Tính giá trị của các biểu thức sau 1) \(A=1+2+2^2+...+2^{2015}\) 2) \(B=\left(\dfrac{1}{4}-1\right)\cdot\left(\dfrac{1}{9}-1\right)\cdot\left(\dfrac{1}{16}-1\right)\cdot\cdot\cdot\cdot\cdot\left(\dfrac{1}{400}-1\right)\) 3) \(C=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\) 4) \(D=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\) 5) \(E=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\) 6) Cho 13+23+...+103=3025 Tính S= 23+43+63+...+203
Tính giá trị của các biểu thức sau
1) \(A=1+2+2^2+...+2^{2015}\)
2) \(B=\left(\dfrac{1}{4}-1\right)\cdot\left(\dfrac{1}{9}-1\right)\cdot\left(\dfrac{1}{16}-1\right)\cdot\cdot\cdot\cdot\cdot\left(\dfrac{1}{400}-1\right)\)
3) \(C=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)
4) \(D=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
5) \(E=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\)
6) Cho 13+23+...+103=3025
Tính S= 23+43+63+...+203
Tính giá trị của các biểu thức sau
1) \(A=1+2+2^2+...+2^{2015}\)
2) \(B=\left(\dfrac{1}{4}-1\right)\cdot\left(\dfrac{1}{9}-1\right)\cdot\left(\dfrac{1}{16}-1\right)\cdot\cdot\cdot\cdot\cdot\left(\dfrac{1}{400}-1\right)\)
3) \(C=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)
4) \(D=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
5) \(E=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\)
6) Cho 13+23+...+103=3025
Tính S= 23+43+63+...+203
1) \(A=1+2+2^2+2^3+......+2^{2015}\)
\(\Leftrightarrow2A=2+2^2+2^3+......+2^{2016}\)
\(\Leftrightarrow2A-A=\left(2+2^2+2^3+......+2^{2016}\right)-\left(1+2+2^2+2^3+......+2^{2015}\right)\)
\(\Leftrightarrow A=2^{2016}-1\)
Vậy \(A=2^{2016}-1\)
6)Ta có: \(13+23+33+43+.......+103=3025\)
\(\Leftrightarrow2.13+2.23+2.33+2.43+.......+2.103=2.3025\)
\(\Leftrightarrow26+46+66+86+.......+206=6050\)
\(\Leftrightarrow\left(23+3\right)+\left(43+3\right)+\left(63+3\right)+\left(83+3\right)+.......+\left(203+3\right)=6050\)
\(\Leftrightarrow23+43+63+83+.......+203+3.10=6050\)
\(\Leftrightarrow23+43+63+83+.......+203+=6050-30\)
\(\Leftrightarrow23+43+63+83+.......+203+=6020\)
Vậy S=6020
b, B có 19 thừa số
=> \(-B=(1-\frac{1}{4})(1-\frac{1}{9})(1-\frac{1}{16})...(1-\frac{1}{400}) \)
<=>\(-B=\frac{(2-1)(2+1)(3-1)(3+1)(4-1)(4+1)...(20-1)(20+1)}{4.9.16...400} \)
<=>\(-B=\frac{(1.2.3.4...19)(3.4.5...21)}{(2.3.4.5.6...20)(2.3.4.5...20)} \)
<=>\(-B=\frac{21}{20.2} =\frac{21}{40} \)
<=>\(B=\frac{-21}{40} \)
Cho :S=\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+.......+\dfrac{1}{2013}-\dfrac{1}{2014}+\dfrac{1}{2015}\) và P=\(\dfrac{1}{1008}+\dfrac{1}{1009}+......+\dfrac{1}{2014}+\dfrac{1}{2015}\) Tính \(\left(S-P\right)^{2016}\)
Ta có:
\(S=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2013}-\dfrac{1}{2014}\)
\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{2015}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2014}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2015}\right)-2\left(1+\dfrac{1}{2}+...+\dfrac{1}{2014}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2015}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{2017}\right)\)
\(=\dfrac{1}{1008}+\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2015}\)
Mà \(P=\dfrac{1}{1008}+\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2015}\)
\(\Rightarrow S=P\Rightarrow S-P=0\)
\(\Rightarrow\left(S-P\right)^{2016}=0^{2016}=0\)
Vậy \(\left(S-P\right)^{2016}=0\)
Bài 1: a) Cho \(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)...\left(\dfrac{1}{2015}-1\right)\left(\dfrac{1}{2016}-1\right)\). So sánh A với \(\dfrac{-1}{2015}\)
b) Cho biểu thức \(A=\dfrac{3x^3-x^2-3x+2015}{3x^4-x^3+3x+2014}\). Tính giá trị của biểu thức với x=\(\dfrac{1}{3}\)
A=\(\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot\cdot\cdot\dfrac{-2015}{2016}\)
=\(-\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\cdot\cdot\dfrac{2015}{2016}\)
=\(\dfrac{-1}{2016}>\dfrac{-1}{2015}\)
Vậy\(A>\dfrac{-1}{2015}\)
2.Cho đa thức R(x)=x2-2x.Tính giá trị của biểu thức
\(S=\dfrac{1}{R\left(3\right)}+\dfrac{1}{R\left(4\right)}+\dfrac{1}{R\left(5\right)}+...+\dfrac{1}{R\left(2022\right)}+\dfrac{1}{2.2023}\)
Giaỉ chi tiết
cái cuối là \(R\left(2023\right)\) hay 2.2023 vậy bạn ?
Sửa đề: 1/R(2023)
R(3)=1*3
R(4)=2*4
R(5)=3*5
...
R(2022)=2020*2022
R(2023)=2021*2023
=>\(S=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{2021\cdot2023}+\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{2020\cdot2022}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2021\cdot2023}+\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{2020\cdot2022}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2021}-\dfrac{1}{2023}+\dfrac{1}{2}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2022}\right)\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{2022}{2023}+\dfrac{505}{1011}\right)\simeq0.7496\)
Cuối đề là \(\dfrac{1}{2.2023}\) đó , ko pk là \(\dfrac{1}{R\left(2023\right)}\) đâu
Cho \(S=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{2011}-\dfrac{1}{2012}+\dfrac{1}{2013}\)
và \(P=\dfrac{1}{1007}+\dfrac{1}{1008}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\)
Giá trị biểu thức \(\left(S-P\right)^{2013}\) là
Ta có :
\(S=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+..........+\dfrac{1}{2011}-\dfrac{1}{2012}+\dfrac{1}{2013}\)
\(=\left(1+\dfrac{1}{3}+..........+\dfrac{1}{2013}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+......+\dfrac{1}{2012}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2013}\right)-2\left(\dfrac{1}{2}+......+\dfrac{1}{2012}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2013}\right)-\left(1+\dfrac{1}{2}+......+\dfrac{1}{1006}\right)\)
\(=\dfrac{1}{1007}+\dfrac{1}{1008}+..........+\dfrac{1}{2013}\)
\(\Leftrightarrow S-P=\left(\dfrac{1}{1007}+\dfrac{1}{1008}+....+\dfrac{1}{2013}\right)-\left(\dfrac{1}{1007}+\dfrac{1}{1008}+....+\dfrac{1}{2013}\right)\)
\(\Leftrightarrow S-P=0\)
\(\Leftrightarrow\left(S-P\right)^{2013}=0^{2013}=0\)
\(1+\dfrac{1}{2}+...+\dfrac{1}{2012}+\dfrac{1}{2013}-2\times\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2010}+\dfrac{1}{2012}\right)\)
\(\Rightarrow1+\dfrac{1}{2}+...+\dfrac{1}{2012}+\dfrac{1}{2013}-\left(1+\dfrac{1}{2}+...+\dfrac{1}{1005}+\dfrac{1}{1006}\right)\)
\(\Rightarrow\dfrac{1}{1007}+\dfrac{1}{1008}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\)
\(\Rightarrow S=P\Rightarrow S-P=0\Rightarrow\left(S-P\right)^{2013}=1\)
A=\(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}+\left(\dfrac{1-3-5-7-...-49}{89}\right)\)
B=\(\dfrac{212.3^5.4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\dfrac{5^{10}.7^3-25^4.49^2}{\left(125.71^3+59.14^3\right)}\)
C=\(\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\)
D=\(\left(\dfrac{1,5+1-0,75}{2,5+\dfrac{5}{3}-1,25}\right)+\left(\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}}\right):\dfrac{1890}{2005}+115\)
E=13+23+...+103=3025
Tính F=23+42+63+...+203=?
Soa sánh 1.\(\left(\dfrac{1}{2}\right)^{27}và\left(\dfrac{1}{3}\right)^{18}\) 2.\(\left(\dfrac{-4}{7}\right)^{205}và\left(\dfrac{-5}{8}\right)^{205}\) 3,\(\left(\dfrac{-1053}{2317}\right)^{2016}và\left(\dfrac{2342}{-3426}\right)^{2015}\)
1: \(\left(\dfrac{1}{2}\right)^{27}=\left(\dfrac{1}{8}\right)^9\)
\(\left(\dfrac{1}{3}\right)^{18}=\left(\dfrac{1}{9}\right)^9\)
mà 1/8>1/9
nên \(\left(\dfrac{1}{2}\right)^{27}>\left(\dfrac{1}{3}\right)^{18}\)
2: \(\dfrac{-4}{7}=\dfrac{-32}{56}\)
\(\dfrac{-5}{8}=\dfrac{-35}{56}\)
mà -32>-35
nên \(\dfrac{-4}{7}>\dfrac{-5}{8}\)
hay \(\left(\dfrac{-4}{7}\right)^{205}>\left(-\dfrac{5}{8}\right)^{205}\)
