1:
Biết M= \(\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^7}-\frac{1}{2^{10}}+...+\frac{1}{2^{43}}-\frac{1}{2^{46}}+\frac{1}{2^{49}}-\frac{1}{2^{52}}\)
Hãy so sánh M và \(\frac{9}{4}\)
M=\(\frac{1}{2}\)-\(\frac{1}{2^4}\)+\(\frac{1}{2^7}\)-\(\frac{1}{2^{10}}\)................+\(\frac{1}{2^{43}}\)-\(\frac{1}{2^{46}}\)+\(\frac{1}{2^{49}}\)
So sánh M và \(\frac{9}{4}\)
\(M=\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^7}-\frac{1}{2^{10}}+......+\frac{1}{2^{\text{49}}}-\frac{1}{2^{52}}\)
so sanh m voi 9/4
nhờ giải đầy đủ nha ai làm đc mình tick cho\
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 \(\frac{1}{2}>\frac{1}{10};\frac{2}{3}>\frac{1}{10};...;\frac{9}{10}>\frac{1}{10}\)
\(\Rightarrow M>1\)
bài khó nhất nhé
2. Ta có :
\(P=\frac{1}{49}+\frac{2}{48}+\frac{3}{47}+...+\frac{48}{2}+\frac{49}{1}\)
cộng vào 48 phân số đầu với 1, trừ phân số cuối đi 48 ta được :
\(P=\left(\frac{1}{49}+1\right)+\left(\frac{2}{48}+1\right)+\left(\frac{3}{47}+1\right)+...+\left(\frac{48}{2}+1\right)+\left(\frac{49}{1}-48\right)\)
\(P=\frac{50}{49}+\frac{50}{48}+\frac{50}{47}+...+\frac{50}{2}+\frac{50}{50}\)
\(P=\frac{50}{50}+\frac{50}{49}+\frac{50}{48}+...+\frac{50}{2}\)
\(P=50.\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+...+\frac{1}{2}\right)\)
\(\Rightarrow\frac{S}{P}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{48}+\frac{1}{49}+\frac{1}{50}}{50.\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+...+\frac{1}{2}\right)}=\frac{1}{50}\)
Bài 1 :Chứng tỏ rằng :
\(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{49}+\frac{1}{50}=\frac{99}{50}-\frac{97}{49}+...+\frac{7}{4}\)\(-\frac{5}{3}+\frac{3}{2}-1\)
Bài 2 : Cho
\(A=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{4998}{4999}\)
Hãy so sánh A và 0,02
Câu hỏi của Lê Thị Minh Trang - Toán lớp 6 - Học toán với OnlineMath
Xem bài 1 nhé !
Bài 1:
Xét vế phải :
\(P=\frac{99}{50}-\frac{97}{49}+...+\frac{7}{4}-\frac{5}{3}+\frac{3}{2}\)\(-1=2\)\(\left(\frac{99}{100}-\frac{97}{98}+...+\frac{7}{8}-\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)\)
\(=2\left(\left(1-\frac{1}{100}\right)-\left(1-\frac{1}{98}\right)+...+\left(1-\frac{1}{4}\right)-\left(1-\frac{1}{2}\right)\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{25}+\frac{1}{26}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{49}+\frac{1}{50}\)
Đẳng thức được chứng tỏ là đúng
Bài 2 :
Đặt \(A'=\frac{3}{4}.\frac{4}{5}.\frac{7}{8}...\frac{4999}{5000}\)
Rõ ràng \(A< A'\)
SUY RA \(A^2< AA'=\frac{2}{50000}=\frac{1}{2500}=\left(\frac{1}{50}\right)^2\)
Nên \(A< \frac{1}{50}=0,02\)
Chúc bạn học tốt ( -_- )
Cho A = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
B = \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\)
a) So sánh A và B
b) Chứng minh A = \(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\)
\(182.\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}}:\frac{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right):\frac{919191}{808080}\)
\(A=\frac{1}{4}.\frac{3}{6}.\frac{5}{8}....\frac{43}{46}.\frac{45}{48}\)
\(B=\frac{2}{5}.\frac{4}{7}.\frac{6}{9}....\frac{44}{47}.\frac{46}{49}\)
a) So sánh A và B
b) Chứng minh A<133
\(A=\frac{1}{4}.\frac{3}{6}.\frac{5}{8}....\frac{43}{46}.\frac{45}{48}\)
\(B=\frac{2}{5}.\frac{4}{7}.\frac{6}{9}....\frac{44}{47}.\frac{46}{49}\)
a) So sánh A và B
b) Chứng minh A<133
a)Ta có:A:B=\(\left(\frac{1}{4}.\frac{3}{6}.\frac{5}{8}....\frac{43}{46}.\frac{45}{48}\right):\left(\frac{2}{5}.\frac{4}{7}.\frac{6}{9}....\frac{44}{47}.\frac{46}{49}\right)=\frac{\left(1.3.5...45\right).\left(2.4.6...46\right)}{\left(4.6.8...48\right)\left(5.7.9...49\right)}=\frac{3.2}{47.48.49}