tìm x: 4/2.4 + 4/4.6 +4/6.8...+ 4/x(x+2) = 2011/2013 x > hoặc bằng 2
Tìm x, biết:
1) 1/3. x + 2/5. (x - 1) = 0
2) (2 x - 3).(6 - 2 x) =0
3) (3 x -1).(-1/2 x + 5) = 0
4) (2 x + 3/5)2 - 9/25 = 0
Tính:
1) 4/2.4 + 4/4.6 + 4/6.8 + ... + 4/2008.2010
2) 1/18 + 1/54 + 1/108 + ... +1/990
\(a,\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0\)
\(\frac{11}{15}x=\frac{2}{5}\)
\(x=\frac{6}{11}\)
b,\(\left(2x-3\right).\left(6-2x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-3=0\\6-2x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{3}{2}\\x=3\end{cases}}\)
Vậy
\(\left(3x-1\right).\left(-\frac{1}{2}x+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-1=0\\-\frac{1}{2}x+5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}}\)\
Vậy
1. tìm x thuộc N
a) hỗn số: x 3/4 = 21989/7996
b) hỗn số: x 2/x = 83/x
2. tính
S1 = 1 / 2.4 + 1 / 4.6 + 1 / 6.8 + ............ + 1 / 100.102
S2 = 3 / 1.3 + 3 / 3.5 + ....... + 3 / 27.29
2
\(S1=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{100.102}\)
\(S1=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{100.102}\right)\)
\(S1=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{102}\right)\)
\(S1=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{102}\right)\)
\(S1=\frac{1}{2}.\left(\frac{51}{102}-\frac{1}{102}\right)\)
\(S1=\frac{1}{2}.\frac{25}{51}\)
\(S1=\frac{25}{102}\)
2/2.4+4/4.6+4/6.8+...+4/2018.2020+4/2020.2022
Sửa đề: \(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2018\cdot2020}+\dfrac{4}{2020\cdot2022}\)
Ta có: \(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2018\cdot2020}+\dfrac{4}{2020\cdot2022}\)
\(=2\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2018\cdot2020}+\dfrac{2}{2020\cdot2022}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2018}-\dfrac{1}{2020}+\dfrac{1}{2020}-\dfrac{1}{2022}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{2022}\right)\)
\(=2\cdot\dfrac{505}{1011}\)
\(=\dfrac{1010}{1011}\)
Thực hiện tính :
a) A = 1+1/2(1+2)+1/3(1+2+3)+1/4(1+2+3+4)+...+1/2013(1+2+3+..+2013)
b) B = 1-3/7.3+2-4/2.4+3-5/3.5+4-6/4.6+....+2011-2013/2011.2013+2012-2014/2012.2014-2013+2014/2013.2014
tim x 2/2.4+2/4.6+...+2/x.(x+2)=4/9
2/2.4+2/4.6+.....+2/x.(x+2) =4/9
\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{x.\left(x+2\right)}=\dfrac{4}{9}\)
=\(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x-2}\)
=\(\dfrac{1}{2}-\dfrac{1}{x+2}\)=\(\dfrac{4}{9}\)
=>\(\dfrac{1}{x+2}=\dfrac{1}{2}-\dfrac{4}{9}=\dfrac{1}{18}\)
=>\(x+2=18\)
=>x=16
K=4/2.4+4/4.6+4/6.8+...+4/2008.2010
K=2.(2/2.4+2/4.6+2/6.8+...+2/2008.2010)
K=2.(4-2/2.4+6-4/4.6+8-6/6.8+...+2010-2008/2008.2010)
K=2.(1/2-1/4+1/4-1/6+1/6-1/8+...+1/2008-1/2010)
K=2.(1.2-1.2010)
K=2.502/1005
K=1004/1005
F=4/2.4+4/4.6+4/6.8+..........+4/2008.2010
F=4/2.4+4/4.6+4/6.8+..........+4/2008.2010
F=2/2-2/4+2/4-2/6+2/6-2/8+......+2/2008-2/2010
F=2/2- 2/4+2/4-2/6+2/6-2/8+......+2/2008-2/2010
F=2/2-2/2010
=>F=2008/2010=1004/1005
F = 4/2.4 + 4/4.6 + 4/6.8 + ....+ 4/2008.2010
Gọi F= 4/2.4+4/4.6+4/6.8+...+4/2008.2010
F/2= 2/2.4+2/4.6+...+2/2008.2010
Mà 2/2.4=1/2-1/4; 2/4.6=1/4-1/6 ....
Vậy F/2= (1/2-1/4)+(1/4-1/6)+....+(1/2008-1/2010)
F/2=1/2-1/2010=2010/4020-2/4020=2008/4...
F= 2008.2/4020=1004/1005