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[1-(3/4-2/3)] - [1-(5/3-1/4)] - [1-(4/3+3/4)]
\(=\left[1-\left(\frac{3}{4}-\frac{2}{3}\right)\right]-\left[1-\left(\frac{5}{3}-\frac{1}{4}\right)\right]-\left[1-\left(\frac{4}{3}+\frac{3}{4}\right)\right]\)
\(=\left[1-\frac{1}{12}\right]-\left[1-\frac{17}{12}\right]-\left[1-\frac{25}{12}\right]\)
\(=\frac{11}{12}-\left(-\frac{5}{12}\right)-\left(-\frac{13}{12}\right)\)
\(=\frac{11}{12}+\frac{5}{12}+\frac{13}{12}\)
\(=\frac{29}{12}\)
\(a,2010:\left(-5\right)+400-1\\ =-402+400-1\\ =-3\\ b,\dfrac{2}{3}+\dfrac{3}{4}.\left(-\dfrac{4}{9}\right)\\ =\dfrac{2}{3}-\dfrac{1}{3}\\ =\dfrac{1}{3}\\ c,\left(1-\dfrac{2}{3}-\dfrac{1}{4}\right)\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\\ =\dfrac{1}{12}.\left(\dfrac{1}{20}\right)^2\\ =\dfrac{1}{12}.\dfrac{1}{400}\\ =\dfrac{1}{4800}\)
a) \(2010:\left(-5\right)+400-1=-400+400-1=-1\)
b) \(\dfrac{2}{3}+\dfrac{3}{4}\cdot\dfrac{-4}{9}=\dfrac{2}{3}+\dfrac{-1}{3}=\dfrac{1}{3}\)
c) \(\left(1-\dfrac{2}{3}-\dfrac{1}{4}\right)\cdot\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2=\dfrac{1}{12}\cdot\dfrac{1}{400}=\dfrac{1}{4800}\)
Ta sẽ dùng quy tắc đổi dấu :
-( 3/5 + 3/4 ) - ( -3/4 + 2/5 )
= -3/5 - 3/4 + 3/4 - 2/5
= -3/5 - ( -2/5 )
= -1/5
-3/5-3/4+3/4-2/5
=(-3/4+3/4)-3/5+2/5
=0-3/5+2/5
=1/5
nhớ tk nhe mn
b: \(\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
a) Đề thiếu nhé. sửa đề:
\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{195}\)
\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\)
\(=1-\frac{1}{15}\)
\(=\frac{14}{15}\)
\(\left(\frac{4}{77}+\frac{4}{165}+\frac{4}{285}\right):\left(\frac{5}{84}+\frac{3}{180}+\frac{4}{285}\right)\)
\(=\frac{4}{77}+\frac{4}{165}+\frac{4}{285}:\frac{5}{84}+\frac{3}{180}+\frac{4}{285}\)
\(=\frac{12}{133}:\frac{12}{133}\)
\(=1.\)
Chúc bạn học tốt!