So sánh A với \(\frac{5}{6}\)biết: A = \(\frac{1}{4}\)+ \(\frac{1}{9}\)+\(\frac{1}{16}\)+\(\frac{1}{25}\)+\(\frac{1}{36}\)
Hãy so sánh \(A\) với \(\frac{31}{6}\)biết rằng:\(A=\frac{1}{7}+\frac{1}{4}+\frac{2005}{2006}+\frac{1}{13}+\frac{1}{9}+\frac{2006}{2007}+\frac{1}{25}+\frac{1}{16}+\frac{2007}{2008}+\frac{1}{49}+\frac{1}{25}+\frac{2008}{2005}+\frac{1}{97}+\frac{1}{36}\).
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So sánh \(A\)với\(13\),biết rằng:
\(A=\frac{13}{15}+\frac{7}{5}+\frac{3}{4}+\frac{1}{5}+\frac{1}{7}+\frac{19}{20}+\frac{5}{4}+\frac{1}{3}+\frac{1}{6}+\frac{1}{13}+\frac{17}{23}+\frac{9}{8}+\frac{2}{5}+\frac{1}{7}+\frac{1}{25}+\frac{3}{2}+\frac{1}{8}+\frac{1}{19}+\frac{1}{9}+\frac{1}{97}\)
So sánh D với \(\frac{3}{4}\)
\(D=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+...+\frac{1}{100}+\frac{1}{121}\)
so sánh các số sau: a,\(0,5\sqrt{100}-\sqrt{\frac{4}{25}}và\left(\sqrt{1\frac{1}{9}-\sqrt{\frac{9}{16}}}\right):5\)
\(0,5\sqrt{100}-\sqrt{\frac{4}{25}}=0,5.10-\frac{\sqrt{4}}{\sqrt{25}}=5-\frac{2}{5}=\frac{23}{5}=\frac{138}{30}\)
\(\left(\sqrt{1\frac{1}{9}-\sqrt{\frac{9}{16}}}\right):5=\left(\sqrt{\frac{10}{9}-\frac{3}{4}}\right):5=\sqrt{\frac{13}{36}}:5=\frac{\sqrt{13}}{6}:5=\frac{\sqrt{13}}{30}\)
Vì 13 < 138 nên \(\sqrt{13}< 138\Rightarrow\frac{\sqrt{13}}{30}< \frac{138}{30}\)
Vậy \(0,5\sqrt{100}-\sqrt{\frac{4}{25}}>\left(\sqrt{1\frac{1}{9}-\sqrt{\frac{9}{16}}}\right):5\).
So sánh A với 1.
Biết: \(A=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\frac{4}{5!}+...+\frac{8}{9!}+\frac{9}{10!}\)
\(\frac{\frac{8}{5}:\left(\frac{8}{5}.\frac{5}{4}\right)}{\frac{16}{25}-\frac{1}{25}}+\frac{1:\frac{4}{7}}{\left(\frac{50}{9}-\frac{9}{4}\right).\frac{36}{17}}+0,6.0,5:\frac{2}{5}\)
8/5:(8/5.5/4) / 16/25-1/25 + 1:4/7 / (50/9-9/4).36/17) + 0,6.0,5:2/5
= 8/5:8/5:5/4 / 3/5 + 7/4 / 50/9.36/17-9/4.36/17 + 0,3.5/2
= 4/5:3/5 + 7/4 / 200/17-81/17 + 3/10.5/2
= 4/5.5/3 + 7/4 / 119/17 + 3/4
= 4/3 + 7/4 : 7 + 3/4
= 4/3 + 4 + 3/4
= 16/12 + 48/12 + 9/12
= 73/12
☆★☆★☆
Cho A=\(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+\frac{1}{100}+\frac{1}{144}+\frac{1}{196}\)
So sánh A với \(\frac{1}{2}\)
Ta có : \(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+.....+\frac{1}{196}\)
=>A=\(\frac{1}{2^2}+\frac{1}{4^2}+......+\frac{1}{13^2}\)
=>A<\(\frac{1}{1.2}+\frac{1}{3.4}+......+\frac{1}{12.13}\)=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{12}-\frac{1}{13}\)
Ta thấy 1/2>1/3;1/4>1/5;........;1/12>1/13
mà các số lớn hơn được xếp vào nhóm số trừ lớn hơn các số được cộng
nên A>1/2
Tính:
a, \(\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{16}\right).\left(1-\frac{1}{25}\right).\left(1-\frac{1}{36}\right)\)
b, \(\left(2-\frac{3}{2}\right).\left(2-\frac{4}{3}\right).\left(2-\frac{5}{4}\right).\left(2-\frac{6}{5}\right)\)
x. (x^2)^3 = x^5
x^7 ≠ x^5
Nếu,
x^7 - x^5 = 0
mủ lẻ nên phương trình có 3 nghiệm
Đáp số:
x = -1
hoặc
x = 0
hoặc
x = 1
a, \(\left(1-\frac{1}{4}\right)\cdot\left(1-\frac{1}{9}\right)\cdot\left(1-\frac{1}{16}\right)\cdot\left(1-\frac{1}{25}\right)\cdot\left(1-\frac{1}{36}\right)\)
\(=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot\frac{24}{25}\cdot\frac{35}{36}\)
\(=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot\frac{3.5}{4.4}\cdot\frac{4.6}{5.5}\cdot\frac{5.7}{6.6}\)
\(=\frac{1.2.3.4.5}{2.3.4.5.6}\cdot\frac{3.4.5.6.7}{2.3.4.5.6}=\frac{1}{6}\cdot\frac{7}{2}\)
\(=\frac{7}{12}\)
b, \(\left(2-\frac{3}{2}\right)\cdot\left(2-\frac{4}{3}\right)\cdot\left(2-\frac{5}{4}\right)\cdot\left(2-\frac{6}{5}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}=\frac{1.2.3.4}{2.3.4.5}\)
\(=\frac{1}{5}\)
câu 1: so sánh A và B
A=\(\frac{10^{15}+1}{10^{16}+1}\)
B=\(\frac{10^{16}+1}{10^{17}+1}\)
Câu 2:so sánh 637 và 1612
( \(\frac{1}{32}\))7 và( \(\frac{1}{16}\))9
câu 3: so sánh
A=\(\frac{10^{1992}+1}{10^{1991}+1}\), B=\(\frac{10^{1993}+1}{10^{1992}+1}\)
câu 4 : CMR :\(\frac{1}{4}\)+\(\frac{1}{16}\)+\(\frac{1}{36}\)+\(\frac{1}{64}\)+.....+\(\frac{1}{10000}\)<\(\frac{1}{2}\)
câu 5 A=1+\(\frac{2^2}{3^2}\)+\(\frac{2^2}{5^2}\)+\(\frac{2^2}{7^2}\)+.......+\(\frac{2^2}{2009^2}\)
So sanh A với 3
câu 6 cho S = \(\frac{3}{4}\)+\(\frac{8}{9}\)+\(\frac{15}{16}\)+......+\(\frac{n^2-1}{n^2}\)
CMR với mọi số tự nhiên n\(\ge\)2 thì 3 không thể là số nguyên