So sánh: M=\(\frac{\left(2^3+1\right)\left(3^3+1\right)...\left(100^3+1\right)}{\left(2^3-1\right)\left(3^3-1\right)...\left(100^3-1\right)}\) với N=\(\frac{3}{2}\)
So sánh:
\(C=\frac{\left(2^3+1\right)\left(3^3+1\right)\left(4^3+1\right)...\left(100^3+1\right)}{\left(2^3-1\right)\left(3^3-1\right)\left(4^3-1\right)...\left(100^3-1\right)}\)và \(D=1,5\)
1, Tính \(\frac{1}{2}-\left(\frac{1}{3}+\frac{2}{3}\right)+\left(\frac{1}{4}+\frac{2}{4}+\frac{3}{4}\right)-\left(\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\right)+...+\left(\frac{1}{100}+\frac{2}{100}+\frac{3}{100}+...+\frac{99}{100}\right)\)2,Tính \(\left(1-\frac{1}{2^2}\right)x\left(1-\frac{1}{3^2}\right)x\left(1-\frac{1}{4^2}\right)x...x\left(1-\frac{1}{n^2}\right)\)
cho A=\(\frac{\left(2^3+1\right)\left(3^3+1\right)\left(4^3+1\right)...\left(10^3+1\right)}{\left(2^3-1\right)\left(3^3-1\right)\left(4^3-1\right)...\left(10^3-1\right)}\) so sánh A với \(\frac{3}{2}\)
Tính:
\(\left[\frac{1}{100}-1^2\right].\left[\frac{1}{100}-\left(\frac{1}{2}\right)^2\right].\left[\frac{1}{100}-\left(\frac{1}{3}\right)^2\right].....\left[\frac{1}{100}-\left(\frac{1}{20}\right)^2\right]\)
Giải nhanh lên giúp mk với! Rồi mk tick cho 3 cái
đây có chắc là toán lớp 7 không đấy
nếu có bài hình nào khó thì cho lên đấy nhé mình chuyên về toán lớp 7 hơn
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right).....\left(\frac{1}{100^2}-1\right)\).So sánh A với \(-\frac{1}{2}\)
1. tính A= \(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{899}{30^2}\)
2. tính B= \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}...\frac{30}{62}.\frac{31}{64}\)
3. So sánh C= \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{20}\right)\)với \(\frac{1}{21}\)
4. So sánh D= \(\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{100}\right)\)với \(\frac{11}{19}\)
\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{899}{30^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{29.31}{30.30}=\frac{1.2.3.....29}{2.3.4.....30}.\frac{3.4.5.....31}{2.3.4.....30}\)
\(=\frac{1}{2}.\frac{31}{30}=\frac{31}{60}\)
Cho \(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)......\left(\frac{1}{100^2}-1\right)\)
Hãy so sánh A với 1/2
A>1/2
Xin lỗi mình đang bận để lúc khác mình sẽ giải chi tiết
A=\(\left[\frac{1}{100}-1^2\right].\left[\frac{1}{100}-\left(\frac{1}{2}\right)^2\right].\left[\frac{1}{100}-\left(\frac{1}{3}\right)^2\right]....\left[\frac{1}{100}-\left(\frac{1}{20}\right)^2\right]\)
Mọi người giúp em với ạ :'(
A=(1/100- 1^2). (1/100-(1/2)^2).....(1/100- (1/510)^2).....(1/100-(1/20)^2)
A=(1/100- 1^2). (1/100-(1/2)^2).....(1/100- 1/100).....(1/100-(1/20)^2)
A=(1/100- 1^2). (1/100-(1/2)^2).....0.....(1/100-(1/20)^2)
A=0
Mình ko biết gõ ngoặc vuông bạn thông cảm nha! Chúc bạn học tốt!!!
Cho M =\(\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)....\left(\frac{1}{100^2}-1\right)\)
So sánh: M và \(-\frac{1}{2}\)
M=-(\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{1-100^2}{100^2}\))
=-(\(\frac{1.3}{2.2}.\frac{2.4}{3.3}\frac{3.5}{4.4}...\frac{99.100}{100.100}\))
=-(\(\frac{1.2.3...99}{2.3.4...100}.\frac{3.4.5...100}{2.3.4..100}\))
=-(\(\frac{1}{100}.\frac{1}{2}\))
=\(\frac{-1}{200}\)