Tính :\(\frac{\left(\frac{-5}{7}\right)^{n+1}}{\left(\frac{-5}{7}\right)^n}\)\(\left(n\ge1\right)\)
Những câu hỏi liên quan
tính \(\frac{\left(\frac{-5}{7}^{ }\right)n+1^{ }}{\left(\frac{-5}{7}\right)^{ }n}\)(n>=1)
frac{left(-7right)^n}{left(-7right)^{n-1}}(nge1) Tính GTBT Bài 2 Tính GTBT theo cách hợp lí nếu có thểc) frac{5^3times3^3}{5^3times0,5+125times2,5}d)frac{5times7^1+7^3times25}{7^5125-7^3times50}e)frac{8^5timesleft(-5right)^8+left(-2right)^5times10^9}{2^{16}times5^7+20^8}h)frac{left(-0,25right)^{-5}times9^4timesleft(-2right)^{-3}-2^{-2}times6^3}{2^9times3^6+6^6times40}Bài 3 Chứng tỏ rằnga)
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\(\frac{\left(-7\right)^n}{\left(-7\right)^{n-1}}\)(n\(\ge1\)) Tính GTBT
Bài 2 Tính GTBT theo cách hợp lí nếu có thể
c) \(\frac{5^3\times3^3}{5^3\times0,5+125\times2,5}\)d)\(\frac{5\times7^1+7^3\times25}{7^5125-7^3\times50}\)e)\(\frac{8^5\times\left(-5\right)^8+\left(-2\right)^5\times10^9}{2^{16}\times5^7+20^8}\)
h)\(\frac{\left(-0,25\right)^{-5}\times9^4\times\left(-2\right)^{-3}-2^{-2}\times6^3}{2^9\times3^6+6^6\times40}\)
Bài 3 Chứng tỏ rằng
a)
Tính :a, frac{frac{left(-5right)^n}{left(7right)}}{frac{left(-5right)^{n-1}}{7}}left(n1right) b,frac{frac{left(-1right)^{2n}}{2}}{frac{left(-1right)^n}{2}}left(nin Nright)Phân số frac{-5}{7}và frac{-1}{2}nằm trong ngoặc nhưng mình chỉ đóng ngoặc đc tử nên đừng hiểu sai nhaAi nhanh mình tick
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Tính :
a, \(\frac{\frac{\left(-5\right)^n}{\left(7\right)}}{\frac{\left(-5\right)^{n-1}}{7}}\left(n>=1\right)\) b,\(\frac{\frac{\left(-1\right)^{2n}}{2}}{\frac{\left(-1\right)^n}{2}}\left(n\in N\right)\)
Phân số \(\frac{-5}{7}\)và \(\frac{-1}{2}\)nằm trong ngoặc nhưng mình chỉ đóng ngoặc đc tử nên đừng hiểu sai nha
Ai nhanh mình tick
Tính:
\(\frac{\left(\frac{-5}{7}\right)^{n+1}}{\left(\frac{-5}{7}\right)^n}\)
\(\frac{\left(\frac{-5}{7}\right)^{n+1}}{\left(\frac{-5}{7}\right)^n}\)
\(=\frac{-5}{7}\)
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#)Giải :
\(\frac{\left(\frac{-5}{7}\right)^{n+1}}{\left(\frac{-5}{7}\right)^n}=\frac{\left(\frac{-5}{7}\right)^n\times\left(\frac{-5}{7}\right)}{\left(\frac{-5}{7}\right)^n}=\frac{-5}{7}\)
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\(\frac{\left(\frac{-5}{7}\right)^{n+1}}{\left(\frac{-5}{7}\right)^n}=\left(\frac{-5}{7}\right)^{n+1-n}=\left(\frac{-5}{7}\right)^1=\frac{-5}{7}\)
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Xem thêm câu trả lời
Tính :
a) \(\frac{\left(\frac{-5}{7}\right)^n}{\left(\frac{-5}{7}\right)^{n-1}}\)( n\(\ge\)1 )
b) \(\frac{\frac{-1}{2}^{2n}}{\left(\frac{-1}{2}\right)^n}\) ( n \(\in\)N )
a: \(=\dfrac{\left(-\dfrac{5}{7}\right)^n}{\left(-\dfrac{5}{7}\right)^n\cdot\dfrac{-7}{5}}=1:\dfrac{-7}{5}=-\dfrac{5}{7}\)
b: \(=\dfrac{\dfrac{1}{4}^n}{\left(-\dfrac{1}{2}\right)^n}=\left(-\dfrac{1}{2}\right)^n\)
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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
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\(\frac{1}{3}-\frac{3}{5}+\frac{5}{7}-\frac{7}{9}+\frac{9}{11}-\frac{11}{13}+\frac{11}{15}+\frac{11}{13}-\frac{9}{11}+\frac{7}{9}-\frac{5}{7}+\frac{3}{5}-\frac{1}{3}\)\(\frac{-3}{4}.31\frac{11}{23}-0,75.8\frac{12}{28}\)\(\left(2\frac{1}{3}+3\frac{1}{2}\right):\left(-4\frac{1}{6}+3\frac{1}{7}\right)+7\frac{1}{2}\)\(4\frac{5}{9}:\left(-\frac{5}{7}\right)+5\frac{4}{9}:\left(-\frac{5}{7}\right)\)\(\frac{2}{3}-4\left(\frac{1}{2}+\frac{3}{4}\right)\)\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{n+1}\right)n\in Z\)
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?
Rút gọn \(A=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
Ta có:
\(A=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
\(=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{2n+1}{n^2\left(n+1\right)^2}\)
\(=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{2n+1}{n^2\left(n+1\right)^2}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{2n+1}{n^2}-\frac{2n+1}{\left(n+1\right)^2}\)
\(=1-\frac{2n+1}{\left(n+1\right)^2}\)
Vậy \(A=\frac{2n+1}{\left(n+1\right)^2}\)
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\(\sin^3\frac{x}{3}+3\sin^3\frac{x}{3^2}+...+3^{n-1}\sin^3\frac{x}{3}=\frac{1}{4}\left(3^n\sin^3\frac{x}{3^n}-\sin x\right)\)\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{2n+1}{2n+2}<\frac{1}{\sqrt{3n+4}}\left(n\ge1\right)\)\(\left(n!\right)^2\ge n^2\ge\left(n+1\right)^{n-1}cho\left(n\ge1\right)\)