c) (0,125 + 40% − \(\dfrac{3}{40}\)) : [ 11\(\dfrac{3}{7}\) + 8\(\dfrac{1}{2}\) − ( \(\dfrac{13}{12}\) − 5\(\dfrac{4}{7}\))]
a) \(\dfrac{5}{7}\)+\(\dfrac{3}{4}\).\(\dfrac{-11}{2}\)
b) (\(\dfrac{12}{17}\)+\(\dfrac{19}{7}\)) - (\(\dfrac{-5}{17}\)-\(\dfrac{3}{7}\))
c) (0,125)\(^{12}\).(-8)\(^{12}\)-\(\dfrac{45^3}{15^3}\)
d) \(5\dfrac{2}{7}\).(\(-\dfrac{1}{3}\))-\(2\dfrac{2}{7}\).(\(-\dfrac{1}{3}\))
e) \(\dfrac{9^2.9^3}{3^9}\)
\(a,=\dfrac{5}{7}-\dfrac{33}{8}=-\dfrac{191}{56}\\ b,=\left(\dfrac{12}{17}+\dfrac{5}{17}\right)+\left(\dfrac{19}{7}+\dfrac{3}{7}\right)=1+3=4\\ c,=\left(0,125\cdot8\right)^{12}-\left(\dfrac{45}{15}\right)^3=1-3^3=-26\\ d,=\left(-\dfrac{1}{3}\right)\left(5\dfrac{2}{7}-2\dfrac{2}{7}\right)=-\dfrac{1}{3}\cdot3=-1\\ e,=\dfrac{3^4\cdot3^6}{3^9}=3\)
1, A= \(\dfrac{-3}{4}.\left(0,125-1\dfrac{1}{2}\right):\dfrac{33}{16}-25\%\)
2, B= \(1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+25\%\right):\dfrac{7}{3}\)
3, C= \(\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)
4, D= \(6\dfrac{5}{12}:2\dfrac{5}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\)
\(1,A=-\dfrac{3}{4}.\left(0,125-1\dfrac{1}{2}\right):\dfrac{33}{16}-25\%\)
\(A=-\dfrac{3}{4}.\left(0,125-\dfrac{3}{2}\right):\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=-\dfrac{3}{4}.\left(-\dfrac{11}{8}\right):\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=\dfrac{33}{32}:\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=\dfrac{33}{32}.\dfrac{16}{33}-\dfrac{1}{4}\)
\(A=\dfrac{1}{2}-\dfrac{1}{4}\)
\(A=\dfrac{2}{4}-\dfrac{1}{4}\)
\(A=\dfrac{1}{4}\)
\(D=6\dfrac{5}{12}:2\dfrac{5}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\)
\(D=\dfrac{77}{12}:\dfrac{13}{4}+\dfrac{45}{4}.\dfrac{2}{15}\)
\(D=\dfrac{77}{39}+\dfrac{3}{2}\)
\(D=\dfrac{271}{78}\)
\(C=\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)
\(C=\dfrac{5}{16}:0,125-\left(\dfrac{9}{4}-0,6\right).\dfrac{10}{11}\)
\(C=\dfrac{5}{16}:0,125-\dfrac{33}{20}.\dfrac{10}{11}\)
\(C=\dfrac{5}{2}-\dfrac{3}{2}\)
\(C=1\)
Tính hợp lý:
\(A=\dfrac{7}{12}+\dfrac{5}{12}:6-\dfrac{11}{36}\) \(B=\left(\dfrac{4}{5}+\dfrac{1}{2}\right):\left(\dfrac{3}{13}-\dfrac{8}{13}\right)\)
\(C=\left(\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}\right):\left(\dfrac{5}{12}+1-\dfrac{7}{11}\right)\)
a: \(A=\dfrac{7}{12}+\dfrac{5}{72}-\dfrac{11}{36}=\dfrac{42}{72}+\dfrac{5}{72}-\dfrac{22}{72}=\dfrac{25}{72}\)
b: \(B=\dfrac{8+5}{10}:\dfrac{-5}{13}=\dfrac{13}{10}\cdot\dfrac{13}{-5}=-\dfrac{169}{100}\)
c: \(C=\left(\dfrac{88}{132}-\dfrac{33}{132}+\dfrac{60}{132}\right):\left(\dfrac{55}{132}+\dfrac{132}{132}-\dfrac{84}{132}\right)\)
\(=\dfrac{88-33+60}{55+132-84}=\dfrac{115}{103}\)
5. \(3-1\dfrac{1}{2}-x+\dfrac{5}{4}=2-\left|1\dfrac{1}{8}-\dfrac{5}{12}\right|\) 6. \(3\dfrac{1}{14}-5\dfrac{1}{3}-\dfrac{4}{7}+\dfrac{11}{21}=-\dfrac{1}{2}\) 7. \(\dfrac{11}{-40}+\dfrac{4}{5}-\left|\dfrac{3}{4}-1\dfrac{5}{12}\right|=\dfrac{3}{20}-X\)
5. \(3-1\dfrac{1}{2}-x+\dfrac{5}{4}=2-\left|1\dfrac{1}{8}-\dfrac{5}{12}\right|\) 6. \(3\dfrac{1}{14}-5\dfrac{1}{3}-\dfrac{4}{7}+\dfrac{11}{21}=x-\dfrac{1}{2}\) 7. \(\dfrac{11}{-40}+\dfrac{4}{5}-\left|\dfrac{3}{4}-1\dfrac{5}{12}\right|=\dfrac{3}{20}-x\)
chắc h có mấy thành cay r nên ko làm bn lên mạng tải phẩn mêm có cánh iair đó :D
\(3-1\dfrac{1}{2}-x+\dfrac{5}{4}=2-\left|1\dfrac{1}{8}-\dfrac{5}{12}\right|\)
\(3-1\dfrac{1}{2}-x+\dfrac{5}{4}=2-\dfrac{17}{24}\)
\(x=3-1\dfrac{1}{2}+\dfrac{5}{4}-2+\dfrac{17}{24}\)
\(x=\dfrac{35}{24}\)
bn làm từng bước nhé :D
\(a,\dfrac{-6}{11}:\left(\dfrac{3}{5}:\dfrac{4}{11}\right)\) b,\(\dfrac{7}{12}+\dfrac{5}{12}:6-\dfrac{11}{36}\)
\(c,\left(\dfrac{4}{5}+\dfrac{1}{2}\right):\left(\dfrac{3}{13}-\dfrac{8}{13}\right)\) \(d,\left(\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}\right):\left(\dfrac{5}{12}-1-\dfrac{7}{11}\right)\)
Giúp mik nha:>>
a: \(=\dfrac{-6}{11}:\dfrac{3\cdot11}{4\cdot5}=\dfrac{-6}{11}\cdot\dfrac{20}{33}=\dfrac{-2}{11}\cdot\dfrac{20}{11}=\dfrac{-40}{121}\)
b: \(=\dfrac{7}{12}+\dfrac{5}{72}-\dfrac{11}{36}=\dfrac{42}{72}+\dfrac{5}{72}-\dfrac{22}{72}=\dfrac{25}{72}\)
c: \(=\dfrac{13}{10}:\dfrac{-5}{13}=\dfrac{-169}{50}\)
Tính (Tính hợp lí nếu có thể)
a) \(\dfrac{-7}{12}\)-\(\dfrac{3}{36}\)
b) (4-\(\dfrac{5}{12}\)):2+\(\dfrac{5}{24}\)
c) \(\dfrac{8}{9}\)+\(\dfrac{1}{9}\).\(\dfrac{2}{13}\)+\(\dfrac{1}{9}\).\(\dfrac{11}{13}\)
d) \(\dfrac{3}{4}\).\(\dfrac{8}{9}\).\(\dfrac{15}{16}\). ... .\(\dfrac{9999}{10000}\)
e) \(\dfrac{3}{1.4}\)+\(\dfrac{3}{4.7}\)+\(\dfrac{3}{7.10}\)+...+\(\dfrac{3}{97.100}\)
*Lưu ý: Mong các anh chị trình bày chi tiết để em có thể hiểu bài, em xin các anh chị đừng viết mỗi kết quả xong em chả biết một cái gì ;-;
a: =-21/36-3/36=-24/36=-2/3
b: =43/12*1/2+5/24=43/24+5/24=2
c: =8/9+1/9=1
e: =1-1/4+1/4-1/7+...+1/97-1/100
=1-1/100=99/100
Tính nhanh .
a, \(P=\dfrac{\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}}{\dfrac{5}{12}+1-\dfrac{7}{11}}\)
b, \(Q=\dfrac{0,125-\dfrac{1}{5}+\dfrac{1}{7}}{0,375-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0,2}{\dfrac{3}{4}+0,5-\dfrac{3}{10}}\)
a) Hình như nhầm đề thì phải :v
\(P=\dfrac{\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}}{\dfrac{5}{12}+1-\dfrac{6}{11}}\)
\(=\dfrac{\dfrac{5}{12}+\dfrac{5}{11}}{\dfrac{5}{12}+\dfrac{5}{11}}=1\)
b) \(Q=\dfrac{0,125-\dfrac{1}{5}+\dfrac{1}{7}}{0,375-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0,2}{\dfrac{3}{4}+0,5-\dfrac{3}{10}}\)
\(Q=\dfrac{0,125-\dfrac{1}{5}+\dfrac{1}{7}}{3\left(0,125-\dfrac{1}{5}+\dfrac{1}{7}\right)}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0,2}{\dfrac{3}{4}+0,5-\dfrac{3}{10}}\)
\(Q=\dfrac{1}{3}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0,2}{\dfrac{3}{4}+0,5-\dfrac{3}{10}}\)
\(Q=\dfrac{1}{3}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{2}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{3}\right)}=\dfrac{1}{3}+\dfrac{1}{\dfrac{3}{2}}\)
\(Q=\dfrac{1}{3}+\dfrac{2}{3}=1\)
a,\(P=\dfrac{\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}}{\dfrac{5}{12}+1-\dfrac{7}{11}}=\dfrac{\left(\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}\right).132}{\left(\dfrac{5}{12}+1-\dfrac{7}{11}\right).132}=\dfrac{88-33+60}{55+132-84}=\dfrac{115}{103}\)
b, Ta có : 0,125 = \(\dfrac{1}{8}\) ; 0,375 = \(\dfrac{3}{8}\) ; 0,2 = \(\dfrac{1}{5}\) ; 0,5 = \(\dfrac{3}{6}\)
\(Q=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{\dfrac{3}{8}-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{4}+\dfrac{3}{6}-\dfrac{3}{10}}\)
\(Q=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{3\cdot\left(\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}\right)}+\dfrac{2\cdot\left(\dfrac{1}{4}+\dfrac{1}{6}-\dfrac{1}{10}\right)}{3\cdot\left(\dfrac{1}{4}+\dfrac{1}{6}-\dfrac{1}{10}\right)}\)
\(Q=\dfrac{1}{3}+\dfrac{2}{3}=1\)
bài 1: Tính
a) \(\dfrac{-5}{7}\) . \(\dfrac{2}{11}\) + \(\dfrac{-5}{7}\) . \(\dfrac{9}{11}\) + 1\(\dfrac{5}{7}\)
b) \(\dfrac{-3}{14}\) +\(\dfrac{5}{8}\) -\(\dfrac{1}{2}\)
c) ( \(\dfrac{1}{4}\) +\(\dfrac{-5}{13}\))+( \(\dfrac{2}{11}\) + \(\dfrac{-8}{13}\) + \(\dfrac{3}{4}\) )
d) ( \(\dfrac{21}{31}\) + \(\dfrac{-16}{7}\) )+( \(\dfrac{44}{53}\) +\(\dfrac{10}{31}\) )+\(\dfrac{9}{53}\)
e) \(\dfrac{1}{9}\) - \(\dfrac{5}{6}\) -4
___________
\(\dfrac{7}{12}\) - \(\dfrac{1}{36}\) -10
giúp em
a: =-5/7(2/11+9/11)+12/7
=12/7-5/7
=7/7=1
b: \(=\dfrac{-12}{56}+\dfrac{35}{56}-\dfrac{28}{56}=\dfrac{-5}{56}\)
c: \(=\dfrac{1}{4}-\dfrac{5}{13}+\dfrac{2}{11}-\dfrac{8}{13}+\dfrac{3}{4}\)
=1-1+2/11
=2/11
d: \(=\dfrac{21}{31}+\dfrac{-16}{7}+\dfrac{44}{53}+\dfrac{10}{31}+\dfrac{9}{53}\)
=1+1-16/7
=-2/7
e: \(=\dfrac{\dfrac{4}{36}-\dfrac{30}{36}-\dfrac{144}{36}}{\dfrac{21}{36}-\dfrac{1}{36}-\dfrac{360}{36}}=\dfrac{-160}{-340}=\dfrac{8}{17}\)