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b: \(1024:2^5+140:\left(38+2^5\right)+2\cdot3^2-7^{23}:7^{21}\)
\(=1024:32+140:\left(38+32\right)+2\cdot9-7^2\)
=32+140:70+18-49
=50-49+2
=1+2
=3
c: \(100:\left\lbrace300:\left\lbrack450-\left(4\cdot5^3-2^3\cdot25\right)\right\rbrack\right\rbrace\)
\(=100:\left\lbrace300:\left\lbrack450-\left(4\cdot125-8\cdot25\right)\right\rbrack\right\rbrace\)
\(=100:\left\lbrace300:\left\lbrack450-\left(500-200\right)\right\rbrack\right\rbrace\)
=100:{300:[450-300]}
=100:{300:150}
=100:2
=50
\(\dfrac{4}{5}\)+\(\dfrac{-5}{4}\)=\(\dfrac{0}{4}\)
\(\dfrac{-1}{3}\)+\(\dfrac{2}{5}-\dfrac{5}{6}\)=\(\dfrac{-10}{30}+\dfrac{12}{30}-\dfrac{25}{30}\)=\(\dfrac{-23}{30}\)
\(\dfrac{2}{3}-\dfrac{5}{7}.\dfrac{14}{25}\)=\(\dfrac{2}{3}-\dfrac{2}{5}\)=\(\dfrac{10}{15}-\dfrac{6}{15}\)=\(\dfrac{4}{15}\)
a: \(=\dfrac{16-25}{20}=\dfrac{-9}{20}\)
b: \(=\dfrac{-10}{30}+\dfrac{12}{30}-\dfrac{25}{30}=-\dfrac{23}{30}\)
c: \(=\dfrac{2}{3}-\dfrac{1}{5}\cdot2=\dfrac{2}{3}-\dfrac{2}{5}=\dfrac{10-6}{15}=\dfrac{4}{15}\)
d: \(=\dfrac{12-25}{30}\cdot\dfrac{15}{8}-1=\dfrac{-13}{16}-1=\dfrac{-29}{16}\)
e: \(=\dfrac{-3}{13}-\dfrac{10}{13}-\dfrac{7}{17}+\dfrac{24}{17}-\dfrac{5}{19}-\dfrac{14}{19}=-1\)
1: \(=-145\cdot13+145\cdot57+57\cdot10-57\cdot145=-1315\)
2: \(=17\cdot15-17\cdot16+16\cdot17-16\cdot20=255-320=-65\)
3: \(=-38\cdot25+38\cdot4-25\cdot4+25\cdot38=13\cdot4=52\)
4: \(=23\cdot145-23\cdot17-145\cdot23+145\cdot6=479\)
5: \(=24\cdot15-24\cdot4+4\cdot24-4\cdot15=360-60=300\)
6: \(=199\left(15-17+17-5\right)=199\cdot10=1990\)
7: \(=-39\cdot5+39\cdot99+99\cdot10-99\cdot39=795\)
1: =−145⋅13+145⋅57+57⋅10−57⋅145=−1315=−145⋅13+145⋅57+57⋅10−57⋅145=−1315
2: =17⋅15−17⋅16+16⋅17−16⋅20=255−320=−65=17⋅15−17⋅16+16⋅17−16⋅20=255−320=−65
3: =−38⋅25+38⋅4−25⋅4+25⋅38=13⋅4=52=−38⋅25+38⋅4−25⋅4+25⋅38=13⋅4=52
4: =23⋅145−23⋅17−145⋅23+145⋅6=479=23⋅145−23⋅17−145⋅23+145⋅6=479
5: =24⋅15−24⋅4+4⋅24−4⋅15=360−60=300=24⋅15−24⋅4+4⋅24−4⋅15=360−60=300
6: =199(15−17+17−5)=199⋅10=1990=199(15−17+17−5)=199⋅10=1990
7: =−39⋅5+39⋅99+99⋅10−99⋅39=795
\(\dfrac{5}{3}\cdot\dfrac{7}{25}+\dfrac{5}{3}\cdot\dfrac{21}{25}-\dfrac{5}{3}\cdot\dfrac{7}{25}\)
\(=\dfrac{5}{3}\cdot\left(\dfrac{7.}{25}+\dfrac{21}{25}-\dfrac{7}{25}\right)\)
\(=\dfrac{5}{3}\cdot\dfrac{21}{25}=\dfrac{7}{5}\)
b) \(250\%+19\dfrac{3}{11}\cdot\dfrac{7}{26}-6\dfrac{3}{11}\cdot\dfrac{7}{26}\)
\(=\dfrac{5}{2}+\dfrac{212}{11}\cdot\dfrac{7}{26}-\dfrac{69}{11}\cdot\dfrac{7}{26}\)
\(=\dfrac{7}{26}\cdot\left(\dfrac{212}{11}-\dfrac{69}{11}\right)+\dfrac{5}{2}\)
\(=\dfrac{7}{26}\cdot13+\dfrac{5}{2}\)
\(=\dfrac{7}{2}+\dfrac{5}{2}\)
\(=\dfrac{12}{2}=6\)