\(\left(3,5-70,84:23+4-3,375.\frac{4}{9}\right):0,78\)
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\(1)\dfrac{179}{50}-\left(\dfrac{59}{30}+\dfrac{3}{5}\right)\)
\(=\dfrac{179}{50}-\left(\dfrac{118}{60}+\dfrac{36}{60}\right)\)
\(=\dfrac{179}{50}-\dfrac{77}{30}\)
\(=\dfrac{537}{150}-\dfrac{385}{150}\)
\(=\dfrac{152}{150}=\dfrac{76}{75}=1\dfrac{1}{75}\)
Câu 1 : Ta có : 179/50 - ( 59/30 + 3/5 ) .
= 179/50 - ( 59/30 + 18/30 ) .
= 179/50 - 77/30 .
= 537/150 - 385/150.
= 76/75 .
Câu 2 : Ta có : ( 3,5 - 70,84 : 23 + 4 ) - 3,375 . 4/9 : 0,78 .
= ( 3,5 - 3,08 + 4 ) - 27/8 . 4/9 : 39/2 .
= 4,42 - 3/2 . 2/39 .
= 221/50 - 1/13 .
= 2873/650 - 50/650 .
= 2823/50 .
= 56,46 .
\(1\times3,5-70,84:23+4-37,375\times\frac{4}{9}.\)
\(=3,5-\frac{77}{25}+4-\frac{299}{18}\)
\(=\left(3,5-\frac{77}{25}\right)+\left(4-\frac{299}{18}\right)\)
\(=\frac{21}{50}+\left(\frac{-277}{18}\right)\)
\(=\frac{-2743}{225}\)
\(\begin{array}{l}a)\left| { - 3,5} \right| = 3,5;\\b)\left| {\frac{{ - 4}}{9}} \right| = \frac{4}{9};\\c)\left| 0 \right| = 0;\\d)\left| {2,0(3)} \right| = 2,0(3)\end{array}\)
Chú ý:
Nếu \(a \ge 0\) thì \(\left| a \right| = a\)
Nếu \(a < 0\) thì \(\left| a \right| = - a\)
\(a,\frac{-8}{15}.\left(-30\right).\frac{15}{-8}.\frac{9}{10}\)
\(=-\left(\frac{8}{15}.\frac{15}{8}\right).\left(30.\frac{9}{10}\right)\)
\(=-1.27
=-27\)
\(b,2\frac{1}{18}.\frac{23}{24}.\frac{9}{37}.\frac{48}{-15}\)
\(=\frac{-37.23.9.48}{18.24.37.15}=\frac{23}{15}\)
c, chịu rồi
a)
\(\begin{array}{l}\frac{{13}}{{23}}.\frac{7}{{11}} + \frac{{10}}{{23}}.\frac{7}{{11}}\\ = \frac{7}{{11}}.\left( {\frac{{13}}{{23}} + \frac{{10}}{{23}}} \right)\\ = \frac{7}{{11}}.\frac{23}{23}\\ = \frac{7}{{11}}.1\\ = \frac{7}{{11}}\end{array}\)
b)
\(\begin{array}{l}\frac{5}{9}.\frac{{23}}{{11}} - \frac{1}{{11}}.\frac{5}{9} + \frac{5}{9}\\ = \frac{5}{9}.\left( {\frac{{23}}{{11}} - \frac{1}{{11}} + 1} \right)\\ = \frac{5}{9}.\left( {2 + 1} \right)\\ = \frac{5}{9}.3 = \frac{5}{3}\end{array}\)
c)
\(\begin{array}{l}\left[ {\left( { - \frac{4}{9} + \frac{3}{5}} \right):\frac{{13}}{{17}}} \right] + \left( {\frac{2}{5} - \frac{5}{9}} \right):\frac{{13}}{{17}}\\ = \left( { - \frac{4}{9} + \frac{3}{5}} \right).\frac{{17}}{{13}} + \left( {\frac{2}{5} - \frac{5}{9}} \right).\frac{{17}}{{13}}\\ = \frac{{17}}{{13}}.\left( { - \frac{4}{9} + \frac{3}{5} + \frac{2}{5} - \frac{5}{9}} \right)\\ = \frac{{17}}{{13}}.\left[ {\left( { - \frac{4}{9} - \frac{5}{9}} \right) + \left( {\frac{3}{5} + \frac{2}{5}} \right)} \right]\\ =\frac{{17}}{{13}}. (\frac{-9}{9}+\frac{5}{5})\\= \frac{{17}}{{13}}.\left( { - 1 + 1} \right)\\ = \frac{{17}}{{13}}.0 = 0\end{array}\)
d)
\(\begin{array}{l}\frac{3}{{16}}:\left( {\frac{3}{{22}} - \frac{3}{{11}}} \right) + \frac{3}{{16}}:\left( {\frac{1}{{10}} - \frac{2}{5}} \right)\\ = \frac{3}{{16}}:\left( {\frac{3}{{22}} - \frac{6}{{22}}} \right) + \frac{3}{{16}}:\left( {\frac{1}{{10}} - \frac{4}{{10}}} \right)\\ = \frac{3}{{16}}:\frac{{ - 3}}{{22}} + \frac{3}{{16}}:\frac{{ - 3}}{{10}}\\ = \frac{3}{{16}}.\frac{{ - 22}}{3} + \frac{3}{{16}}.\frac{{ - 10}}{3}\\ = \frac{3}{{16}}.\left( {\frac{{ - 22}}{3} + \frac{{ - 10}}{3}} \right)\\ = \frac{3}{{16}}.\frac{{ - 32}}{3}\\ = - 2\end{array}\)
\(P=\frac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)\(=\frac{\left(1+4\right)\left(4^2+1\right)\left(6^2+1\right)\left(8^2+1\right)\left(10^2+1\right)...\left(20^2+1\right)\left(\cdot22^2+1\right)}{\left(2^2+1\right)\left(4^2+1\right)\left(6^2+1\right)\left(8^2+1\right)\left(10^2+1\right)\left(12^2+1\right)...\left(22^2+1\right)\left(24^2+1\right)}\)
\(=\frac{1+4}{\left(2^2+1\right)\left(24^2+1\right)}=\frac{5}{5\left(24^2+1\right)}=\frac{1}{24^2+1}=\frac{1}{577}\)
cái bước tách ra bn nhân lại là có kết quả y chang, VD:
\(\left(5^4+4\right)=\left(4^2+1\right)\left(6^2+1\right)=629\)
\(\frac{209}{156}\)