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22 tháng 3 2020
\(A=\frac{1}{3}-\frac{3}{4}-\frac{-3}{5}+\frac{1}{73}-\frac{1}{26}+\frac{1}{15}+\frac{-2}{9}\)
\(\Rightarrow A=\left(\frac{1}{3}-\frac{-3}{5}+\frac{1}{15}\right)-\frac{3}{4}+\frac{1}{73}-\frac{1}{26}+\frac{-2}{9}\)
\(\Rightarrow A=\left(\frac{5}{15}-\frac{-9}{15}+\frac{1}{15}\right)-\frac{39}{52}-\frac{2}{52}+\frac{1}{73}+\frac{-2}{9}\)
\(\Rightarrow A=\left(\frac{5}{15}+\frac{9}{15}+\frac{1}{15}\right)-\left(\frac{39}{52}+\frac{2}{52}\right)+\frac{9}{657}+\frac{-146}{657}\)
\(\Rightarrow A=1-\frac{41}{52}+\frac{-137}{657}=\frac{11}{52}+\frac{-137}{657}=\frac{7227}{34164}+\frac{-7124}{34164}=\frac{103}{34163}\)
Vậy \(A=\frac{103}{34164}\)
Chúc bn học tốt
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
11 tháng 6
a; \(\dfrac{3}{11}\) + \(\dfrac{5}{-9}\) + \(\dfrac{4}{11}\) - \(\dfrac{4}{9}\) + \(\dfrac{3}{17}\) + \(\dfrac{15}{11}\)
= (\(\dfrac{3}{11}\) + \(\dfrac{4}{11}\) + \(\dfrac{15}{11}\)) - (\(\dfrac{5}{9}\) + \(\dfrac{4}{9}\)) + \(\dfrac{3}{17}\)
= 2 - 1 + \(\dfrac{3}{17}\)
= 1 + \(\dfrac{3}{17}\)
= \(\dfrac{20}{17}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
11 tháng 6
c; N = \(\dfrac{\dfrac{5}{7}-\dfrac{5}{9}-\dfrac{5}{11}}{\dfrac{15}{7}+\dfrac{15}{9}+\dfrac{15}{11}}\)
Phải là - \(\dfrac{5}{7}\) chỗ tử số mới đúng em nhé!
29 tháng 4 2019
Ta có: \(A=\frac{1}{3}-\frac{3}{4}+\frac{3}{5}+\frac{1}{73}-\frac{1}{36}+\frac{1}{15}-\frac{2}{9}\)
\(A=\left(\frac{1}{3}-\frac{2}{9}\right)+\left(\frac{3}{5}+\frac{1}{15}\right)-\frac{3}{4}-\frac{1}{36}+\frac{1}{73}\)
\(A=\left(\frac{3}{9}-\frac{2}{9}\right)+\left(\frac{9}{15}+\frac{1}{15}\right)-\left(\frac{3}{4}+\frac{1}{36}\right)+\frac{1}{73}\)
\(A=\frac{1}{9}+\frac{10}{15}-\frac{7}{9}+\frac{1}{73}\)
\(A=\frac{1}{9}+\frac{2}{3}-\frac{7}{9}+\frac{1}{73}\)
\(A=\frac{1}{9}+\frac{6}{9}-\frac{7}{9}+\frac{1}{73}\)
\(A=\frac{7}{9}-\frac{7}{9}+\frac{1}{73}\)
\(A=\frac{1}{73}\)
Vậy: \(A=\frac{1}{73}\)
HH
7 tháng 6 2020
\(1)\frac{1}{3}.\frac{4}{5}+\frac{1}{3}.\frac{6}{5}-\frac{1}{3}\)
\(=\frac{1}{3}\left(\frac{4}{5}+\frac{6}{5}\right)\)
\(=\frac{1}{3}.2\)
\(=\frac{2}{3}\)
G
7 tháng 6 2020
\(1)\)\(\frac{1}{3}.\frac{4}{5}+\frac{1}{3}.\frac{6}{5}-\frac{1}{3}\)
\(=\frac{1}{3}.\left(\frac{4}{5}+\frac{6}{5}-1\right)\)
\(=\frac{1}{3}.\left(\frac{4}{5}+\frac{6}{5}-\frac{5}{5}\right)\)
\(=\frac{1}{3}.1\)
\(=\frac{1}{3}\)
\(2)\)\(\frac{3}{7}.\frac{9}{26}-\frac{1}{14}.\frac{1}{13}\)
\(=\frac{1}{7}.\frac{27}{26}-\frac{1}{7}.\frac{1}{26}\)
\(=\left(\frac{27}{26}-\frac{1}{26}\right).\frac{1}{7}\)
\(=1.\frac{1}{7}\)
\(=\frac{1}{7}\)
\(3)\)\(15\left(\frac{1}{5}-\frac{7}{15}+\frac{17}{3}\right)\)
\(=3-7+85\)
\(=81\)
\(4)\)\(38\left(2\frac{1}{19}-5\frac{7}{38}\right)\)
\(=38\left(\frac{39}{19}-\frac{197}{38}\right)\)
\(=78-197\)
\(=-119\)
\(5)\)\(75\left(\frac{-4}{25}+7\frac{1}{15}-5\frac{7}{75}\right)\)
\(=75\left(\frac{-4}{25}+\frac{106}{15}-\frac{382}{75}\right)\)
\(=-12+530-382\)
\(=136\)
Ddaeng
MM
25 tháng 8 2018
\(a,\frac{47}{9}+\left(\frac{41}{3}-\frac{26}{9}\right)\)
\(=\frac{47}{9}+\left(\frac{82}{9}-\frac{26}{9}\right)\)
\(=\frac{47}{9}+\frac{56}{9}\)
\(=\frac{103}{9}\)
\(b,\frac{26}{15}:\frac{13}{5}-\frac{1}{4}\)
\(=\frac{26.5}{15.3}-\frac{1}{4}\)
\(=\frac{12}{3}-\frac{1}{4}\)
\(=\frac{4}{1}-\frac{1}{4}=\frac{16}{4}-\frac{1}{4}=\frac{15}{4}\)
HN
19 tháng 7 2016
a) \(\frac{15}{16}\cdot\frac{4}{3}-\frac{1}{2}:\frac{5}{4}+3\)
\(=\frac{5}{4}-\frac{2}{5}+3\)
\(=\frac{25}{20}-\frac{8}{20}+\frac{60}{20}\)
\(=\frac{77}{20}=3\frac{17}{20}\)
b) \(\left(\frac{2}{3}-\frac{3}{8}\right):\left(\frac{3}{5}+\frac{1}{4}\right)\)
\(=\frac{7}{24}:\frac{17}{20}\)
\(=\frac{35}{102}\)
c) \(\frac{15}{8}\cdot\left(\frac{1}{3}+\frac{1}{8}\right)-\frac{3}{8}:\frac{3}{4}\)
\(=\frac{15}{8}\cdot\frac{11}{24}-\frac{1}{2}\)
\(=\frac{55}{64}-\frac{1}{2}\)
\(=\frac{23}{64}\)
d) \(\frac{20}{21}:\left(\frac{4}{5}-\frac{1}{10}\right)+\frac{13}{15}:\frac{5}{26}\)
\(=\frac{20}{21}:\frac{7}{10}+\frac{52}{15}\)
\(=\frac{200}{147}+\frac{52}{15}\)
\(=4\frac{608}{735}\)
A = 1/3 -3/4--3/5+1/73-1/26+1/15-2/9
A = (1/3+3/5+1/15)-3/4+1/73-1/26-2/9
A = 15/15 - 472/468 +1/73
A = 1 - 34924/34164
A = -760/34164
A = -190/8541