Cho A = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\)
CMR:\(\frac{4}{11}\) <A<1
Những câu hỏi liên quan
CMR: \(\frac{1}{10^2}+\frac{1}{11^2}+\frac{1}{12^2}+...+\frac{1}{100^2}>\frac{3}{4}\)
1 CMR:
B=\(\frac{4}{3}+\frac{7}{3^2}+\frac{10}{3^3}+.....+\frac{3n+1}{3^n}< \frac{11}{4}\)(n thuộc N*;n>3)
A=\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}< \frac{3}{4}\)
C=\(\frac{2}{3}+\frac{8}{9}+\frac{26}{27}+...+\frac{3^{20}-1}{3^{20}}>19\frac{1}{2}\)
Có : \(3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(3A-A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow2A< 1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
Có: \(6A< 3+1+\frac{1}{3}+...+\frac{1}{3^{98}}\)
\(6A-2A< 3-\frac{1}{3^{99}}< 3\)
\(\Rightarrow4A< 3\Rightarrow A< \frac{3}{4}\)(đpcm)
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1) Cho \(A=\frac{9}{10!}+\frac{9}{11!}+\frac{9}{12!}+...+\frac{9}{1000!}.CMR:A< \frac{1}{9!}\)
2) \(CMR:\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2}\)
Ai giúp mk sẽ đc thưởng 3 tick , phải ghi chép đầy đủ nha
bài 1: cho x, y thuộc Q. cmr:
|x + y| =< |x| + |y|
bài 2: tính:
\(A=\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{7}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
bài 3: cho a + b + c = a^2 + b^2 + c^2 = 1 và x : y : z = a : b : c.
cmr: (x + y + z)^2 = x^2 + y^2 + z^2
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Ez lắm =)
Bài 1:
Với mọi gt \(x,y\in Q\) ta luôn có:
\(x\le\left|x\right|\) và \(-x\le\left|x\right|\)
\(y\le\left|y\right|\) và \(-y\le\left|y\right|\Rightarrow x+y\le\left|x\right|+\left|y\right|\) và \(-x-y\le\left|x\right|+\left|y\right|\)
Hay: \(x+y\ge-\left(\left|x\right|+\left|y\right|\right)\)
Do đó: \(-\left(\left|x\right|+\left|y\right|\right)\le x+y\le\left|x\right|+\left|y\right|\)
Vậy: \(\left|x+y\right|\le\left|x\right|+\left|y\right|\)
Dấu "=" xảy ra khi: \(xy\ge0\)
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Bài 3:
Ta có: \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\frac{x+y+z}{a+b+c}=x+y+z\) (vì a + b + c = 1)
Do đó: \(\left(x+y+z\right)^2=\frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac{z^2}{c^2}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=x^2+y^2+z^2\) (vì a2 + b2 + c2 = 1)
Vậy: (x + y + z)2 = x2 + y2 + z2
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Xem thêm câu trả lời
\(\frac{10+\frac{9}{2}+\frac{8}{3}+\frac{7}{4}+ \frac{6}{5}+\frac{5}{6}+\frac{4}{7}+\frac{3}{8}+\frac{2}{9}+\frac{1}{10}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}+\frac{1}{11}}\)
a,A=\(2\frac{1}{2}:\left(\frac{-1}{2}\right)^2-\frac{1}{-3}.\left(\frac{-1}{2}-\frac{4}{3}:\frac{-8}{9}\right)\)
b,B=\(\left(3\frac{10}{99}+4\frac{11}{99}-\frac{58}{299}\right).\left(\frac{1}{2}-\frac{4}{3}-\frac{1}{6}\right)\)
Trả lời
B=(3 10/99+4 11/99-58/299).(1/2-4/3-1/6)
=(......................................).0
=0.
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Cho A=\(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{70}CMR:\frac{4}{3}< A< \frac{5}{2}\)
Bạn vô câu hỏi tương tự để tham khảo nha!!!
\(A=\left(6:\frac{3}{5}-1\frac{1}{6}x\frac{6}{7}\right):\left(4\frac{1}{5}x\frac{10}{11}+5\frac{2}{11}\right)\)\(B=\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{4}\right)x.......x\left(1-\frac{1}{2015}\right)x\left(1-\frac{1}{2016}\right)\)
\(C=5\frac{9}{10}:\frac{3}{2}-\left(2\frac{1}{3}x4\frac{1}{2}-2x2\frac{1}{3}\right):\frac{7}{4}\)
\(\frac{10+\frac{9}{2}+\frac{8}{3}+\frac{7}{4}+ \frac{6}{5}+\frac{5}{6}+\frac{4}{7}+\frac{3}{8}+\frac{2}{9}+\frac{1}{10}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}+\frac{1}{11}}\)