Giúp mình với
Tính nhanh
\(\dfrac{1}{3}\) - \(\dfrac{3}{4}\) - \(\left(\dfrac{-3}{5}\right)\) + \(\dfrac{1}{64}\) - \(\dfrac{2}{9}\) -\(\dfrac{1}{36}\) + \(\dfrac{1}{15}\)
BT2: Tính nhanh
9) (\(\left(5+\dfrac{1}{5}-\dfrac{2}{9}\right)-\left(2-\dfrac{1}{23}-2\dfrac{3}{35}+\dfrac{5}{6}\right)-\left(8+\dfrac{2}{7}-\dfrac{1}{18}\right)\)
10)\(\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)
9:
\(=5+\dfrac{1}{5}-\dfrac{2}{9}-2+\dfrac{1}{23}+\dfrac{73}{35}-\dfrac{5}{6}-8-\dfrac{2}{7}+\dfrac{1}{18}\)
\(=\left(5-2-8\right)+\left(\dfrac{1}{5}+\dfrac{73}{35}-\dfrac{2}{7}\right)+\left(-\dfrac{2}{9}+\dfrac{1}{18}-\dfrac{5}{6}\right)+\dfrac{1}{23}\)
\(=\left(-5\right)+\dfrac{7+73-10}{35}+\dfrac{-4+1-15}{18}+\dfrac{1}{23}\)
\(=-5+2-1+\dfrac{1}{23}=-4+\dfrac{1}{23}=-\dfrac{91}{23}\)
10: \(=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)+\left(\dfrac{-3}{4}-\dfrac{2}{9}-\dfrac{1}{36}\right)+\dfrac{1}{64}\)
\(=\dfrac{5+9+1}{15}+\dfrac{-27-8-1}{36}+\dfrac{1}{64}\)
=1/64
BT2: Tính nhanh
3) \(\dfrac{2}{3}-\left(-\dfrac{1}{4}\right)+\dfrac{3}{5}-\dfrac{7}{45}-\left(-\dfrac{5}{9}\right)+\dfrac{1}{12}+\dfrac{1}{90}\)
4) B = \(\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)
Giúp mk nhé các bạn!
3) \(\dfrac{2}{3}-\left(-\dfrac{1}{4}\right)+\dfrac{3}{5}-\dfrac{7}{45}-\left(-\dfrac{5}{9}\right)+\dfrac{1}{12}+\dfrac{1}{90}\)
= \(\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{7}{45}+\dfrac{5}{9}+\dfrac{1}{12}+\dfrac{1}{90}\)
= \(\left(\dfrac{2}{3}+\dfrac{3}{5}-\dfrac{7}{45}+\dfrac{5}{9}+\dfrac{1}{90}\right)+\left(\dfrac{1}{4}+\dfrac{1}{12}\right)\)
= \(\left(\dfrac{60}{90}+\dfrac{54}{90}-\dfrac{14}{90}+\dfrac{50}{90}+\dfrac{1}{9}\right)+\left(\dfrac{4}{12}+\dfrac{1}{12}\right)\)
= \(\dfrac{151}{90}+\dfrac{1}{3}=\dfrac{151}{90}+\dfrac{30}{90}=\dfrac{181}{90}\)
Tính:
a, \(\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)
b, \(\left(3-\dfrac{1}{4}+\dfrac{2}{3}\right)-\left(5-\dfrac{1}{3}-\dfrac{6}{5}\right)-\left(6-\dfrac{7}{4}-\dfrac{3}{2}\right)\)
tinh nhanh
A=\(\dfrac{1}{3}-\dfrac{3}{4}-\left(\dfrac{-3}{5}\right)+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)
B=\(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{3}+\dfrac{13}{15}+\dfrac{11}{15}-\dfrac{9}{11}+\dfrac{7}{9}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{3}-\dfrac{1}{3}\)
giup minh nhe minh dang can gap
\(A=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)+\dfrac{1}{64}\)
\(=\dfrac{5+9+1}{15}-\dfrac{27+8+1}{36}+\dfrac{1}{64}\)
=1/64
tinh nhanh
A=\dfrac{1}{3}-\dfrac{3}{4}-\left(\dfrac{-3}{5}\right)+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}
B=\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{3}+\dfrac{13}{15}+\dfrac{11}{15}-\dfrac{9}{11}+\dfrac{7}{9}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{3}-\dfrac{1}{3}
giup minh nhe minh dang can gap
Tính:
\(\left(\dfrac{9}{25}-2.18\right):\left(3\dfrac{4}{5}+0,2\right)\)
\(\dfrac{3}{8}.19\dfrac{1}{3}\dfrac{3}{8}.33\dfrac{1}{3}\)
\(15.\left(-\dfrac{2}{3}\right)^2-\dfrac{7}{3}\)
\(\dfrac{1}{2}\sqrt{64}-\sqrt{\dfrac{4}{25}}+\left(-1\right)^{2007}\)
\(\left(-\dfrac{5}{2}\right)^2:\left(-15\right)-\left(0,45+\dfrac{3}{4}\right).\left(-1\dfrac{5}{9}\right)\)
\(\left(\dfrac{-1}{3}\right)-\left(\dfrac{-3}{5}\right)^0+\left(1-\dfrac{1}{2}\right)^2:2\)
\(\left(\dfrac{1}{2}\right)^{15}.\left(\dfrac{1}{4}\right)^{20}\)
\(\dfrac{5^4.20}{25^5.4^5}\)
a) Ta có: \(\left(\dfrac{9}{25}-2\cdot18\right):\left(3\dfrac{4}{5}+0.2\right)\)
\(=\left(\dfrac{9}{25}-36\right):\left(\dfrac{19}{5}+\dfrac{1}{5}\right)\)
\(=\left(\dfrac{9}{25}-\dfrac{900}{25}\right):\dfrac{20}{5}\)
\(=\dfrac{-891}{25}\cdot\dfrac{1}{4}\)
\(=-\dfrac{891}{100}\)
b) Ta có: \(\dfrac{3}{8}\cdot19\dfrac{1}{3}+\dfrac{3}{8}\cdot33\dfrac{1}{3}\)
\(=\dfrac{3}{8}\cdot\dfrac{58}{3}+\dfrac{3}{8}\cdot\dfrac{100}{3}\)
\(=\dfrac{58}{8}+\dfrac{100}{8}\)
\(=\dfrac{158}{8}=\dfrac{79}{4}\)
c) Ta có: \(15\cdot\left(-\dfrac{2}{3}\right)^2-\dfrac{7}{3}\)
\(=15\cdot\dfrac{4}{9}-\dfrac{7}{3}\)
\(=\dfrac{20}{3}-\dfrac{7}{3}\)
\(=\dfrac{13}{3}\)
d) Ta có: \(\dfrac{1}{2}\sqrt{64}-\sqrt{\dfrac{4}{25}}+\left(-1\right)^{2007}\)
\(=\dfrac{1}{2}\cdot8-\dfrac{2}{5}-1\)
\(=4-1-\dfrac{2}{5}\)
\(=3-\dfrac{2}{5}\)
\(=\dfrac{15}{5}-\dfrac{2}{5}=\dfrac{13}{5}\)
e) Ta có: \(\left(-\dfrac{5}{2}\right)^2:\left(-15\right)-\left(0.45+\dfrac{3}{4}\right)\cdot\left(-1\dfrac{5}{9}\right)\)
\(=\dfrac{25}{4}\cdot\dfrac{-1}{15}-\left(\dfrac{9}{20}+\dfrac{15}{20}\right)\cdot\dfrac{-14}{9}\)
\(=\dfrac{-25}{60}-\dfrac{24}{20}\cdot\dfrac{-14}{9}\)
\(=\dfrac{-25}{60}+\dfrac{28}{15}\)
\(=\dfrac{-25}{60}+\dfrac{112}{60}\)
\(=\dfrac{87}{60}=\dfrac{29}{20}\)
f) Ta có: \(\left(-\dfrac{1}{3}\right)-\left(-\dfrac{3}{5}\right)^0+\left(1-\dfrac{1}{2}\right)^2:2\)
\(=-\dfrac{1}{3}-1+\left(\dfrac{1}{2}\right)^2\cdot\dfrac{1}{2}\)
\(=\dfrac{-4}{3}+\dfrac{1}{4}\cdot\dfrac{1}{2}\)
\(=\dfrac{-4}{3}+\dfrac{1}{8}\)
\(=\dfrac{-32}{24}+\dfrac{3}{24}=\dfrac{-29}{24}\)
g) Ta có: \(\left(\dfrac{1}{2}\right)^{15}\cdot\left(\dfrac{1}{4}\right)^{20}\)
\(=\left(\dfrac{1}{2}\right)^{15}\cdot\left(\dfrac{1}{2}\right)^{40}\)
\(=\left(\dfrac{1}{2}\right)^{55}\)
\(=\dfrac{1}{2^{55}}\)
h) Ta có: \(\dfrac{5^4\cdot20}{25^5\cdot4^5}\)
\(=\dfrac{5^4\cdot5\cdot2^2}{5^{10}\cdot2^{10}}\)
\(=\dfrac{5^5}{5^{10}}\cdot\dfrac{2^2}{2^{10}}\)
\(=\dfrac{1}{5^5}\cdot\dfrac{1}{2^8}\)
\(=\dfrac{1}{800000}\)
\(\dfrac{1}{3}\) - \(\dfrac{3}{4}\) - \(\left(-\dfrac{3}{5}\right)\) + \(\dfrac{1}{75}\) - \(\dfrac{2}{9}\) -\(\dfrac{1}{36}\) + \(\dfrac{1}{15}\)
\(\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{75}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{1}{75}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)
\(=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)+\dfrac{1}{75}\)
\(=\left(\dfrac{5}{15}+\dfrac{9}{15}+\dfrac{1}{15}\right)-\left(\dfrac{27}{36}+\dfrac{8}{36}+\dfrac{1}{36}\right)+\dfrac{1}{75}\)
\(=1-1+\dfrac{1}{75}\)
\(=0+\dfrac{1}{75}\)
\(=\dfrac{1}{75}\)
#AvoidMe
=1/3+3/5+1/15-3/4-2/9-1/36+1/75
=5/15+9/15+1/15-27/36-8/36-1/36+1/75
=1/75
Tính: \(E=\dfrac{\left(\dfrac{1}{2}-1\right).\left(\dfrac{1}{3}-1\right)...\left(\dfrac{1}{2002}-1\right).\left(\dfrac{1}{2003}-1\right)}{\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}...\dfrac{9999}{10000}}\)
Giải chi tiết giúp mình nha. Thanks
\(E=\dfrac{\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{2002}-1\right)\left(\dfrac{1}{2003}-1\right)}{\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot...\cdot\dfrac{9999}{10000}}\)
\(=\dfrac{\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{2002}\right)\left(1-\dfrac{1}{2003}\right)}{\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{100^2}\right)}\)
\(=\dfrac{\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{2002}\right)\left(1-\dfrac{1}{2003}\right)}{\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{100}\right)\left(1+\dfrac{1}{100}\right)}\)
\(=\dfrac{\dfrac{100}{101}\cdot\dfrac{101}{102}\cdot...\cdot\dfrac{2002}{2003}}{\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1+\dfrac{1}{100}\right)}\)
\(=\dfrac{100}{2003}:\left(\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\right)\)
\(=\dfrac{100}{2003}:\left(\dfrac{101}{2}\right)=\dfrac{100}{2003}\cdot\dfrac{2}{101}=\dfrac{200}{202303}\)
tìm x biết
\(\dfrac{7}{9}:\left(2+\dfrac{3}{4}.x\right)+\dfrac{5}{9}=\dfrac{23}{27}\)
|x|\(-\dfrac{3}{4}=\dfrac{5}{3}\)
\(\left|2.x-\dfrac{1}{3}\right|+\dfrac{5}{6}=1\)
giúp mk vs nhanh lên mình đang bận
b) Ta có: \(\left|x\right|-\dfrac{3}{4}=\dfrac{5}{3}\)
\(\Leftrightarrow\left|x\right|=\dfrac{5}{3}+\dfrac{3}{4}=\dfrac{20}{12}+\dfrac{9}{12}=\dfrac{29}{12}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{12}\\x=-\dfrac{29}{12}\end{matrix}\right.\)
c) Ta có: \(\left|2x-\dfrac{1}{3}\right|+\dfrac{5}{6}=1\)
\(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{1}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{1}{3}=\dfrac{1}{6}\\2x-\dfrac{1}{3}=\dfrac{-1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{1}{6}+\dfrac{1}{3}=\dfrac{1}{2}\\2x=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{1}{12}\end{matrix}\right.\)