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\(\dfrac{4}{5}.\dfrac{6}{2}-\dfrac{4}{5}.\dfrac{8}{9}-\dfrac{1}{2}=\dfrac{12}{5}-\dfrac{32}{45}-\dfrac{1}{2}=\dfrac{76}{45}-\dfrac{1}{2}=\dfrac{107}{90}\)
\(=\dfrac{4}{5}.\dfrac{6}{2}-\dfrac{4}{5}.\dfrac{8}{9}-\dfrac{1}{2}.1\)
\(=\dfrac{4}{5}.\left(\dfrac{6}{2}-\dfrac{1}{2}\right).1\)
\(=\dfrac{4}{5}.\dfrac{5}{2}.1\)
\(=2.1\)
\(=2\)
Sửa đề : \(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{67.68}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{67}-\frac{1}{68}\)
\(=\frac{1}{4}-\frac{1}{68}=\frac{4}{17}\)
a: \(\Leftrightarrow\dfrac{16}{x}=\dfrac{-4}{5}\)
hay x=-20
b: =>x+7=-10
hay x=-17
\(35x4=\left(25+15\right)x4=100+60\)\(=160\)
\(25x36=25x4x9=100x9=900\)
\(35\times4=140\)
\(25\times36=900\)
\(36\times12=432\)
\(75\times11=825\)
\(67\times101=6767\)
\(-16+24+16-34=\left(24-34\right)+\left(16-16\right)=-10\\ 25+37-48-25-37=\left(25-25\right)+\left(37-37\right)-48=-48\\ 2575+37-2576-29=\left(2575-2576\right)+\left(37-29\right)=-1+6=5\\ 34+35+36+37-14-15-16-17=\left(34-14\right)+\left(35-15\right)+\left(36-16\right)+\left(37-17\right)=0\)
(-16)+24+16-34=24-34=-10
25+37-48-25-37=(25-25)+(37-37)-48=-48
2575+37-2576-29=8-1=7
34+35+36+37-14-15-16-17
=(34-14)+(35-15)+(36-16)+(37-17)
=20+20+20+20=20.4=80
Bài làm :
\(a,35.34+35.38+65.75+65.45\)
\(=35.\left(34+38\right)+65.\left(74+45\right)\)
\(=35.72+65.120\)
\(=2520+780\)
\(=10320\)
\(b,39.8+60.2+21.8\)
\(=\left(39+21\right).8+60.2\)
\(=60.8+60.2\)
\(=60.\left(8+2\right)\)
\(=60.10=600\)
\(c,36.28+36.82+64.69+64.4\)
\(=36.\left(28+82\right)+64.\left(69+4\right)\)
\(=36.110+64.73\)
\(=3960+4672\)
\(=8632\)
\(d,123.1001\)
\(=123123\)
Học tốt
a, \(\dfrac{1}{2}\) - ( - \(\dfrac{1}{3}\) ) + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)
= \(\dfrac{5}{6}\) + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)
= 1 + \(\dfrac{1}{23}\)
= \(\dfrac{24}{23}\)
b, \(\dfrac{11}{24}\) - \(\dfrac{5}{41}\) + \(\dfrac{13}{24}\) + 0,5 - \(\dfrac{36}{41}\)
= (\(\dfrac{11}{24}\) + \(\dfrac{13}{24}\)) - ( \(\dfrac{5}{41}\) + \(\dfrac{36}{41}\)) + 0,5
= 1 - 1 + 0,5
= 0,5
c,\(-\dfrac{1}{12}-\left(\dfrac{1}{6}-\dfrac{1}{4}\right)\)
=\(-\dfrac{1}{12}-\left(-\dfrac{1}{12}\right)\)
=0
d, \(\dfrac{1}{6}-\left[\dfrac{1}{6}-\left(\dfrac{1}{4}+\dfrac{9}{12}\right)\right]\)
= \(\dfrac{1}{6}-\left[\dfrac{1}{6}-1\right]\)
= \(\dfrac{1}{6}-\left(-\dfrac{5}{6}\right)\)
= 1