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\(A=\dfrac{3}{1^2.2^2}+\dfrac{7}{3^2.4^2}+\dfrac{11}{5^2.6^2}+\dfrac{15}{7^2.8^2}+\dfrac{19}{9^2.10^2}\)
\(A=\dfrac{1+2}{1^2.2^2}+\dfrac{3+4}{3^2.4^2}+\dfrac{5+6}{5^2.6^2}+\dfrac{7+8}{7^2.8^2}+\dfrac{9+10}{9^2.10^2}\)
\(A=\dfrac{1}{1.2^2}+\dfrac{1}{1^2.2}+\dfrac{1}{3.4^2}+\dfrac{1}{3^2.4}+\dfrac{1}{5.6^2}+\dfrac{1}{5^2.6}+...+\dfrac{1}{9^2.10}\)
\(A=\dfrac{1}{4}+\dfrac{1}{2}+\dfrac{1}{48}+\dfrac{1}{36}+\dfrac{1}{180}+\dfrac{1}{150}+....+\dfrac{1}{900}\)
\(\left\{{}\begin{matrix}\dfrac{1}{48}< \dfrac{3}{32}\\\dfrac{1}{36}< \dfrac{1}{32}\\........\\\dfrac{1}{900}< \dfrac{1}{32}\end{matrix}\right.\)
Nên \(A< \dfrac{1}{4}+\dfrac{1}{2}+\dfrac{1}{32}.8=1\)
\(LINH=\dfrac{3}{1^2.2^2}+\dfrac{7}{3^2.4^2}+\dfrac{11}{5^2.6^2}+\dfrac{15}{7^2.8^2}+\dfrac{19}{9^2.10^2}\)
\(LINH=\dfrac{1+2}{1^2.2^2}+\dfrac{3+4}{3^2.4^2}+\dfrac{5+6}{5^2.6^2}+\dfrac{7+8}{7^2.8^2}+\dfrac{9+10}{9^2.10^2}\)
\(LINH=\dfrac{1}{1^2.2^2}+\dfrac{2}{1^2.2^2}+\dfrac{3}{3^2.4^2}+\dfrac{4}{3^2.4^2}+\dfrac{5}{5^2.6^2}+\dfrac{6}{5^2.6^2}+\dfrac{7}{7^2.8^2}+\dfrac{8}{7^2.8^2}+\dfrac{9}{9^2.10^2}+\dfrac{10}{9^2.10^2}\)
\(LINH=\dfrac{1}{1.2^2}+\dfrac{1}{1^2.2}+\dfrac{1}{3.4^2}+\dfrac{1}{3^2.4}+\dfrac{1}{5.6^2}+\dfrac{1}{5^2.6}+\dfrac{1}{7.8^2}+\dfrac{1}{7^2.8}+\dfrac{1}{9.10^2}+\dfrac{1}{9^2.10}\)\(LINH=\dfrac{1}{4}+\dfrac{1}{2}+\dfrac{1}{48}+\dfrac{1}{36}+\dfrac{1}{180}+\dfrac{1}{150}+\dfrac{1}{448}+\dfrac{1}{392}+\dfrac{1}{900}+\dfrac{1}{810}\)Vì:
\(\left\{{}\begin{matrix}\dfrac{1}{48}< \dfrac{1}{32}\\\dfrac{1}{36}< \dfrac{1}{32}\\...............\\\dfrac{1}{810}< \dfrac{1}{32}\end{matrix}\right.\)
Nên:
\(\dfrac{1}{48}+\dfrac{1}{36}+.....+\dfrac{1}{810}< \dfrac{1}{32}+\dfrac{1}{32}+....+\dfrac{1}{32}\)
\(\Rightarrow\dfrac{1}{48}+\dfrac{1}{36}+....+\dfrac{1}{810}< \dfrac{1}{32}.8=\dfrac{1}{4}\)
Nên:
\(LINH=\dfrac{1}{4}+\dfrac{1}{2}+\dfrac{1}{48}+\dfrac{1}{36}+....+\dfrac{1}{810}< \dfrac{1}{4}+\dfrac{1}{2}+\dfrac{1}{4}=1\)
Nên \(LINH< 1\left(đpcm\right)\)
a,\(0,5-\frac{5}{41}+\frac{1}{2}-\frac{36}{41}\)
=\(\frac{1}{2}-\frac{5}{41}+\frac{1}{2}-\frac{36}{41}\)
=\(\left(\frac{1}{2}+\frac{1}{2}\right)-\left(\frac{5}{41}+\frac{36}{41}\right)\)
= \(1-1=0\)
b,\(\left(\frac{-1}{3}\right)^2.\frac{4}{11}+\frac{7}{11}.\left(\frac{1}{3}\right)^2\)
=\(\frac{1}{9}.\frac{4}{11}+\frac{7}{11}.\frac{1}{9}\)
=\(\frac{1}{9}\left(\frac{4}{11}+\frac{7}{11}\right)\)
=\(\frac{1}{9}.1=\frac{1}{9}\)
c,\(\left(\frac{-2}{3}+\frac{3}{7}\right):\frac{4}{5}+\left(\frac{-1}{3}+\frac{4}{7}\right):\frac{4}{5}\)
=\(\left(\frac{-2}{3}+\frac{3}{7}-\frac{1}{3}+\frac{4}{7}\right):\frac{4}{5}\)
=\(\left(-1+1\right).\frac{5}{4}=0.\frac{5}{4}\)=0
A = 13/21.2/11 + 13/21.9/11 + 8/21
= (13/21) + (13/21) + (8/21)
= (13 + 13 + 8)/21
= 34/21
B = (1 - 1/5)(1 - 2/5)(1 - 3/5)...(1 - 9/5)
= (4/5)(3/5)(2/5)(1/5)(0/5)(-1/5)(-2/5)(-3/5)(-4/5)
= 0
C = (1 - 1/2)(1 - 1/3)(1 - 1/4)...(1 - 1/50)
= (1/2)(2/3)(3/4)(4/5)...(49/50)
= 1/50
D = (2^2/1.3) * (3^2/2.4) * (4^2/3.5) * (5^2/4.6) * (6^2/5.7)
= (4/3) * (9/8) * (16/15) * (25/23) * (36/35)
= 0.979
a, (-0.25) * 7.9 * 40 = [ (-0.25) * 40] * 7.9 = -10 * 7.9 = -79
b, ( 3/2 )^3 * 2^3 = ( 3/2*2) ^3 = 3^3 = 27
c, (1.75 / \(\frac{7}{2}\)) * \(\frac{4}{5}\)= ( 7/4 * 7/2 ) *4/5 = \(\frac{49}{10}\)
d, \(\frac{11}{2}\cdot4\frac{5}{3}-2\frac{5}{3}\cdot\frac{11}{2}\)= \(\frac{11}{2}\cdot\left(\frac{17}{3}-\frac{11}{3}\right)\)= \(\frac{11}{2}\cdot\frac{6}{3}\)= \(\frac{11}{2}\cdot2\)=11
\(\dfrac{22}{7}\cdot\dfrac{-2}{11}+\dfrac{3}{2}\cdot\dfrac{4}{9}:\left(-4\right)^2\\ =\dfrac{22}{7}\cdot\dfrac{-2}{11}+\dfrac{3}{2}\cdot\dfrac{4}{9}:16\\ =\dfrac{22}{7}\cdot\dfrac{-2}{11}+\dfrac{3}{2}\cdot\dfrac{4}{9}\cdot\dfrac{1}{16}\\ =\dfrac{-4}{7}+\dfrac{2}{3}\cdot\dfrac{1}{16}\\ =\dfrac{-4}{7}+\dfrac{1}{24}\\ =\dfrac{-96}{168}+\dfrac{7}{168}\\ =\dfrac{91}{168}\)
Bạn làm sai rồi.
\(\dfrac{22}{7}.\dfrac{-2}{11}=\dfrac{-4}{7}\) chứ ko phải bằng \(\dfrac{4}{7}\) .
\(C=...\)
\(=\frac{1}{3^2}+\frac{1}{4^2}-\frac{1}{4^2}+\frac{1}{5^2}-\frac{1}{5^2}+...-\frac{1}{14^2}+\frac{1}{14^2}-\frac{1}{15^2}\)
\(=\frac{1}{3^2}-\frac{1}{15^2}\)
\(=\frac{1}{9}-\frac{1}{225}\)
Do \(\frac{1}{9}-\frac{1}{225}\)<\(\frac{1}{5}\)
\(=>C< \frac{1}{5}\)( ĐPCM )
C = ...
=> C = \(\frac{1}{3^2}-\frac{1}{4^2}+\frac{1}{4^2}-\frac{1}{5^2}+...+\frac{1}{14^2}-\frac{1}{15^2}\)
C = \(\frac{1}{3^2}-\frac{1}{15^2}\)
Ta thấy : \(\frac{1}{3^2}< \frac{1}{5}\Leftrightarrow\frac{1}{3^2}-\frac{1}{15^2}< \frac{1}{5}\)
=> C < \(\frac{1}{5}\)
1/9 hoặc ra số thập phân 0,1111111111