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LV
22 tháng 10 2017
\(a,\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{5}{6}+\frac{1}{6}=1.\)
\(b,\frac{1}{2}-\frac{1}{3}-\frac{1}{6}=\frac{1}{6}-\frac{1}{6}=0.\)
\(c,\frac{1}{2}x\frac{2}{3}x\frac{3}{4}=\frac{1}{3}x\frac{3}{4}=\frac{1}{4}.\)
\(d,\frac{1}{2}:\frac{2}{3}:\frac{3}{4}=\frac{3}{4}:\frac{3}{4}=1.\)
N
1 tháng 9 2017
a) \(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{3}{6}+\frac{2}{6}+\frac{1}{6}=1\)
b) \(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}=\frac{3}{6}-\frac{2}{6}-\frac{1}{6}=0\)
c) \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}=\frac{1}{4}\)
d) \(\frac{1}{2}:\frac{2}{3}:\frac{3}{4}=\frac{1}{2}.\frac{3}{2}.\frac{4}{3}=1\)
Chú ý : Dấu \(.\)là dấu x nha
JB
1
N
14 tháng 6 2018
Câu 1 : \(\frac{10}{3}\cdot x-\frac{1}{2}=\frac{1}{10}\)
\(\Rightarrow\frac{10}{3}\cdot x=\frac{3}{5}\Rightarrow x=\frac{9}{50}\)
Câu 2 : \(\frac{1}{3}\cdot x+\frac{1}{3}\cdot x=\frac{1}{8}\Rightarrow x\cdot\left(\frac{1}{3}+\frac{1}{3}\right)=\frac{1}{8}\)
\(x\cdot\frac{2}{3}=\frac{1}{8}\Rightarrow x=\frac{3}{16}\)
Câu 3 : \(\left[3x-1\right]\left[\frac{1}{2}x-\frac{2}{5}\right]=0\)
\(\Rightarrow3x-1=0\)hoặc \(\frac{1}{2}x-\frac{2}{5}=0\)
( vô lí ) x = 4/5
9 tháng 7 2018
\(a)\) \(S=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}\)
\(S=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)
\(3S=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\)
\(3S-S=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\right)\)
\(2S=3+\frac{1}{3^7}\)
\(2S=\frac{3^8+1}{3^7}\)
\(S=\frac{3^8+1}{3^7}.\frac{1}{2}\)
\(S=\frac{3^8+1}{2.3^7}\)
Vậy \(S=\frac{3^8+1}{2.3^7}\)
Chúc bạn học tốt ~
23 tháng 4 2016
= 1/1x2 + 1/2x3 + 1/3x4 ...... +1/9x10
= 1-1/2+1/2-1/3+1/3-1/4+........+1/9-1/10
=1-1/10=9/10
TN
23 tháng 4 2016
đặt A=1/1 x 1/2 + 1/2 x 1/3 + 1/3 + 1/4 + .......... + 1/9 x 1/10
\(A=\frac{1}{1}\cdot\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{3}+...+\frac{1}{9}\cdot\frac{1}{10}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
đặt B=2/1 x 2 + 2/2 x 3 + 2/3 x4 + .............. + 2/98 x 99 + 2/99 x 100
\(B=2\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=2\left(1-\frac{1}{100}\right)\)
\(=2\times\frac{99}{100}\)
\(=\frac{99}{50}\)
HT
28 tháng 9 2020
\(a,\frac{1}{3}+\frac{1}{6}+x=\frac{3}{4}\) \(b,\frac{1}{2}+\frac{1}{3}-x=\frac{2}{3}\) \(c,\frac{2}{3}\cdot\frac{3}{4}\cdot x=\frac{2}{5}\)
\(\left(\frac{1}{3}+\frac{1}{6}\right)+x=\frac{3}{4}\) \(\left(\frac{1}{2}+\frac{1}{3}\right)-x=\frac{2}{3}\) \(\left(\frac{2}{3}\cdot\frac{3}{4}\right)\cdot x=\frac{2}{5}\)
\(\frac{1}{2}+x=\frac{3}{4}\) \(\frac{5}{6}-x=\frac{2}{3}\) \(\frac{1}{2}\cdot x=\frac{2}{5}\)
\(x=\frac{3}{4}-\frac{1}{2}\) \(x=\frac{5}{6}-\frac{2}{3}\) \(x=\frac{2}{5}:\frac{1}{2}\)
\(x=\frac{1}{4}\) \(x=\frac{1}{6}\) \(x=\frac{4}{5}\)
\(d,x:\left(\frac{1}{4}:\frac{1}{2}\right)=\frac{2}{3}\) \(e,\frac{1}{3}:\frac{1}{2}:x=\frac{3}{4}\)
\(x:\frac{1}{2}=\frac{2}{3}\) \(\frac{1}{3}:\frac{1}{2}=\frac{3}{4}\cdot x\)
\(x=\frac{2}{3}\cdot\frac{1}{2}\) \(\frac{2}{3}=\frac{3}{4}\cdot x\)
\(x=\frac{1}{3}\) \(x=\frac{2}{3}:\frac{3}{4}\)
\(x=\frac{8}{9}\)
28 tháng 9 2020
a) 1/3 + 1/6 +x = 3/4 b) 1/2 + 1/3 - x = 2/3
5/6 - x = 2/3
x =5/6 -2/3
x = 1/6
1/2+ x =3/4
x = 3/4 -1/2
x = 1/4
\(\left(x-\dfrac{2}{3}\right):1\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{3}{2}\)
\(\left(x-\dfrac{2}{3}\right):\dfrac{4}{3}+\dfrac{1}{2}=\dfrac{3}{2}\)
\(\left(x-\dfrac{2}{3}\right):\dfrac{4}{3}=\dfrac{3}{2}-\dfrac{1}{2}\)
\(\left(x-\dfrac{2}{3}\right):\dfrac{4}{3}=1\)
\(x-\dfrac{2}{3}=1\times\dfrac{4}{3}\)
\(x-\dfrac{2}{3}=\dfrac{4}{3}\)
\(x=\dfrac{4}{3}+\dfrac{2}{3}\)
\(x=2\)
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