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a) ( 1/2-1/3-1/6).(1/2+2/3+3/4+...+2017/2018) + 3/4.x = 9/10
0.(1/2+2/3+3/4+...+2017/2018) + 3/4.x = 9/10
0+3/4.x = 9/10
3/4.x = 9/10
x = 9/10: 3/4
x = 6/5
b) x + ( 3/1.3+3/3.5+...+3/13.15) = 11/5
x + 3/2. ( 1-1/3 + 1/3 - 1/5 + ...+ 1/13 - 1/15) = 11/5
x + 3/2. ( 1-1/15) = 11/5
x + 3/2.14/15 = 11/5
x + 7/5 = 11/5
x = 11/5 - 7/5
x = 4/5
=17/6:(1-2/3)
=17/6:1/3
=17/2
=13/6×9/2-6/7
=39/4-6/7
=249/28
a) \(\left(\frac{5}{2}+\frac{1}{3}\right):\left(1-\frac{2}{3}\right)=\left(\frac{15}{6}+\frac{2}{6}\right):\frac{1}{3}\)
\(=\frac{17}{6}:\frac{1}{3}=\frac{17}{6}\cdot\frac{3}{1}=\frac{17}{2}\cdot\frac{1}{1}=\frac{17}{2}\)
b) \(\left(\frac{5}{2}-\frac{1}{3}\right)\cdot\frac{9}{2}-\frac{6}{7}=\left(\frac{15}{6}-\frac{2}{6}\right)\cdot\frac{9}{2}-\frac{6}{7}\)
\(=\frac{13}{6}\cdot\frac{9}{2}-\frac{6}{7}=\frac{13}{2}\cdot\frac{3}{2}-\frac{6}{7}=\frac{39}{4}-\frac{6}{7}=\frac{273}{28}-\frac{24}{28}=\frac{249}{28}\)
Ta có
\(\frac{1}{2}=\frac{1\times3\times5}{2\times3\times5}=\frac{15}{30}\)
\(\frac{1}{3}=\frac{1\times2\times5}{3\times2\times5}=\frac{10}{30}\)
\(\frac{2}{5}=\frac{2\times2\times3}{5\times2\times3}=\frac{12}{30}\)
Hok tốt !!!!!!!!!!!!!!!!!!!
\(3\dfrac{2}{5}\cdot1\dfrac{4}{7}=\dfrac{17}{5}\cdot\dfrac{11}{7}=\dfrac{187}{35}\)
\(\frac{1}{3}.\frac{1}{x}+\left(\frac{1}{x}+1\right).\frac{2}{5}=\frac{9}{5}\)
\(\Rightarrow\frac{1}{3x}+\left(\frac{1}{x}+\frac{x}{x}\right).\frac{2}{5}=\frac{9}{5}\)
\(\Rightarrow\frac{1}{3x}+\frac{1+x}{x}.\frac{2}{5}=\frac{9}{5}\)
\(\Rightarrow\frac{1}{3x}+\frac{2\left(1+x\right)}{5x}=\frac{9}{5}\)
\(\Rightarrow\frac{1}{3x}+\frac{2+2x}{5x}=\frac{9}{5}\)
\(\frac{5}{15x}+\frac{6+6x}{15x}=\frac{9}{5}\)
\(\Rightarrow\frac{5+6+6x}{15x}=\frac{3.9x}{15x}\)
\(\Rightarrow\frac{11+6x}{15x}=\frac{27x}{15x}\)
\(\Rightarrow11+6x=27x\)
\(\Rightarrow21x=11\)
\(\Rightarrow x=\frac{11}{21}\)
Vậy \(x=\frac{11}{21}\)
_Hok tốt_
đề = \(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{50.51}\)( áp dụng c.thức tính tổng )
= ..........
= 2 .( \(\frac{1}{2}-\frac{1}{51}\))
= dễ
\(D=\frac{5}{1+2+3}+\frac{5}{1+2+3+4}+...+\frac{5}{1+2+...+100}\)
\(\Rightarrow D=5\left(\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+...+100}\right)\)
\(\Rightarrow D=5\left(\frac{1}{\frac{4.3}{2}}+\frac{1}{\frac{5.4}{2}}+...+\frac{1}{\frac{101.100}{2}}\right)\)
\(\Rightarrow D=5\left(\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{100.101}\right)\)
\(\Rightarrow D=10\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(\Rightarrow D=10\left(\frac{1}{3}-\frac{1}{101}\right)\)
\(\Rightarrow D=\frac{10}{3}-\frac{10}{101}=\frac{980}{303}\)
I don't now
or no I don't
..................
sorry