cho m=1/2+2/3+3/4+4/5+...+9/10 so sánh m với 1
cho m=1/2!+2/3!+3/4!+4/5!+5/6!+6/7!+7/8!+8/9!+9/10!. so sánh m với 1
so sánh với 1
2.tính M=4/1*5+4/5*9+4/13*17+4/17*21
so sánh M với 1
3.so sánh Q với 1
Q=1/1*2+1/2*3+...+1/99*100
Cho \(M=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\frac{4}{5}+\frac{5}{6}+\frac{6}{7}+\frac{7}{8}+\frac{8}{9}+\frac{9}{10}\)
So sánh M với 1
Ta có:
1 = \(\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+............+\frac{1}{10}\)(10 phân số \(\frac{1}{10}\))
Mà \(\frac{1}{2}>\frac{1}{10};\frac{2}{3}>\frac{1}{10};............;\frac{9}{10}>10\)
\(\Rightarrow M>1\)
Vậy M > 1
Ta có:
1/2=0,5
2/3>0,6
<=>1/2+2/3>1,1>1
<=>1/2+2/3+3/4+...+9/10>1
Vì 1 = \(\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}\)
\(\Rightarrow\)M > 1 vì \(\frac{1}{2}>\frac{1}{10};\frac{2}{3}>\frac{1}{10};...;\frac{9}{10}>\frac{1}{10}\)
\(\Rightarrow M>1\)
m = 1-(1/2^2+1/3^2+1/4^2+1/5^2+...+1/10^2) . so sánh m với 0
Bài 1 :
a) Cho : S = 1 + 2 + 2^2 + 2^ 3 +... + 2^ 9
So sánh S với 5 * 2 ^8
b) Cho M = 1+2 + 2^2 +2^3 + 2^4
N = 2^5-1
So sánh M và N
a)S=1+2+2^2+2^3+...+2^9
2S=2+2^2+2^3+...+2^10
2S-S=(2+2^2+2^3+2^4+...+2^10)-(1+2+2^2+2^3+...+2^9)
S=2^10-1
S=1024-1
S=1023
Ta có:5.2^8=5.256=1280
Mà 1280>1023
=>S<5.2^8
b)Ta có:M=1+2+2^2+2^3+2^4
=>2M=2+2^2+2^3+2^4+2^5
=>2M-M=(2+2^2+2^3+2^4+2^5)-(1+2+2^2+2^3+2^4)
=>M=2^5-1
Mà N=2^5-1
=>M=N
Không biết có bị sai lỗi nào hay không,nhớ kiểm tra đó
Cho M = \(1-\dfrac{1}{2}-\dfrac{1}{2^2}-\dfrac{1}{2^3}-\dfrac{1}{2^4}-....-\dfrac{1}{2^{10}}\) . So sánh M với \(\dfrac{1}{2^{11}}\)
\(M=1-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\right)\)
Đặt \(N=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\)
\(2N=1+\dfrac{1}{2}+...+\dfrac{1}{2^9}\)
\(\Rightarrow2N-N=1-\dfrac{1}{2^{10}}\)
\(\Rightarrow N=1-\dfrac{1}{2^{10}}\)
\(\Rightarrow M=1-\left(1-\dfrac{1}{2^{10}}\right)=\dfrac{1}{2^{10}}>\dfrac{1}{2^{11}}\)
Vậy \(M>\dfrac{1}{2^{11}}\)
Cho A=1/2!+2/3!+3/4!+...+9/10!.So sánh A với 1
Ta có :
\(A=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{9}{10!}\)
\(A=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{10-1}{10!}\)
\(A=\left(\frac{2}{2!}-\frac{1}{2!}\right)+\left(\frac{3}{3!}-\frac{1}{3!}\right)+\left(\frac{4}{4!}-\frac{1}{4!}\right)+...+\left(\frac{10}{10!}-\frac{1}{10!}\right)\)
\(A=\left(1-\frac{1}{2!}\right)+\left(\frac{1}{2!}-\frac{1}{3!}\right)+\left(\frac{1}{3!}-\frac{1}{4!}\right)+...+\left(\frac{1}{9!}-\frac{1}{10!}\right)\)
\(A=1-\frac{1}{10!}< 1\)
vậy A < 1 vì \(0< \frac{1}{10!}< 1\)
Cho \(M=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+...+\frac{2014}{4^{2014}}\). So sánh M với 4/9
\(4.M=4.\left(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+...+\frac{2014}{4^{2014}}\right)=1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2014}{4^{2013}}\)
=> 4M - M = \(1+\left(\frac{2}{4}-\frac{1}{4}\right)+\left(\frac{3}{4^2}-\frac{2}{4^2}\right)+...+\left(\frac{2014}{4^{2013}}-\frac{2013}{4^{2013}}\right)-\frac{2014}{4^{2014}}\)
=> 3.M = \(1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2013}}-\frac{2014}{4^{2014}}\)
Tính \(N=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2013}}\)
=> \(4.N=4+1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2012}}\)
=> 4N - N = 4 - \(\frac{1}{4^{2013}}\)=> N = \(\frac{4}{3}-\frac{1}{3.4^{2013}}\)=> N < 4/3
Ta có: 3M < N => M < N/3 => M < (4/3)/3 = 2/9
vậy M < 4/9
Cho 0<m<4 ,so sánh P=(m+1)(m+2)(m+3)(m+4)(m-5)với 1 kết quả là...............................
Cho 0<m<4 ,so sánh P=(m+1)(m+2)(m+3)(m+4)(m-5)với 1 kết quả là P < 1.