Thực hiện phép tính: \(\frac{1994.1993-2}{1992+1992.1994}\)
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1, a) ( 42 - 69 + 17 ) - ( 42 + 17 )
= 42 - 69 + 17 - 42 - 17
= - 69
a)
\(\begin{array}{l}M = \frac{1}{2} + \frac{2}{3} + \left( { - \frac{1}{2}} \right) + \frac{1}{3}\\ = \frac{3}{6} + \frac{4}{6} + \left( {\frac{{ - 3}}{6}} \right) + \frac{2}{6}\\ = \frac{{3 + 4 + \left( { - 3} \right) + 2}}{6}\\ = \frac{6}{6} = 1\end{array}\)
b)
\(\begin{array}{l}M = \frac{1}{2} + \frac{2}{3} + \left( { - \frac{1}{2}} \right) + \frac{1}{3}\\ = \left[ {\frac{1}{2} + \left( {\frac{{ - 1}}{2}} \right)} \right] + \left[ {\frac{2}{3} + \frac{1}{3}} \right]\\ = 0 + 1 = 1\end{array}\)
\(\dfrac{xy}{x-y}-\dfrac{2x^2}{y-2x}\)
\(=\dfrac{xy}{x-y}+\dfrac{2x^2}{2x-y}\)
\(=\dfrac{xy\left(2x-y\right)+2x^2\left(x-y\right)}{\left(x-y\right)\left(2x-y\right)}\)
\(=\dfrac{2x^2y-xy^2+2x^3-2x^2y}{\left(x-y\right)\left(2x-y\right)}\)
\(=\dfrac{2x^3-xy^2}{\left(x-y\right)\left(2x-y\right)}=\dfrac{x\left(2x^2-y^2\right)}{\left(x-y\right)\left(2x-y\right)}\)
\(\begin{array}{l} - \left( { - \frac{3}{4}} \right) - \left( {\frac{2}{3} + \frac{1}{4}} \right)\\ = \frac{3}{4} - \frac{2}{3} - \frac{1}{4}\\ = \left( {\frac{3}{4} - \frac{1}{4}} \right) - \frac{2}{3}\\ = \frac{2}{4} - \frac{2}{3}\\= \frac{1}{2} - \frac{2}{3}\\ = \frac{3}{6} - \frac{4}{6}\\ = \frac{{ - 1}}{6}\end{array}\)
a)\(\frac{{ - 2}}{5} + \frac{3}{7} = \frac{{ - 14}}{{35}} + \frac{{15}}{{35}} = \frac{1}{{35}}\)
b)\(0,123 - 0,234 = - \left( {0,234 - 0,123} \right) = - 0,111.\)
a) Mẫu số chung = BCNN(11, 7) = 77
Thừa số phụ: 77: 11= 7; 77:7 = 11.
Ta có:
\(\begin{array}{l}\frac{7}{{11}} + \frac{5}{7} = \frac{{7.7}}{{11.7}} + \frac{{5.11}}{{7.11}}\\ = \frac{{49}}{{77}} + \frac{{55}}{{77}} = \frac{{104}}{{77}}\end{array}\).
b) Mẫu số chung = BCNN(20, 15)= 60
Thừa số phụ: 60:20 = 3; 60:15 = 4
Ta có:
\(\begin{array}{l}\frac{7}{{20}} - \frac{2}{{15}} = \frac{{7.3}}{{20.3}} - \frac{{2.4}}{{15.4}}\\ = \frac{{21}}{{60}} - \frac{8}{{60}} = \frac{{13}}{{60}}\end{array}\).
\(\dfrac{x+1}{x-5}+\dfrac{x-18}{x-5}+\dfrac{x+2}{x-5}\)
\(=\dfrac{x+1+x-18+x+2}{x-5}\)
\(=\dfrac{3x-15}{x-5}\)
\(=\dfrac{3\left(x-5\right)}{x-5}\)
\(=\dfrac{3}{1}\)
\(=3\)
\(\dfrac{x+1}{x-5}+\dfrac{x-18}{x-5}+\dfrac{x+2}{x-5}\\ =\dfrac{x+1+x-18+x+2}{x-5}\\ =\dfrac{\left(x+x+x\right)+\left(1-18+2\right)}{x-5}\\ =\dfrac{3x-15}{x-5}=\dfrac{3\left(x-5\right)}{x-5}=3\)
\(\frac{1994.1993-2}{1992+1992.1994}=\frac{1994.1992+1994-2}{1992+1992.1994}=\frac{1994.1992+1992}{1992+1992.1994}=1\)
\(\frac{1994.1993-2}{1992+1991.1994}\)
= \(\frac{1994.1992-1994-2}{1992-1992.1994}\)
= \(\frac{1994.1992-1992}{1992-1992.1994}\)
= \(1\)