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\(B=\frac{1}{4}+\left(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}\right)+\left(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}\right)\)
Xét \(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}>\frac{1}{9}+\frac{1}{9}+...+\frac{1}{9}=\frac{1}{9}.5=\frac{5}{9}>\frac{1}{2}\)
và \(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}>\frac{1}{19}+\frac{1}{19}+...+\frac{1}{19}=\frac{1}{19}.10=\frac{10}{19}>\frac{1}{2}\)
Do đó \(B>\frac{1}{4}+\frac{1}{2}+\frac{1}{2}=\frac{5}{4}>1\)
Ta có: \(1+2+3+...+n=\frac{n.\left(n+1\right)}{2}\)
\(Q=\frac{1}{1+2}+\frac{1}{1+2+3}+....+\frac{1}{1+2+3+...+10}\)
\(Q=\frac{1}{\frac{2.\left(2+1\right)}{2}}+\frac{1}{\frac{3.\left(3+1\right)}{2}}+....+\frac{1}{\frac{10.\left(10+1\right)}{2}}\)
\(Q=\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+....+\frac{1}{\frac{10.11}{2}}\)
\(Q=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{10.11}\)
\(\frac{1}{2}Q=\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{10.11}\)
\(\frac{1}{2}Q=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{10}-\frac{1}{11}=\frac{1}{2}-\frac{1}{11}=\frac{9}{22}\)
=>\(Q=\frac{9}{22}.2=\frac{9}{11}\)
\(Q=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{55}\\ \Rightarrow\frac{1}{2}Q=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{110}\)
Tiếp theo tự tính nhé
2. \(\frac{1995.1994-1}{1993.1995+1994}=\frac{1995.\left(1993+1\right)-1}{1993.1995+1994}=\frac{1995.1993+1995-1}{1993.1995+1994}=\frac{1995.1993+1994}{1993.1995+1994}\)
1. \(\frac{4}{3.7}+\frac{5}{7.12}+\frac{1}{12.13}+\frac{7}{13.20}+\frac{3}{20.23}\)
\(=\frac{7-3}{3.7}+\frac{12-7}{7.12}+\frac{13-12}{12.13}+\frac{23-20}{20.23}\)
\(=\left[\frac{7}{3.7}-\frac{3}{3.7}\right]+\left[\frac{12}{7.12}-\frac{7}{7.12}\right]+\left[\frac{13}{12.13}-\frac{12}{12.13}\right]+\left[\frac{20}{13.20}-\frac{13}{13.20}\right]+\left[\frac{23}{20.23}-\frac{20}{20.23}\right]\) \(=\left[\frac{1}{3}-\frac{1}{7}\right]+\left[\frac{1}{7}-\frac{1}{12}\right]+\left[\frac{1}{12}-\frac{1}{13}\right]+\left[\frac{1}{13}-\frac{1}{20}\right]+\left[\frac{1}{20}-\frac{1}{23}\right]\) \(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{20}+\frac{1}{20}-\frac{1}{23}\) \(=\frac{1}{3}-\frac{1}{23}\\ =\frac{20}{69}\)
\(1+\frac{3}{15}+\frac{3}{35}+\frac{3}{63}+\frac{3}{99}+\frac{3}{143}\)
Đặt : \(A=\frac{3}{15}+\frac{3}{35}+\frac{3}{63}+\frac{3}{99}+\frac{3}{143}\)
\(B=1\)
\(\Rightarrow A=5.\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\right)\)
\(\Rightarrow A=\frac{5}{2}.\left(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\right)\)
\(\Rightarrow A=\frac{5}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(\Rightarrow A=\frac{5}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(\Rightarrow A=\frac{5}{2}.\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(\Rightarrow A=\frac{5}{2}.\frac{10}{39}\)
\(\Rightarrow A=\frac{50}{78}=\frac{25}{39}\)
Thay vào , ta có :
\(=1+\frac{25}{39}=\frac{39}{39}+\frac{25}{39}=\frac{64}{39}\)
Vậy giá trị biểu thức trên là \(\frac{64}{39}\)
b) Ta có:
\(B=\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{1}{2016}\)
\(\Rightarrow B=\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{1}{2016}+1\right)+1\)
\(\Rightarrow B=\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2016}+\frac{2017}{2017}\)
\(\Rightarrow B=2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}\right)}=\frac{1}{2017}\)
Vậy \(\frac{A}{B}=\frac{1}{2017}\)
a) \(\frac{x-2}{3}=\frac{x+1}{4}\)
=> (x - 2).4 = 3.(x + 1)
=> 4x - 8 = 3x + 3
=> 4x - 3x = 3 + 8
=> x = 11
Vậy x = 11
b) \(2.\left(x+3\right)-\frac{1}{2}=x-1\)
=> \(2x+6-\frac{1}{2}=x-1\)
=> \(2x+\frac{11}{2}=x-1\)
=> \(2x-x=-1-\frac{11}{2}\)
=> \(x=-\frac{13}{2}\)
Vậy \(x=-\frac{13}{2}\)
a) \(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
\(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
\(x-\frac{1}{2}=\frac{1}{3}\)
\(x=\frac{1}{3}+\frac{1}{2}\)
\(x=\frac{5}{6}\)
b)\(\left(2x-3\right)^3=343\)
\(\left(2x-3\right)^3=7^3\)
\(2x-3=7\)
\(2x=7+3\)
\(2x=10\)
\(x=10:2\)
\(x=5\)
a) Ta có: \(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
<=> \(x-\frac{1}{2}=\sqrt[3]{\frac{1}{27}}=\frac{1}{3}\)
<=> \(x=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)
Vậy x=5/6
b)\(\left(2x-3\right)^3=343\)
<=>\(2x-3=\sqrt[3]{343}=7\)
<=> 2x=10 <=> x=5
c) \(\left(\frac{1}{3}\right)^{2x}+1=\frac{1}{7}\)
<=>\(\left(\frac{1}{3}\right)^{2x}=\frac{-6}{7}\)
<=> \(\left(\frac{1}{3^x}\right)^2=-\frac{6}{7}\)(vô lí vì \(\left(\frac{1}{3^x}\right)^2\ge0\))
Vậy ko tìm được x thỏa mãn.
d)\(\left(2x-3\right)^2=9\)
=>\(\left[\begin{array}{nghiempt}2x-3=3\\2x-3=-3\end{array}\right.\)<=> \(\left[\begin{array}{nghiempt}x=3\\x=0\end{array}\right.\)
Vậy x=3 hoặc x=0.
e) \(\left(x-3\right)^6=\left(x-3\right)^7\)
<=> \(\left(x-3\right)^7-\left(x-3\right)^6=0\)
<=> \(\left(x-3\right)^6\left(x-3-1\right)=0\)
<=>\(\left(x-3\right)^6\left(x-4\right)=0\)
<=> \(\left[\begin{array}{nghiempt}x-3=0\\x-4=0\end{array}\right.\)=> \(\left[\begin{array}{nghiempt}x=3\\x=4\end{array}\right.\)
Vậy x \(\in\left\{3;4\right\}\)
\(\left(\frac{1}{1}+\frac{1}{2}\right).\left(\frac{1}{1}+\frac{1}{3}\right).\left(\frac{1}{1}+\frac{1}{4}\right)...\left(\frac{1}{1}+\frac{1}{2016}\right).\left(\frac{1}{1}+\frac{1}{2017}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{2017}{2016}.\frac{2018}{2017}\)
\(=\frac{3.4.5...2017.2018}{2.3.4...2016.2017}\)
\(=\frac{2018}{2}=1009\)