(1-1/2).(1-1/3).(1-1/4)...(1-1/2002).x=1-1/1.2-1/2.3-1/3.4-...-1/2002.2003
so sanh (a+1).(a+2).(a+3)-a.(a+1).(a+2) va 3(a+1).(a+2)
b,tinh
M=1.2+2.3+3.4+...+ 2002.2003
GIUP MINH NHANH LEN NHE CAC BAN!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Chứng tỏ rằng: 1/1 - 1/2 =1/1.2 ; 1/2 -1/3 = 1/2.3 ; 1/3 -1/4 =1/3.4
Tính S = 1/1.2 + 1/2.3 + 1/3.4+... 1/49.50
Phần chứng tỏ quy đồng lên rồi tính là ra
Còn phần tính S thì áp dụng tính chất vừa chứng tỏ để tách ra
Kết quả là 49/50
1/1.2+1/2.3+1/3.4+...+1/x(x+1)=2/3
Ta có: 1/1.2+1/2.3+1/3.4+...+1/x(x+1)=2/3
=> 1-1/2+1/2-1/3+1/3-1/4+...+1/x-1/x+1=2/3
=>1-1/x+1=2/3
=>1/x+1=1/3
=>3=x+1
=>x=2
Ta có\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{x\left(x+1\right)}=\frac{2}{3}\)
=>\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2}{3}\)
=>\(1-\frac{1}{x+1}=\frac{2}{3}\)
=>\(\frac{1}{x+1}=1-\frac{2}{3}\)
=>\(\frac{1}{x+1}=\frac{1}{3}\)
=>\(x+1=3\)
=>\(x=2\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{2}{3}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x-1}=\frac{2}{3}\)
\(\Rightarrow1-\frac{1}{x-1}=\frac{2}{3}\)
\(\Rightarrow\frac{1}{3}=\frac{1}{x-1}\)
\(\Rightarrow x=3+1=4\)
1.Tính
A= (1-1/22).(1-1/32)...(1-1/1002)
B= -1/1.2-1/2.3-1/3.4-...-1/100.101
C= 1.2+2.3+3.4+...+100.101
Lời giải :
Đặt S=1.2+2.3+3.4+4.5+…+99.100+100.101
3S=1.2.3+2.3.3+3.4.3+4.5.3+…+99.100.3+100.101.3
=1.2(3−0)+2.3(4−1)+3.4(5−2)+4.5(6−3)+…+99.100(101−98)+100.101(102−99)
=0.1.2-1.2.3+1.2.3-2.3.4+...+99.100.101-100.101.102
=100.101.102
S=100.101.34=343400
1.Tính
a) Ta có:
A=(1-1/22).(1-1/32)...(1-1/1002)
=>A=3/22.8/32.....9999/1002
=>A=(1.3/2.2).(2.4/3.3).....(99.101/100.100)
=>A=(1.2.3.....99/2.3.4.....100).(3.4.5.....101/2.3.4.....100)
=>A=1/100.101/2
=>A=101/200
b) Ta có:
B=-1/1.2-1/2.3-1/3.4-...-1/100.101
=>B=-(1/1.2+1/2.3+1/3.4+...+1/100.101)
=>B=-(1-1/2+1/2-1/3+1/3-1/4+...+1/100-1/101)
=>B=-(1-1/101)
=>B=-100/101
c) Ta có:
C=1.2+2.3+3.4+...+100.101
=>3C=1.2.3+2.3.3+3.4.3+...+100.101.3
=>3C=1.2.3+2.3.(4-1)+3.4.(5-2)+...+100.101.(102-99)
=>3C=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-3.4.5+...+100.101.102
=>3C=100.101.102
=>3C=1030200
=>C=343400
Chúc bạn hok tốt nhé >:)!!!!!
(1-1/2).(1-1/3). ... .(1-1/2022).x=1-1/1.2-1/2.3-...-1/2002*2003
(1 - \(\dfrac{1}{2}\)).(1 - \(\dfrac{1}{3}\))....(1- \(\dfrac{1}{2022}\)).\(x\) = 1 - \(\dfrac{1}{1.2}\) - \(\dfrac{1}{2.3}\)-...-\(\dfrac{1}{2002.2003}\)
(\(\dfrac{2-1}{2}\)).(\(\dfrac{3-1}{3}\))...(\(\dfrac{2022-1}{2022}\)).\(x\) = 1 - (\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{2002.2003}\))
\(\dfrac{1}{2}\).\(\dfrac{2}{3}\)...\(\dfrac{2021}{2022}\).\(x\) = 1 - (\(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\)+ ... + \(\dfrac{1}{2002}\) - \(\dfrac{1}{2003}\))
\(\dfrac{1}{2022}\).\(x\) = 1 - (\(\dfrac{1}{1}\) - \(\dfrac{1}{2003}\))
\(\dfrac{1}{2022}\).\(x\) = \(\dfrac{1}{2003}\)
\(x\) = \(\dfrac{1}{2003}\) : \(\dfrac{1}{2022}\)
\(x\) = \(\dfrac{2022}{2003}\)
Tìm số hữu tỉ x biết
1.2-1/2!+2.3-1/3!+3.4-1/4!+.....+2016.2017-1/2017!
tìm x biết :(1.2+2.3+3.4+...+2017.2018)/(2018.2019.x)=1/(1+2)+1/(1+2+3)+....+1/(1+2+....+2018)
Chứng minh rằng:
a) 1.2 - 1 phần 2! + 2.3 -1 phần 3! + 3.4 -1/4! + ... + 99.100 -1 /100! < 2
b) 1/1.2 + 1/3.4 + 1/5.6 + ... + 1/49.50 = 1/26 + 1/27 + 1/28 + ... + 1/50
cmr 1.2-1/2!+2.3-1/3!+3.4-1/4!+....+99.100-1/100!<2