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\(\left(1+\dfrac{1}{1.3}\right).\left(1+\dfrac{1}{2.4}\right).\left(1+\dfrac{1}{3.5}\right).........\left[1+\dfrac{1}{x.\left(x+2\right)}\right]=\dfrac{31}{16}\)
\(\Rightarrow\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}........\dfrac{\left(x+1\right)^2}{x.\left(x+2\right)}=\dfrac{31}{16}\)
\(\Rightarrow\dfrac{\left[2.3.4.............\left(x+1\right)\right].\left[2.3.4.............\left(x+1\right)\right]}{\left(1.2.3...................x\right).\left(3.4.5..........................\left(x+2\right)\right)}=\dfrac{31}{16}\)
\(\Rightarrow\dfrac{\left(x+1\right).2}{1.\left(x+2\right)}=\dfrac{31}{16}\)
\(\Leftrightarrow16.2\left(x+1\right)=31.\left(x+2\right)\)
\(\Rightarrow32x+32=31x+62\)
\(\Rightarrow x=30\)
Vậy x=30
Chúc bn học tốt
\(C=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{3}{3.5}\right)...\left(1+\dfrac{2014}{2016}\right)\)
\(C=\dfrac{4}{1.3}.\dfrac{9}{2.4}.\dfrac{16}{3.5}.....\dfrac{4060225}{2014.2016}\)
\(C=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}.\dfrac{4.4}{3.5}.....\dfrac{2015.2015}{2014.2016}\)
\(C=\dfrac{2.2.3.3.4.4.....2015.2015}{1.3.2.4.3.5.....2014.2016}\)
\(C=\dfrac{2.\left(3.2\right)\left(4.3\right).....\left(2015.2014\right).2015}{1.\left(3.2\right)\left(4.3\right).....\left(2015.2014\right).2016}\)\(\)
\(C=\dfrac{2.2015}{1.2016}\)
\(C=\dfrac{4030}{2016}\)\(=1\dfrac{2014}{2016}\).
\(P=\left(1+\dfrac{1}{1\cdot3}\right)\left(1+\dfrac{1}{2\cdot4}\right)\cdot...\cdot\left(1+\dfrac{1}{2023\cdot2025}\right)\)
\(=\left(1+\dfrac{1}{2^2-1}\right)\left(1+\dfrac{1}{3^2-1}\right)\cdot...\cdot\left(1+\dfrac{1}{2024^2-1}\right)\)
\(=\dfrac{2^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(3-1\right)\left(3+1\right)}\cdot...\cdot\dfrac{2024^2}{\left(2024-1\right)\left(2024+1\right)}\)
\(=\dfrac{2\cdot2}{1\cdot3}\cdot\dfrac{3\cdot3}{2\cdot4}\cdot...\cdot\dfrac{2024\cdot2024}{2023\cdot2025}\)
\(=\dfrac{2\cdot3\cdot...\cdot2024}{1\cdot2\cdot3\cdot...\cdot2023}\cdot\dfrac{2\cdot3\cdot...\cdot2024}{3\cdot4\cdot5\cdot...\cdot2025}\)
\(=2024\cdot\dfrac{2}{2025}=\dfrac{4048}{2025}\)
Bài 1a) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2018.2019}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{2018}-\dfrac{1}{2019}\)
\(=1-\dfrac{1}{2019}=\dfrac{2018}{2019}\)
b) \(S=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2017.2019}\)
\(2S=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2017.2019}\)
\(2S=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2017}-\dfrac{1}{2019}\)
\(2S=1-\dfrac{1}{2019}=\dfrac{2018}{2019}\)
\(S=\dfrac{1009}{2019}\)
Còn lại bạn làm tương tự hết nhé .
Dat A=\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{13.15}\)
2A=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{13.15}\)
= 1-\(\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-....+\dfrac{1}{13}-\dfrac{1}{15}\)
= 1-\(\dfrac{1}{15}=\dfrac{14}{15}\)
=> A=\(\dfrac{7}{15}\)
Ta co : \(\dfrac{7}{15}\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)
=> \(\dfrac{7}{15}x-\dfrac{7}{15}+\dfrac{7}{15}=\dfrac{3}{5}x\)
=> \(\dfrac{7}{15}x-\dfrac{3}{5}x=0\)
=> x\(\left(\dfrac{7}{15}-\dfrac{3}{5}\right)=0\)
=> x\(\left(-\dfrac{2}{15}\right)=0\)
=> x=0
\(\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{13.15}\right)\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)
<=>\(\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{13.15}\right)\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)
<=>\(\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)
<=>\(\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{15}\right)\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)
<=> \(\dfrac{7}{15}\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)
<=>\(\dfrac{7}{15}x-\dfrac{7}{15}=\dfrac{3}{5}x-\dfrac{7}{15}\)
<=>\(\dfrac{7}{15}x-\dfrac{3}{5}x=\dfrac{-7}{15}+\dfrac{7}{15}\)
<=> \(\dfrac{-2}{15}x=0\)
<=> \(x=0\)
Vậy: \(s=\left\{0\right\}.\)
1)\(2a^4+1\ge2a^3+a^2\)
\(\Leftrightarrow2a^4-2a^3-a^2+1\ge0\)
\(\Leftrightarrow\left(a^4-2a^3+a^2\right)+\left(a^4-2a^2+1\right)\ge0\)
\(\Leftrightarrow\left(a^2-a\right)^2+\left(a^2-1\right)^2\ge0\)(luôn đúng)
"="<=>a=1
Ta có:\(2A=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{9\cdot11}\)
\(2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{11}\)
\(2A=1-\dfrac{1}{11}=\dfrac{10}{11}\)
\(B=\left(1+\dfrac{1}{1\cdot3}\right)\left(1+\dfrac{1}{2\cdot4}\right)\cdot...\cdot\left(1+\dfrac{1}{9\cdot11}\right)\)
\(B=\dfrac{4}{1\cdot3}\cdot\dfrac{9}{2\cdot4}\cdot...\cdot\dfrac{100}{9\cdot11}\)
\(B=\dfrac{2\cdot2\cdot3\cdot3\cdot...\cdot10\cdot10}{1\cdot3\cdot2\cdot4\cdot...\cdot9\cdot11}\)
\(B=\dfrac{20}{11}\)
\(\Rightarrow11< 2x< 20\)
\(\Rightarrow x\in\left\{6;7;8;9\right\}\)
nhóm từng cái vào r` tính ra
Đặt biểu thức cần tính là A, ta có:
A=\(\dfrac{1}{7}\left(\dfrac{7}{3.10}+\dfrac{7}{10.17}+...+\dfrac{7}{73.80}\right)\)
Làm tg tự với những cái khác là ok