\(a,\frac{1}{2}+\frac{1}{3}\)
\(b,\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\)
\(c,\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\)
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a: \(\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}}{1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}:\dfrac{13+\dfrac{13}{2}+\dfrac{13}{3}+\dfrac{13}{4}}{17-\dfrac{17}{2}+\dfrac{17}{3}-\dfrac{17}{4}}\)
\(=\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}}{1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}\cdot\dfrac{17\left(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}{13\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\right)}=\dfrac{17}{13}\)
b: \(\dfrac{0.125-\dfrac{1}{5}+\dfrac{1}{7}}{0.375-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0.2}{\dfrac{3}{4}+0.5-\dfrac{3}{10}}\)
\(=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{\dfrac{3}{8}-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{4}+\dfrac{3}{6}-\dfrac{3}{10}}\)
\(=\dfrac{1}{3}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{2}\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}\right)}=\dfrac{1}{3}+\dfrac{2}{3}=1\)
d,
\(|x-\frac{1}{3}|=\frac{5}{6}\Rightarrow \left[\begin{matrix} x-\frac{1}{3}=\frac{5}{6}\\ x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{7}{6}\\ x=\frac{-1}{2}\end{matrix}\right.\)
e,
\(\frac{3}{4}-2|2x-\frac{2}{3}|=2\)
\(\Leftrightarrow 2|2x-\frac{2}{3}|=\frac{3}{4}-2=\frac{-5}{4}\)
\(\Leftrightarrow |2x-\frac{2}{3}|=-\frac{5}{8}<0\) (vô lý vì trị tuyệt đối của 1 số luôn không âm)
Vậy không tồn tại $x$ thỏa mãn đề bài.
f,
\(\frac{2x-1}{2}=\frac{5+3x}{3}\Leftrightarrow 3(2x-1)=2(5+3x)\)
\(\Leftrightarrow 6x-3=10+6x\)
\(\Leftrightarrow 13=0\) (vô lý)
Vậy không tồn tại $x$ thỏa mãn đề bài.
a,
$0-|x+1|=5$
$|x+1|=0-5=-5<0$ (vô lý do trị tuyệt đối của một số luôn không âm)
Do đó không tồn tại $x$ thỏa mãn điều kiện đề.
b,
\(2-|\frac{3}{4}-x|=\frac{7}{12}\)
\(|\frac{3}{4}-x|=2-\frac{7}{12}=\frac{17}{12}\)
\(\Rightarrow \left[\begin{matrix} \frac{3}{4}-x=\frac{17}{12}\\ \frac{3}{4}-x=\frac{-17}{12}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{3}\\ x=\frac{13}{6}\end{matrix}\right.\)
c,
\(2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\)
\(2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}\)
\(|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}\)
\(\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.\)
a, \(A=\frac{2}{5}+\frac{-1}{6}-\frac{3}{4}-\frac{-2}{3}\)
\(A=\left(\frac{2}{5}-\frac{3}{4}\right)+\left(\frac{-1}{6}-\frac{-2}{3}\right)\)
\(A=\left(\frac{8}{20}-\frac{15}{20}\right)+\left(\frac{-3}{18}-\frac{-12}{18}\right)\)
\(A=\frac{-7}{20}+\frac{1}{2}\)
\(\Rightarrow A=\frac{-7}{20}+\frac{10}{20}=\frac{3}{20}\)
b, \(B=\frac{7}{10}-\frac{-3}{4}+\frac{-5}{6}-\frac{1}{5}+\frac{-2}{3}\)
\(B=\left(\frac{7}{10}-\frac{1}{5}\right)+\left(\frac{-5}{6}+\frac{-2}{3}\right)-\frac{-3}{4}\)
\(B=\left(\frac{7}{10}-\frac{2}{10}\right)+\left(\frac{-5}{6}+\frac{-4}{6}\right)-\frac{-3}{4}\)
\(B=\frac{1}{2}+\frac{-3}{2}-\frac{-3}{4}\)
\(B=\frac{2}{4}+\frac{-6}{4}-\frac{-3}{4}\)
\(\Rightarrow B=\frac{2+-6+3}{4}=\frac{-1}{4}\)
c, \(C=\frac{\left(\frac{1}{2}-0,75\right)\times\left(0,2-\frac{2}{5}\right)}{\frac{5}{9}-1\frac{1}{12}}\)
\(C=\frac{\left(\frac{1}{2}-\frac{3}{4}\right)\times\left(\frac{1}{5}-\frac{2}{5}\right)}{\frac{5}{9}-\frac{1\times12+1}{12}}\)
\(C=\frac{\left(\frac{2}{4}-\frac{3}{4}\right)\times\left(\frac{-1}{5}\right)}{\frac{5}{9}-\frac{13}{12}}\)
\(C=\frac{\left(\frac{-1}{4}\right)\times\left(\frac{-1}{5}\right)}{\frac{60}{108}-\frac{117}{108}}\)
\(C=\frac{\frac{1}{20}}{\frac{-19}{36}}=\frac{1}{20}\div\frac{-19}{36}=\frac{1}{20}\times\frac{36}{-19}\)
\(\Rightarrow C=\frac{36}{-380}=\frac{-9}{95}\)
d, \(D=\frac{\frac{2}{3}+\frac{2}{7}-\frac{1}{4}}{-1-\frac{3}{7}+\frac{3}{28}}\)
\(D=\frac{\frac{56}{84}+\frac{24}{84}-\frac{21}{84}}{\frac{-10}{7}+\frac{3}{28}}\)
\(D=\frac{\frac{59}{84}}{\frac{-40}{28}+\frac{2}{28}}=\frac{59}{84}\div\frac{-37}{28}=\frac{59}{84}\times\frac{28}{-37}\)
\(\Rightarrow D=\frac{1652}{-3108}=\frac{-59}{111}\)
-\(\frac{-2}{3}+\frac{3}{4}-\frac{-1}{6}+\frac{-2}{5}=-\frac{4}{6}+\frac{1}{6}+\frac{3}{4}-\frac{2}{5}=-\frac{2}{4}+\frac{3}{4}-\frac{2}{5}=\frac{1}{4}-\frac{2}{5}=-\frac{3}{20}\)
= \(-\frac{3}{20}\)
Đăt A = \(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+......+\frac{1}{7^{100}}\)
\(\Rightarrow7A=1+\frac{1}{7}+\frac{1}{7^2}+.....+\frac{1}{7^{100}}\)
\(\Rightarrow7A-A=1-\frac{1}{7^{100}}\)
\(\Rightarrow6A=1-\frac{1}{7^{100}}\)
\(\Rightarrow A=\frac{1-\frac{1}{7^{100}}}{6}\)
a)\(\frac{5}{6}\)
b)\(\frac{13}{12}\)
c)\(\frac{77}{60}\)
1/2 + 1/3 = 2/6 + 3/6 = 5/6
1/2 + 1/3 + 1/4 = 5/6 + 1/4 = 20/24 + 6/24 = 13/12
1/2 + 1/3 + 1/4 + 1/5 = 13/12 + 1/5 =65/60 + 12/60 = 77/60