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a, (-8)-(-12-8)=(-8)+12+8= [(-8)+8}]+12= 0+12=12
b, 1765-(312+1765)= 1765-312-1765=(1765-1765)-312=0-312=-312
c. (31-12)-(46+31-12)= 31-12+46-31+12= (31-31)+(12-12)+46=0+0+46=46
d, (5674-97)-5674=5674-97-5674=(5674-5674)-97=0-97=-97
e, (-1075)-(29-1075)=(-1075)+29+1075=[(-1075)+1075]+29=0+29=29
g, (18+29)+(158-18-29)=18+29+158-18-29=(18-18)+(29-29)+158=0+0+158=158
h, (13-135+49)-(13+49)=13-135+49-13-49=(13-13)+(49-49)-135=0+0-149=0-149=-149
a) \(10+11+12+...+99=\dfrac{\left(99+10\right)\left(\dfrac{99-10}{1}+1\right)}{2}=4905\)
b) \(1+6+11+...+51=\dfrac{\left(51+1\right)\left(\dfrac{51-1}{5}+1\right)}{2}=286\)
c) \(\left(1+3+5+...+2017\right)\left(135135.137-135.137137\right)=\left(1+3+5+...+2017\right)\left[1001\left(135,137-135.137\right)\right]=\left(1+3+5+...+2017\right).0=0\)
a)
=10+(11+99)+(12+98)+.....+(54+56)+55
=10+55+110+110+...+110
=10+55+110.(99-11):2
=10+55+110.44
=65+4840=4905
a) \(\left(x+\frac{1}{4}-\frac{1}{3}\right):\left(2+\frac{1}{6}-\frac{1}{4}\right)=\frac{7}{46}\)
\(\left(x-\frac{1}{12}\right):\left(2-\frac{1}{12}\right)=\frac{7}{46}\)
\(\left(x-\frac{1}{12}\right):\frac{23}{12}=\frac{7}{46}\)
\(x-\frac{1}{12}=\frac{7}{46}.\frac{23}{12}\)
\(x-\frac{1}{12}=\frac{7}{24}\)
\(x=\frac{7}{24}+\frac{1}{12}\)
\(x=\frac{3}{8}\)
Vậy \(x=\frac{3}{8}\)
b) \(\frac{13}{15}-\left(\frac{13}{21}+x\right).\frac{7}{12}=\frac{7}{10}\)
\(\frac{13}{15}-\left(\frac{13}{21}+x\right)=\frac{7}{10}:\frac{7}{12}\)
\(\frac{13}{15}-\left(\frac{13}{21}+x\right)=\frac{7}{10}.\frac{12}{7}\)
\(\frac{13}{15}-\left(\frac{13}{21}+x\right)=\frac{6}{5}\)
\(\frac{13}{21}+x=\frac{13}{15}-\frac{6}{5}\)
\(\frac{13}{21}+x=\frac{-1}{3}\)
\(x=\frac{-1}{3}-\frac{13}{21}\)
\(x=\frac{-20}{21}\)
Vậy \(x=\frac{-20}{21}\)
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