Tính \(\dfrac{1^2}{1.3}+\dfrac{2^2}{3.5}+\dfrac{3^3}{5.7}+...+\dfrac{1006^2}{2011.2013}\)
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Tính giá trị biểu thức sau :
\(A=\dfrac{1^2}{1.3}+\dfrac{2^2}{3.5}+\dfrac{3^2}{5.7}+...+\dfrac{1006^2}{2011.2013}+\dfrac{1007^2}{2013.2015}\)
Tìm x nguyên biết:
a)\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2x-1\right).\left(2x+1\right)}=\dfrac{49}{99}\)
b)\(1-3+3^2-3^3+...+\left(-3\right)^x=\dfrac{9^{1006}-1}{2}\)
a, \(\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}\right)=\dfrac{49}{99}\)
\(\Leftrightarrow\dfrac{1}{2}.\left(1-\dfrac{1}{2x+1}\right)=\dfrac{49}{99}\)
\(\Leftrightarrow\dfrac{2x+1-1}{2x+1}=\dfrac{98}{99}\)
\(\Leftrightarrow98\left(2x+1\right)=99.2x\)
\(\Leftrightarrow2x=98\Rightarrow x=49\)
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b: Đặt \(A=1-3+3^2-3^3+...+\left(-3\right)^x\)
\(=\left(-3\right)^0+\left(-3\right)^1+\left(-3\right)^2+...+\left(-3\right)^x\)
\(\Leftrightarrow-3A=\left(-3\right)^1+\left(-3\right)^2+...+\left(-3\right)^{x+1}\)
\(\Leftrightarrow-3A-A=\left(-3\right)^1+\left(-3\right)^2+...+\left(-3\right)^{x+1}-...-1\)
\(\Leftrightarrow-4A=\left(-3\right)^{x+1}-1\)
\(\Leftrightarrow A=\dfrac{\left(-3\right)^{x+1}-1}{-4}=\dfrac{-\left(-3\right)^{x+1}+1}{4}\)
\(\Leftrightarrow\dfrac{-\left(-3\right)^{x+1}+1}{4}=\dfrac{3^{2012}-1}{2}\)
\(\Leftrightarrow-\left(-3\right)^{x+1}+1=2\cdot3^{2012}-2\)
\(\Leftrightarrow-\left(-3\right)^{x+1}=2\cdot3^{2012}-3\)
\(\Leftrightarrow-\left(-3\right)^{x+1}=3\left(2\cdot3^{2011}-1\right)\)
\(\Leftrightarrow-\left(-3\right)^x=2\cdot3^{2011}-1\)
=>x=2010
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\(\dfrac{1^2}{1.3}+\dfrac{2^2}{3.5}+\dfrac{3^2}{5.7}\)
1 tháng 5 2021 lúc 9:04
Tính hợp lí:
A=\(\dfrac{3}{1.3}+\dfrac{3}{3.5}+\dfrac{3}{5.7}+...+\dfrac{3}{49.51}\)
B=\(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^8}\)
Giúp mik nha mik đang cần rất là gấp nha !!!!!!!!!!
A bn lướt xuống dưới mà xem cách làm
nhưng của bn là cho 3 ra ngoài nha
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Giải:
A=3/1.3+3/3.5+3/5.7+...+3/49.51
A=3/2.(2/1.3+2/3.5+2/5.7+...+2/49.51)
A=3/2.(1/1-1/3+1/3-1/5+1/5-1/7+...+1/49-1/51)
A=3/2.(1/1-1/51)
A=3/2.50/51
A=25/17
B=1/3+1/32+1/33+...+1/38
3B=1+1/3+1/32+...+1/37
3B-B=(1+1/3+1/32+...+1/37)-(1/3+1/32+1/33+...+1/38)
2B=1-1/38
B=1-1/38 /2
Chúc bạn học tốt!
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tính tổng A=\(\dfrac{2}{1.3}\)+\(\dfrac{2}{3.5}\)+\(\dfrac{2}{5.7}\)+...+\(\dfrac{2}{99.101}\)
`A=2/[1.3]+2/[3.5]+2/[5.7]+.....+2/[99.101]`
`A=1-1/3+1/3-1/5+1/5-1/7+......+1/99-1/101`
`A=1-1/101=101-1/101=100/101`
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A=2/1.3+2/3.5+2/5.7+...+2/99.101
= 1/1 - 1/3 +1/3 - 1/5 +.... +1/99 - 1/101
= 1-1/101
=101/101-1/101
=100/101
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Xem thêm câu trả lời
17 tháng 3 2023 lúc 20:16
Tính giá trị biểu thức:
B= \(\dfrac{\left(-2\right)^{24}.3^5-4^{12}.9^2}{8^8.3^5}+\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{301.303}\)
\(B=\dfrac{2^{24}\cdot3^5-2^{24}\cdot3^4}{2^{24}\cdot3^5}+1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{301}-\dfrac{1}{303}\)
\(=\dfrac{2^{24}\cdot3^4\left(3-1\right)}{2^{24}\cdot3^5}+\dfrac{302}{303}\)
\(=\dfrac{2}{3}+\dfrac{302}{303}=\dfrac{202+302}{303}=\dfrac{504}{303}\)
=168/101
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Tính nhanh các tổng sau :
S=\(\dfrac{2}{1.3}\)+\(\dfrac{2}{3.5}+\)\(\dfrac{2}{5.7}+\)...+\(\dfrac{2}{47.49}\)
Tính tổng: \(\dfrac{1^2}{1.3}+\dfrac{2^2}{3.5}+\dfrac{3^2}{5.7}+...+\dfrac{2016^2}{4031.4033}\)
\(\sum\limits^{2016}_{x=1}\left(\dfrac{x^2}{\left(2x-1\right)\left(2x+1\right)}\right)\)
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\(\dfrac{2}{1.3}\) + \(\dfrac{2}{3.5}\) + \(\dfrac{2}{5.7}\) + ... + \(\dfrac{2}{95.97}\)
\(\dfrac{2}{1.3}\) + \(\dfrac{2}{3.5}\) + ..... + \(\dfrac{2}{95.97}\)
= 1 - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + .... + \(\dfrac{1}{95}\) - \(\dfrac{1}{97}\)
= \(1-\dfrac{1}{97}\)
= \(\dfrac{96}{97}\)
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\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+...+\dfrac{2}{95\times97}\)
\(=\dfrac{2}{3}\left(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{95\times97}\right)\)
\(=\dfrac{2}{3}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)
\(=\dfrac{2}{3}\left(1-\dfrac{1}{97}\right)\)\(=\dfrac{2}{3}\times\dfrac{96}{97}\)\(=\dfrac{64}{97}\)
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