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NH
21 tháng 8 2017
a) \(x+\frac{3}{2}=\frac{9}{2}\)
\(\Rightarrow x=\frac{9}{2}-\frac{3}{2}=\frac{6}{2}=3\)
b) \(\frac{1}{3}x+\frac{3}{4}x-75\%=-5\frac{1}{4}\)
\(\Leftrightarrow\frac{4x}{12}+\frac{9x}{12}-\frac{9}{12}=-\frac{63}{12}\)
\(\Leftrightarrow\frac{13x-9}{12}=\frac{-63}{12}\)
\(\Rightarrow13x-9=-63\)
\(13x=-54\Rightarrow x=\frac{-54}{13}\)
21 tháng 8 2017
a) \(x+\frac{3}{2}=\frac{9}{2}\)
\(\Rightarrow x=\frac{9}{2}-\frac{3}{2}\)
\(\Rightarrow x=\frac{6}{2}=3\)
29 tháng 5 2017
Ta có :
\(A=\dfrac{10^{50}+2}{10^{50}-1}=\dfrac{10^{50}-1+3}{10^{50}-1}=\dfrac{10^{50}-1}{10^{50}-1}+\dfrac{3}{10^{50}-1}=1+\dfrac{3}{10^{50}-1}\)
\(B=\dfrac{10^{50}}{10^{50}-3}=\dfrac{10^{50}-3+3}{10^{50}-3}=\dfrac{10^{50}-3}{10^{50}-3}+\dfrac{3}{10^{50}-3}=1+\dfrac{3}{10^{50}-3}\)
Vì \(1+\dfrac{3}{10^{50}-1}< 1+\dfrac{3}{10^{50}-3}\Rightarrow A< B\)
TL
2
22 tháng 5 2017
Bai 2: b)
\(B=\dfrac{\left(\dfrac{8^2}{128}+\dfrac{3}{4}\right):1\dfrac{3}{16}}{1\dfrac{11}{19}}\\ =\dfrac{\left(\dfrac{\left(2^3\right)^2}{2^7}+\dfrac{3}{4}\right):\dfrac{19}{16}}{\dfrac{30}{19}}=\dfrac{\left(\dfrac{1}{2}+\dfrac{3}{4}\right).\dfrac{16}{19}}{\dfrac{30}{19}}\\ =\dfrac{\dfrac{5}{4}.\dfrac{16}{19}}{\dfrac{30}{19}}=\dfrac{\dfrac{20}{19}}{\dfrac{30}{19}}=\dfrac{2}{3}\)
22 tháng 5 2017
a) \(A=\left(-1,5\right)^2.2\dfrac{2}{3}.\dfrac{1}{6}+\left(\dfrac{4}{7}-\dfrac{2}{5}\right):1\dfrac{1}{35}\\ =2,25.\dfrac{8}{3}.\dfrac{1}{6}+\dfrac{6}{35}:\dfrac{36}{35}\\ =\dfrac{9}{4}.\dfrac{8}{3}.\dfrac{1}{6}+\dfrac{6}{35}.\dfrac{35}{36}\\ =1+\dfrac{1}{6}=1\dfrac{1}{6}\)
chung minh A<2
26 tháng 4 2017
Dễ thấy:
\(\dfrac{1}{1^2}=\dfrac{1}{1.1}=\dfrac{1}{1}=1\)
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)
\(.....................\)
\(\dfrac{1}{50^2}=\dfrac{1}{50.50}< \dfrac{1}{49.50}\)
Cộng các vế trên với nhau ta được:
\(A< 1+\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)\)
\(\Rightarrow A< 1+\) \(\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
\(\Rightarrow A< 1+\left(1-\dfrac{1}{50}\right)\)
\(\Rightarrow A< 1+1-\dfrac{1}{50}=2-\dfrac{1}{50}\)
Mà \(2-\dfrac{1}{50}< 2\Leftrightarrow A< 2\)
Vậy \(A=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}< 2\) (Đpcm)
21 tháng 6 2016
\(a.x+9=2\frac{1}{3}\)
\(x=2\frac{1}{3}-9\)
\(x=-\frac{20}{3}\)
\(b.\frac{-5}{2}.x=\frac{1}{10}\)
\(x=\frac{1}{10}:\frac{-5}{2}\)
\(x=-\frac{1}{25}\)
21 tháng 6 2016
a)
<=> x=\(\frac{7}{3}-9=-\frac{20}{3}\)
b)
<=> x=\(\frac{1}{10}:\frac{-5}{2}=-\frac{2}{25}\)
8 tháng 11 2017
3S= 3/2.5+3/5.8+....+3/17.20
= (5-2)/2.5+(8-5)/5.8+....+(20-17)/17.20
= 1/2 - 1/5 + 1/5 - 1/8 + ...... + 1/17 - 1/20 = 1/2 - 1/20 = 9/20
=> S = 3/20
8 tháng 11 2017
\(S=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{17.20}\)
\(S=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\right)\)
\(S=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(S=\frac{1}{3}.\frac{9}{20}\)
\(S=\frac{3}{20}\)
2 tháng 5 2017
Ta có :
\(A=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+.........+\dfrac{1}{50^2}\)
Ta thấy :
\(\dfrac{1}{1^2}=1\)
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)
\(..............\)
\(\dfrac{1}{50^2}=\dfrac{1}{50.50}< \dfrac{1}{49.50}\)
\(\Rightarrow A< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+..........+\dfrac{1}{49.50}\)
\(A< 1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...........+\dfrac{1}{49}-\dfrac{1}{50}\)
\(A< 1+1-\dfrac{1}{50}\)
\(A< 2-\dfrac{1}{50}\)
\(\Rightarrow A< 2\rightarrowđpcm\)
21 tháng 6 2016
\(S=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{17.20}\)
\(S=3.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\right)\)
\(S=3.\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(S=3.\frac{9}{20}=\frac{27}{20}\)
NB
1
DN
10 tháng 1 2018
S=1/2.5+1/5.8+1/8.11+...+1/17.20
=1/3.(1/2-1/5/+1/5-1/8+1/8-1/11+...+1/17-1/20)
=1/3.(1/2-1/20)
=1/3.9/20
=3/20
a) \(x+\dfrac{3}{2}=\dfrac{9}{2}\)
\(x=\dfrac{9}{2}-\dfrac{3}{2}\)
\(x=3\)
Vậy x = 3 là giá trị cần tìm
b) \(\dfrac{1}{3}x+\dfrac{3}{4}x-75\%=-5\dfrac{1}{4}\)
\(x\left(\dfrac{1}{3}+\dfrac{3}{4}\right)-\dfrac{75}{100}=\dfrac{-19}{4}\)
\(x\left(\dfrac{4}{12}+\dfrac{9}{12}\right)-\dfrac{3}{4}=\dfrac{-19}{4}\)
\(x.\dfrac{13}{12}-\dfrac{3}{4}=\dfrac{-19}{4}\)
\(x.\dfrac{13}{12}=\dfrac{-19}{4}+\dfrac{3}{4}\)
\(x.\dfrac{13}{12}=-4\)
\(x=-4:\dfrac{13}{12}\)
\(x=-\dfrac{48}{13}\)
Vậy..................................
a) x + \(\dfrac{3}{2}\)= \(\dfrac{9}{2}\)
=> x= \(\dfrac{9}{2}\)- \(\dfrac{3}{2}\)
=> x=3
Vậy x=3
b) \(\dfrac{1}{3}\)x + \(\dfrac{3}{4}\)x - 75%= \(-5\dfrac{1}{4}\)
=> x.(\(\dfrac{1}{3}\)+ \(\dfrac{3}{4}\)) - \(\dfrac{3}{4}\)= \(\dfrac{-21}{4}\)
=> x. \(\dfrac{13}{12}\)- \(\dfrac{3}{4}\)=\(\dfrac{-21}{4}\)
=> x. \(\dfrac{13}{12}\)= \(\dfrac{-21}{4}\)+ \(\dfrac{3}{4}\)
=> x. \(\dfrac{13}{12}\)= \(\dfrac{-9}{2}\)
=> x= \(\dfrac{-9}{2}\): \(\dfrac{13}{12}\)= \(\dfrac{-9}{2}\). \(\dfrac{12}{13}\)
=> x= \(\dfrac{-54}{13}\)
Vậy x= \(\dfrac{-54}{13}\)