Cho \(Q=\frac{5}{7}.\frac{13}{7^2}.\frac{97}{7^4}.....\frac{3^{2^{99}}+2^{2^{99}}}{7^{2^{99}}}\)
Chứng minh rằng: \(Q.\left(7^{2^{100}-1}\right)\in N\).
Bài 1:
a, Cho S=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\) .Chứng minh rằng \(\frac{2}{5}< S< \frac{8}{9}\)
b, Tìm x thuộc z để phân số \(\frac{x^2-5x-1}{x+2}\)có giá trị là số nguyên
c, Chứng minh rằng \(\left(\frac{7}{65}+1\right)\left(\frac{7}{84}+1\right)\left(\frac{7}{105}+1\right)\left(\frac{7}{124}+1\right)...\left(\frac{7}{153+1}\right)\left(\frac{7}{560}+1\right)< 2\)
d, Chứng minh rằng \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\frac{5}{3^5}-...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
bài 1:
tìm n biết: 5n+7 chia hết 3n+2
bài 2:
1, tìm chữ số tận cùng của:
a,57^1999
b,93^1999
2, Cho A= 999993^1999 - 555557^1997
chứng minh rằng: A chia hết cho 5
bài 3:chứng minh rằng:
a) \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
b)\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
Bài 5:Tìm x biết:
a)11.(x-6)=4.x+11
b)\(4\frac{1}{3}.\left(\frac{1}{6}-\frac{1}{2}\right)\le x\le\frac{2}{3}.\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{4}\right)\)với x\(\in\)Z
c)|x-3|+1=x
Tính:
a, A=\(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\)
b, B=\(\left(\frac{1}{2}-\frac{1}{3}\right).\left(\frac{1}{2}-\frac{1}{5}\right).\left(\frac{1}{2}-\frac{1}{7}\right)....\left(\frac{1}{2}-\frac{1}{99}\right)\)
a) \(A=\left(\frac{1}{2}-\frac{1}{3}\right)\left(\frac{1}{2}-\frac{1}{5}\right)\left(\frac{1}{2}-\frac{1}{7}\right)...\left(\frac{1}{2}-\frac{1}{99}\right)\)
b) \(B=\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+...+\frac{1}{2^{99}}-\frac{1}{2^{100}}\)
Giúp mình nhanh nhé, 3 tick đấy!
Chứng minh rằng \(\frac{1}{2}\times\frac{3}{4}\times\frac{5}{6}\times\frac{7}{8}\times...\times\frac{99}{100}< 0,01\)
\(A=\frac{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+......+\frac{1}{999}}{\frac{1}{1.999}+\frac{1}{3.997}+\frac{1}{5.995}+......+\frac{1}{999.1}}\)
\(B=\frac{1+\left(1+2\right)+\left(1+2+3\right)+\left(1+2+3+4\right)+......+\left(1+2+3+...+98\right)}{1.2+2.3+3.4+4.5+......+98.99}\)
\(C=\frac{\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+......+\frac{1}{100.400}}{\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+......+\frac{1}{299.400}}\)
\(D=\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+......+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{100}}:\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{97}-......-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+......+\frac{1}{500}}\)
\(H=\frac{\left(1+97\right)\left(1+\frac{97}{2}\right)\left(1+\frac{97}{3}\right)\left(1+\frac{97}{4}\right)+...+\left(1+\frac{97}{99}\right)}{\left(1+99\right)\left(1+\frac{99}{2}\right)\left(1+\frac{99}{3}\right)\left(1+\frac{99}{4}\right)+...+\left(1+\frac{99}{97}\right)}\)
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