Cộng (1), (2), (3) vế theo vế ta được:
\(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\le\dfrac{4}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{4.4}{16}=1\)
Dấu = xảy ra khi \(x=y=z=\dfrac{3}{4}\)
Chi tiết, và chuẩn đúng toán học. " dãy số hiểu n thuộc N*"
*)với n=1 ta có: \(A=\dfrac{1}{1+1}=\dfrac{1}{2}=B\)
*) với n>1 ta có: \(\dfrac{1}{n+1}>\dfrac{1}{2n}\) {c/m: không quá khó bỏ qua}. áp vào từng số hạng VT:
vậy ta có:\(A=\left(\dfrac{1}{n+1}+..+\dfrac{1}{2n}\right)>n.\dfrac{1}{2n}=\dfrac{1}{2}=B=VP\)
Kết luận:
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}n=1\\A=B\end{matrix}\right.\\\left\{{}\begin{matrix}n\ne1\\A>B\end{matrix}\right.\end{matrix}\right.\) hoặc \(KL:A\ge B..\forall n\in N^o\)
Ta có: \(A=\dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{2n}>\dfrac{1}{2n}+\dfrac{1}{2n}+...+\dfrac{1}{2n}=\dfrac{n}{2n}=\dfrac{1}{2}\)
Vậy \(A>B\)
a/ \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{\left(n-1\right)n}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}=1-\dfrac{1}{n}< 1\)
Vậy A < 1
b/ Dựa vô câu a mà làm câu b nhé
\(B=\dfrac{1}{2^2}+\dfrac{1}{4^2}+...+\dfrac{1}{\left(2n\right)^2}=\dfrac{1}{4}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\right)\)
\(< \dfrac{1}{4}\left(1+1-\dfrac{1}{n}\right)=\dfrac{1}{2}-\dfrac{1}{4n}< \dfrac{1}{2}\)
Vậy \(B< \dfrac{1}{2}\)
\(\dfrac{1}{2}+\dfrac{1}{n}>\dfrac{1}{4}+\dfrac{2}{5}\Leftrightarrow\dfrac{1}{2}+\dfrac{1}{n}>0,65\)
\(\Leftrightarrow\dfrac{1}{n}>\dfrac{3}{20}\Leftrightarrow\dfrac{20}{20n}>\dfrac{3n}{20n}\Rightarrow20>3n\Rightarrow n< 7\)
vậy n = 6
\(\dfrac{1}{2}+\dfrac{1}{n}>\dfrac{1}{4}+\dfrac{2}{5}\\\)
<=> \(0.5+\dfrac{1}{n}>0.25+0.4\) <=> \(0.5+\dfrac{1}{n}>0.65\) <=> 1/n >0.15 <=>n=6