Cho\(D=\dfrac{1}{2^3}+\dfrac{1}{2^6}+\dfrac{1}{2^9}+...+\dfrac{1}{2^{2025}}TínhP=\left(7D+\dfrac{1}{2^{2025}}\right)^{1981}\)
K
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22 tháng 5 2023
Để tính giá trị của biểu thức $A = \frac{1}{5^2-1} \cdot \frac{1}{6^2-1} \cdot \frac{1}{7^2-1} \cdots \frac{1}{2025^2-1}$, ta có thể sử dụng công thức $a^2-b^2=(a+b)(a-b)$ để đơn giản hóa các mẫu số trong từng phân số. Ta có:
\begin{align*}
A &= \frac{1}{(5+1)(5-1)} \cdot \frac{1}{(6+1)(6-1)} \cdot \frac{1}{(7+1)(7-1)} \cdots \frac{1}{(45+1)(45-1)} \
&= \frac{1}{4 \cdot 6} \cdot \frac{1}{5 \cdot 7} \cdot \frac{1}{6 \cdot 8} \cdots \frac{1}{46 \cdot 44} \
&= \frac{1}{4} \cdot \frac{1}{5} \cdot \frac{1}{7} \cdot \frac{1}{8} \cdots \frac{1}{44} \cdot \frac{1}{46} \
&= \frac{1}{4} \cdot \frac{1}{46} \cdot \frac{1}{5} \cdot \frac{1}{44} \cdot \frac{1}{7} \cdot \frac{1}{42} \cdots \frac{1}{23} \cdot \frac{1}{21} \
&= \frac{1}{2} \cdot \frac{1}{23} \cdot \left( \frac{1}{2} - \frac{1}{23} \right) \cdot \frac{1}{3} \cdot \left( \frac{1}{3} - \frac{1}{22} \right) \cdots \frac{1}{20} \cdot \left( \frac{1}{20} - \frac{1}{25} \right) \
&= \frac{1}{2} \cdot \frac{1}{23} \cdot \frac{21}{22} \cdot \frac{1}{3} \cdot \frac{19}{22} \cdots \frac{1}{20} \cdot \frac{5}{25} \
&= \frac{1}{2} \cdot \frac{21}{23} \cdot \frac{19}{22} \cdot \frac{17}{20} \cdots \frac{3}{5} \cdot \frac{1}{5} \
&= \frac{21 \cdot 19 \cdot 17 \cdots 3}{2 \cdot 23 \cdot 22 \cdots 5} \cdot \frac{1}{5} \
&= \frac{21 \cdot 19 \cdot 17 \cdots 3}{2 \cdot 23 \cdot 22 \cdots 6} \
\end{align*}
Vậy giá trị của biểu thức $A$ là $\frac{21 \cdot 19 \cdot 17 \cdots 3}{2 \cdot 23 \cdot 22 \cdots 6}$.
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 1
\(4\left(a+b+c\right)=a^2+\left(b+c\right)^2\ge\dfrac{1}{2}\left(a+b+c\right)^2\)
\(\Rightarrow a+b+c\le8\)
\(a^2+16-16\ge8a-16\)
\(\Rightarrow P\ge8\left(a+b+c\right)-16+\dfrac{8100}{\sqrt{2a+2b+1}+\sqrt{2c+1}}\)
\(\Rightarrow P\ge8\left(a+b+c\right)-16+\dfrac{48600}{6\sqrt{2a+2b+1}+6\sqrt{2c+1}}\)
\(\Rightarrow P\ge8\left(a+b+c\right)-16+\dfrac{24300}{a+b+c+10}\)
\(\Rightarrow P\ge8\left(a+b+c+10+\dfrac{324}{a+b+c+10}\right)+\dfrac{21708}{a+b+c+10}-96\)
\(\Rightarrow P\ge16.\sqrt{324}+\dfrac{21708}{18}-96=1398\)
Dấu "=" xảy ra tại \(\left(a;b;c\right)=\left(4;0;4\right)\)
5 tháng 6 2017
\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{[\left(n+1\right)\sqrt{n}-n\sqrt{n+1}].[\left(n+1\right)\sqrt{n}+n\sqrt{n+1}]}\)
=\(\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)^2-n^2\left(n+1\right)}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\dfrac{\sqrt{n}}{n}-\dfrac{\sqrt{n+1}}{n+1}\)
=\(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
Áp dụng ta có S=\(\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-...+\dfrac{1}{\sqrt{2024}}-\dfrac{1}{\sqrt{2025}}=1-\dfrac{1}{\sqrt{2025}}=1-\dfrac{1}{45}=\dfrac{44}{45}\)
15 tháng 10 2018
Ta có công thức tổng quát:
\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n+1}+\sqrt{n}\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}\left(n+1-n\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
Vậy \(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{2025\sqrt{2024}+2024\sqrt{2025}}=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{3}}-\dfrac{1}{\sqrt{4}}+...+\dfrac{1}{\sqrt{2024}}-\dfrac{1}{\sqrt{2025}}=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2025}}=1-\dfrac{1}{45}=\dfrac{44}{45}\)
TA
12 tháng 5 2021
a) `1/3 - 1/4 : 2/5 = 1/3 - 5/8 = -7/24`
b) `6/7-(5/6+1/3)-(2/3+1/7) = 6/7-5/6-1/3-2/3-1/7`
`=(6/7-1/7)-(1/3+2/3)-5/6`
`=5/7-1-5/6`
`=-47/42`
c) `-5/9 . 2/5 + 4 5/9 + 5/9 . (-3/5)`
`= -5/9 . 2/5 + 4 + 5/9 + (-5/9) . 3/5`
`=-5/9 . (2/5 + 3/5-1) + 4`
`=-5/9 . 0 +4`
`=4`
d) 3 1/2 - (5 4/7 - 1 1/2) : 0,75`
`=7/2 - (39/7 - 3/2) : 3/4`
`= 7/2 - 57/14 : 3/4`
`=7/2 - 38/7`
`=-27/14`
3 tháng 8 2023
` a/`
` 2 - 1 5/6 + 2 2/3 = 2 - 11/6 - 8/3 = 1/6+ 8/3 = 1/6 + 16/6 = 17/6 `
`b/`
`5/9 xx ( 2 5/6 - 1 2/3 ) = 5/9 xx ( 17/6 - 5/3 ) = 5/9 xx 7/6 = 35/54 `
`c/`
` 1 1/3 : ( 2 + 1 1/6 : 2 5/6 ) `
`= 4/3 : ( 2 + 7/6 : 17/6 ) `
`= 4/3 : ( 2 + 7/6 xx 6/17 )`
`= 4/3 : ( 2 + 7/17 ) `
`= 4/3 : ( 34/17 + 7/17 ) `
`= 4/3 : 41/17 `
`= 4/3 xx 17/41 `
`= 68/123`
` d/`
` 2 3/5 : 3/4 xx 1 4/5 = 13/5 xx 4/3 xx 9/5 =52/15 xx 9/5 = 156/25`
HL
1
8 tháng 2 2023
\(1:\dfrac{2}{3}:\dfrac{3}{4}:\dfrac{4}{5}:...:\dfrac{2024}{2025}\)
= \(1\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{2025}{2024}=\dfrac{2025}{2}\)
\(D=\dfrac{1}{2^3}+\dfrac{1}{2^6}+\dfrac{1}{2^9}+\dfrac{1}{2^{2025}}\)
\(8D=1+\dfrac{1}{2^3}+\dfrac{1}{2^6}+\dots+\dfrac{1}{2^{2022}}\)
\(8D-D=\left(1+\dfrac{1}{2^3}+\dfrac{1}{2^6}+\dots+\dfrac{1}{2^{2022}}\right)-\left(\dfrac{1}{2^3}+\dfrac{1}{2^6}+\dfrac{1}{2^9}+\dots+\dfrac{1}{2^{2024}}\right)\)
\(7D=1-\dfrac{1}{2^{2025}}\)
Khi đó: \(P=\left(7D+\dfrac{1}{2^{2025}}\right)^{1981}=\left(1-\dfrac{1}{2^{2025}}+\dfrac{1}{2^{2025}}\right)^{1981}\)
\(=1^{1981}=1\)