\(\frac{19}{2}+\frac{6}{8}\cdot\frac{2}{3}:\left(1-\frac{3}{2}\right)\)
\(a.A=[\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}+\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}]+\frac{1890}{2005}+115\)
b.B=\(\left[\frac{1\frac{11}{31}\cdot4\frac{3}{7}-\left(15-6\cdot\frac{1}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(42-5\frac{1}{3}\right)}\cdot\left(-1\frac{19}{93}\right)\right]\cdot\frac{31}{50}\)
Tính
a)\(\left(-\frac{1}{4}\right)^2+\frac{3}{8}\cdot\left(-\frac{1}{6}\right)-\frac{3}{16}:\left(-\frac{1}{2}\right)\)
b)\(-\frac{1}{2}:\left(1-\frac{3}{4}\right)^2-\frac{2}{3}:\frac{9}{8}-\left(\frac{9}{8}\right)^0\)
c)\(4\cdot\left(-\frac{1}{2}\right)^3+2\cdot\left(-\frac{1}{2}\right)^2-3\cdot\left(-\frac{1}{2}\right)+2006^0\)
a) \(\frac{\left(-1\right)}{4}^2+\frac{3}{8}.\left(\frac{-1}{6}\right)-\frac{3}{16}:\left(\frac{-1}{2}\right)=\left(\frac{-1}{4}\right)^2+\left(\frac{-3}{68}\right)-\left(\frac{-3}{8}\right)=\left(\frac{1}{16}\right)+\left(\frac{-3}{68}\right)-\left(\frac{-3}{8}\right)=\frac{5}{272}-\left(\frac{-3}{8}\right)=\frac{107}{272}\)
Bài 4 :
a) Tính giá trị của biểu thức :
\(A=\left(\frac{1\frac{11}{31}\cdot4\frac{3}{7}-\left(15-6\frac{1}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-5\frac{1}{3}\right)}\cdot\left(-1\frac{14}{93}\right)\right)\cdot\frac{31}{50}\)
b) Chứng tỏ rằng : \(B=1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{3^2}-...-\frac{1}{2004^2}>\frac{1}{2004}\)
CÂU 1:
\(a)A=\frac{\left(\frac{2}{3}\right)^3\cdot\left(-\frac{3}{4}\right)^2\cdot\left(-1\right)^{2019}}{36\cdot\frac{1}{5}\cdot\left(\frac{2}{5}\right)^2\cdot\left(-\frac{5}{12}\right)^3}\)
\(b)B=\frac{1}{19}+\frac{9}{19\cdot29}+\frac{9}{29\cdot39}+\frac{9}{39\cdot49}+....+\frac{9}{2009.2019}\)
HELP ME, AI ĐÚNG MÌNH TICK CHO
Tính nhanh giá trị biểu thức sau:
a) \(-\frac{9}{10}\cdot\frac{5}{14}+\frac{1}{10}\cdot\left(-\frac{9}{2}\right)+\frac{1}{7}\cdot\left(-\frac{9}{10}\right)\)
b)\(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{11}\right)\cdot132\)
c)\(-\frac{2}{3}\cdot\left(\frac{8}{9}\cdot\frac{8}{13}-\frac{8}{27}\cdot\frac{3}{13}+\frac{4}{3}\cdot\frac{22}{39}\right)\)
a/ \(\frac{-9}{10}.\frac{5}{14}+\frac{1}{10}.\left(\frac{-9}{2}\right)+\frac{1}{7}.\left(-\frac{9}{10}\right)\)
= \(-\frac{9}{10}.\left(\frac{5}{14}+\frac{1}{7}\right)+\frac{1}{10}.\left(-\frac{9}{2}\right)\)
= \(-\frac{9}{10}.\frac{1}{2}+\frac{1}{10}.\left(-\frac{9}{2}\right)\)
= \(\frac{-9}{20}+\left(-\frac{9}{20}\right)=\frac{-18}{20}=\frac{-9}{10}\)
b/ \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{11}\right).132\)
\(=\left(\frac{1}{2}.132\right)+\left(\frac{1}{3}.132\right)+\left(\frac{1}{4}.132\right)+\left(\frac{1}{6}.132\right)\)\(+\left(\frac{1}{11}.132\right)\)
\(=66+44+33+22+12=177\)
c/ \(-\frac{2}{3}.\left(\frac{8}{9}.\frac{8}{13}-\frac{8}{27}.\frac{8}{13}+\frac{4}{3}.\frac{22}{39}\right)\)
= \(-\frac{2}{3}.\left[\frac{8}{13}\left(\frac{8}{9}-\frac{8}{27}\right)+\frac{88}{117}\right]\)
= \(-\frac{2}{3}.\left(\frac{8}{13}.\frac{16}{27}+\frac{88}{117}\right)\)
= còn lại làm nốt nha! bận ròy
Thực hiện phép tính hợp lí nếu có thể:
a/ \(\frac{6}{7}+\frac{1}{7}\cdot\frac{2}{7}+\frac{1}{7}\cdot\frac{5}{7}\)
b/\(\frac{2}{3}\cdot\frac{5}{7}\cdot\frac{-3}{8}\cdot\frac{11}{5}\)
c/\(11\frac{4}{7}-\left(2\frac{3}{5}+5\frac{4}{7}\right)\)
d/\(\frac{3}{4}-\frac{3}{4}\cdot\left(\frac{2}{3}+1\right)\)
e/\(0.5\cdot1\frac{1}{3}\cdot75\%:\frac{2}{5}+\frac{3}{5}\)
f/\(\frac{6}{7}+\frac{5}{8}:5-\frac{3}{16}\cdot\left(-2\right)^2\)
g/\(1\frac{3}{8}+\left(\frac{-5}{6}+\frac{7}{12}\right):\frac{2}{3}\)
h/\(1\frac{1}{4}\cdot\frac{-3}{2}+50\%\cdot98\)
i/\(\left(2,09:1,1+4,5\right)\cdot\frac{5}{8}+4,32\)
kazuto kirigaya thật là bt làm ko đó ko bt thì nói đi còn bt thì làm đi
trời ơi bài dễ thế này tự làm đi còn hỏi
Đã biết thì đã không hỏi. Đồ kazuto kirigaya xấu tính!
Tính
a)\(5^2-\left(-5\right)^3+\frac{1}{2}-\left(\frac{1}{2}\right)^2\)
b)\(2\frac{3}{4}-\frac{1}{8}\cdot1\frac{1}{2}\)
c)\(\left(\frac{1}{3^5}+\frac{1}{3^6}+\frac{1}{3^7}+\frac{1}{3^8}\right)\cdot3^9\)
d)\(\left(\frac{1}{2^2}-1\right)\cdot\left(\frac{1}{3^2}-1\right)\cdot\cdot\cdot\cdot\left(\frac{1}{100^2}-1\right)\)
\(\left[6\cdot\left(-\frac{1}{3}\right)^2-3\cdot\left(-\frac{1}{3}\right)+1\right]:\left(-\frac{1}{3}-1\right)\)
\(\frac{\left(\frac{2}{3}\right)^3\cdot\left(-\frac{3}{4}\right)^2\cdot\left(-1\right)^{2003}}{\left(\frac{2}{5}\right)^2\cdot\left(-\frac{5}{12}\right)^3}\)
\(\frac{\left(\frac{2}{3}\right)^3\cdot\left(-\frac{3}{4}^2\right)\cdot\left(-1\right)^{2003}}{\left(\frac{2}{5}\right)^2\cdot\left(-\frac{5}{12}\right)^3}\)
\(=\frac{\frac{8}{27}\cdot\frac{9}{16}\cdot\left(-1\right)}{\frac{4}{25}\cdot\left(-\frac{125}{1728}\right)}\)
\(=\frac{-\frac{1}{6}}{-\frac{5}{432}}=-\frac{1}{6}:\left(-\frac{5}{432}\right)=\frac{72}{5}\)
\(\left[6.\left(\frac{-1}{3}\right)^2-3.\left(\frac{-1}{3}\right)+1\right]:\left(\frac{-1}{3}-1\right)\)
\(=\left[6.\frac{1}{9}-\left(-1\right)+1\right]:\frac{-4}{3}\)
\(=\left[\frac{2}{3}-\left(-1\right)+1\right]:\frac{-4}{3}\)
\(=\frac{8}{3}:\frac{-4}{3}=\frac{-24}{12}=-2\)
~ Hok tốt ~
\(\left[6\cdot\left(-\frac{1}{3}\right)^2-3\cdot\left(-\frac{1}{3}\right)+1\right]:\left(-\frac{1}{3}-1\right)\)
\(=\left[6\cdot\left(-\frac{1}{9}\right)+1+1\right]:\left(-\frac{4}{3}\right)\)
\(=\left(-\frac{2}{3}+2\right):\left(-\frac{4}{3}\right)\)
\(=\frac{4}{3}:\left(-\frac{4}{3}\right)=-1\)
Tính giá trị của biểu thức:
a,(32)2-(-23)2-(-52)3
b,\(\left|\frac{-1}{2}\right|^2\cdot\left(-32\right)-\left(-8\right)+\left|\frac{1}{2}\right|^3\)
c,\(2^3+3\cdot\left(\frac{-5}{86}\right)^0\cdot\left(\frac{1}{2}\right)^2\cdot4+\left[\left(-2\right)^2:\frac{1}{2}\right]:8\)
d,\(\left|\frac{5}{7}\cdot\left(-14\right)\right|-\left(\frac{2}{3}\right)^2\cdot\left(-18\right)+6^2\cdot\frac{-1}{18}\)