\(\frac{\left(\frac{-2}{11}\right)^{n+1}}{\left(\frac{-2}{11}\right)^n}\left(n\ge1\right)\)
Những câu hỏi liên quan
frac{1}{3}-frac{3}{5}+frac{5}{7}-frac{7}{9}+frac{9}{11}-frac{11}{13}+frac{11}{15}+frac{11}{13}-frac{9}{11}+frac{7}{9}-frac{5}{7}+frac{3}{5}-frac{1}{3}frac{-3}{4}.31frac{11}{23}-0,75.8frac{12}{28}left(2frac{1}{3}+3frac{1}{2}right):left(-4frac{1}{6}+3frac{1}{7}right)+7frac{1}{2}4frac{5}{9}:left(-frac{5}{7}right)+5frac{4}{9}:left(-frac{5}{7}right)frac{2}{3}-4left(frac{1}{2}+frac{3}{4}right)left(1-frac{1}{2}right).left(1-frac{1}{3}right).left(1-frac{1}{4}right)....left(1-frac{1}{n+1}right)nin Z
Đọc tiếp
\(\frac{1}{3}-\frac{3}{5}+\frac{5}{7}-\frac{7}{9}+\frac{9}{11}-\frac{11}{13}+\frac{11}{15}+\frac{11}{13}-\frac{9}{11}+\frac{7}{9}-\frac{5}{7}+\frac{3}{5}-\frac{1}{3}\)\(\frac{-3}{4}.31\frac{11}{23}-0,75.8\frac{12}{28}\)\(\left(2\frac{1}{3}+3\frac{1}{2}\right):\left(-4\frac{1}{6}+3\frac{1}{7}\right)+7\frac{1}{2}\)\(4\frac{5}{9}:\left(-\frac{5}{7}\right)+5\frac{4}{9}:\left(-\frac{5}{7}\right)\)\(\frac{2}{3}-4\left(\frac{1}{2}+\frac{3}{4}\right)\)\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{n+1}\right)n\in Z\)
\(\sin^3\frac{x}{3}+3\sin^3\frac{x}{3^2}+...+3^{n-1}\sin^3\frac{x}{3}=\frac{1}{4}\left(3^n\sin^3\frac{x}{3^n}-\sin x\right)\)\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{2n+1}{2n+2}<\frac{1}{\sqrt{3n+4}}\left(n\ge1\right)\)\(\left(n!\right)^2\ge n^2\ge\left(n+1\right)^{n-1}cho\left(n\ge1\right)\)
lim\(\frac{\left(2n+1\right)\left(n+3\right)\left(n^2-11\right)}{\left(n+1\right)\left(n+2\right)\left(n-5\right)}\)
\(=lim\frac{\left(2+\frac{1}{n}\right)\left(1+\frac{n}{3}\right)\left(n-\frac{11}{n}\right)}{\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)\left(1-\frac{5}{n}\right)}=\frac{\infty}{1}=+\infty\)
Thu gọn biểu thức
a) \(C=\frac{7}{9}x^3y^2\left(\frac{6}{11}axy^3\right)+\left(-5bx^2y^4\right)\left(\frac{-1}{2}axz\right)+ax\left(x^2y\right)^3\)
b)\(D=\frac{\left(3x^4y^4\right)^2\left(\frac{6}{11}x^3y\right)\left(8x^{n-7}\right)\left(-2x^{7-n}\right)}{15x^3y^2\left(0,4ax^2y^2z^2\right)^2}\)(với axyz khác 0)
\(C=\frac{7}{9}x^3y^2\left(\frac{6}{11}axy^3\right)+\left(-5bx^2y^4\right)\left(\frac{-1}{2}axz\right)+ax\left(x^2y\right)^3\)
\(\Rightarrow C=\frac{42}{9}ax^4y^5+\frac{5}{2}abx^3y^4z+ax\left(x^6y^3\right)\)
\(\Rightarrow C=\frac{42}{9}ax^4y^5+\frac{5}{2}abx^3y^4z+ax^7y^3\)
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\(D=\frac{\left(3x^4y^4\right)^2\left(\frac{6}{11}x^3y\right)\left(8x^{n-7}\right)\left(-2x^{7-n}\right)}{15x^3y^2\left(0,4ax^2y^2z^2\right)^2}\)
\(D=\frac{\left[3.\frac{6}{11}.8.\left(-2\right)\right]\left(x^8x^3x^{n-7}x^{7-n}\right)\left(y^8y\right)}{15.0,4.\left(x^3x^4\right)\left(y^2y^4\right)z^4a}\)
\(D=\frac{\frac{-188}{11}x^{24}y^9}{6x^7y^6z^4a}\)
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Làm tiếp bài của Song Ngư (๖ۣۜO๖ۣۜX๖ۣۜA)
\(D=\frac{\frac{-188}{11}x^{17}y^3}{6z^4a}\)
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Xem thêm câu trả lời
\(D=\left(1-\frac{4}{1}\right)\left(1-\frac{4}{9}\right)\left(1-\frac{4}{25}\right)...\left(1-\frac{1}{\left(2n-1\right)^2}\right),\)với \(n\in N,n\ge1\)
Cơn Mưa Tình yêu nhận hàng
Ta có:
\(B=\frac{2n+5}{n-3}=\frac{2n+\left(11-6\right)}{n-3}=\frac{2n-6+11}{n-3}=\frac{2n-6}{n-3}+\frac{11}{n-3}=\frac{2.\left(n-3\right)}{n-3}+\frac{11}{n-3}=2+\frac{11}{n-3}\)
Để B là số nguyên thì 11⋮ n-3
\(\Rightarrow n-3\inƯ\left(11\right)=\left\{-11;-1;1;11\right\}\)
\(\Rightarrow n\in\left\{-8;2;4;14\right\}\)
a)Tìm số nguyên dương n thỏa mãn:
\(\frac{1}{2}.\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{n.\left(n+2\right)}\right)=\frac{2013}{2014}\)
b)tìm a sao cho
\(\left(a+\frac{1}{1.3}\right)+\left(a+\frac{1}{3.5}\right)+\left(a+\frac{1}{5.7}\right)+...+\left(a+\frac{1}{23.25}\right)=11.a+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
Cho \(\sqrt{1+\left(1+\frac{1}{n}\right)^2}+\sqrt{1+\left(1-\frac{1}{n}\right)^2}\) \(\left(n\ge1\right)\)
CMR: \(S=\frac{1}{a_{ }\%\%_1}+\frac{1}{a_2}+...+\frac{1}{a_{20}}\in N\)
\(a_n=\sqrt{2+\frac{2}{n}+\frac{1}{n^2}}+\sqrt{2-\frac{2}{n}+\frac{1}{n^2}}\)
\(\Rightarrow\frac{1}{a_n}=\frac{1}{4}\left(\sqrt{\left(n+1\right)^2+n^2}-\sqrt{n^2+\left(n-1\right)^2}\right)\)
\(\Rightarrow S=\frac{1}{4}\left(\sqrt{2^2+1}-\sqrt{1^2+0}+\sqrt{3^2+2^2}-\sqrt{2^2+1}+...+\sqrt{21^2+20^2}-\sqrt{20^2+19^2}\right)\)
\(=\frac{1}{4}\left(\sqrt{21^2+20^2}-\sqrt{1}\right)=7\)
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a, Cm công thức
\(\forall n\ge1\) ta có \(\frac{2}{\left(2n+1\right)\left(\sqrt{n}-\sqrt{n+1}\right)}< \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
b, áp dụng tính
\(\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+...+\frac{1}{4023\cdot\left(\sqrt{2011}+\sqrt{2012}\right)}< \frac{2011}{2013}\)
chỗ \(\sqrt{n}-\sqrt{n+1}\)phải là \(\sqrt{n}+\sqrt{n+1}\)
a, Ta có
\(\frac{2}{\left(2n+1\right)\left(\sqrt{n}-\sqrt{n+1}\right)}=\frac{2\cdot\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(2n+1\right)\left(\sqrt{n}-\sqrt{n+1}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\)
\(=\frac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{2n+1}=\frac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{4n^2+4n+1}}< \frac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{4n^2+4n}}\)
mà \(\frac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{4n^2+4n}}=\frac{2\cdot\left(\sqrt{n+1}-\sqrt{n}\right)}{2\sqrt{n\left(n+1\right)}}=\frac{\sqrt{n+1}}{\sqrt{n}\cdot\sqrt{n+1}}-\frac{\sqrt{n}}{\sqrt{n}\cdot\sqrt{n+1}}\)
\(=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
b, áp dụng bđt ta có
\(\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+...+\frac{1}{4023\cdot\left(\sqrt{2011}+\sqrt{2012}\right)}< \frac{2011}{2013}\)
\(=\frac{1}{\left(2\cdot1+1\right)\left(1+\sqrt{2}\right)}+\frac{1}{\left(2\cdot2+1\right)\left(\sqrt{2}+\sqrt{3}\right)}+...+\frac{1}{\left(2\cdot2011+1\right)\left(\sqrt{2011}-\sqrt{2012}\right)}\)
\(< 1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{2011}}-\frac{1}{\sqrt{2012}}\)..
\(=1-\frac{1}{\sqrt{2012}}=\frac{\sqrt{2012}-1}{\sqrt{2012}}=\frac{2011}{\sqrt{2012}\cdot\left(\sqrt{2012}+1\right)}\)
\(=\frac{2011}{2012+\sqrt{2012}}< \frac{2011}{2013}\)
Bạn Nhật sai đề bài
Câu. a. Dòng thứ nhất xuống dòng thứ 2. Em chú ý mẫu số sai rồi.
b. Công thức có số 2 trên tử số. Mà em ko đưa số 2 vào thì sao áp dụng dc công thức?