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1)\(\dfrac{2}{9}+\dfrac{-3}{4}+\dfrac{5}{30}\)
\(=\dfrac{2.20}{9.20}+\dfrac{-3.45}{4.45}+\dfrac{5.6}{30.6}\)
\(=\dfrac{40}{180}+\dfrac{-135}{180}+\dfrac{30}{180}\)
\(=\dfrac{40+\left(-135\right)+30}{180}\)
\(=\dfrac{-65}{180}\)
\(=\dfrac{-13}{36}\)
2)\(\dfrac{-7}{12}-\dfrac{11}{18}\)
\(=\dfrac{-7.3}{12.3}-\dfrac{11.2}{18.2}\)
\(=\dfrac{-21}{36}-\dfrac{22}{36}\)
\(=\dfrac{-21-22}{36}\)
\(=\dfrac{-43}{36}\)
3)\(\dfrac{7}{8}-\dfrac{-5}{16}\)
\(=\dfrac{7.2}{8.2}-\dfrac{-5}{16}\)
\(=\dfrac{14}{16}-\dfrac{-5}{16}\)
\(=\dfrac{14-\left(-5\right)}{16}\)
\(=\dfrac{19}{16}\)
4)\(\dfrac{3}{8}-\dfrac{-9}{10}-\dfrac{5}{16}\)
\(=\dfrac{3.10}{8.10}-\dfrac{-9.8}{10.8}-\dfrac{5.5}{16.5}\)
\(=\dfrac{30}{80}-\dfrac{-72}{80}-\dfrac{25}{80}\)
\(=\dfrac{30-\left(-72\right)-25}{80}\)
\(=\dfrac{77}{80}\)
\(\dfrac{21}{36}-\left(-\dfrac{11}{30}\right)=\dfrac{7}{12}+\dfrac{11}{30}=\dfrac{7.5+11.2}{60}=\dfrac{57}{60}=\dfrac{19}{20}\\ ----\\\dfrac{-4}{8}+\left(-\dfrac{3}{10}\right)=\dfrac{-1}{2}-\dfrac{3}{10}=\dfrac{-1.5-3}{10}=\dfrac{-8}{10}=-\dfrac{4}{5}\\ ----\\ \dfrac{7}{12}-\left(-\dfrac{9}{20}\right)=\dfrac{7}{12}+\dfrac{9}{20}=\dfrac{7.5+9.3}{60}=\dfrac{62}{60}=\dfrac{31}{30}\\ ---\\ \dfrac{-2}{5}+\left(-\dfrac{11}{30}\right)=-\dfrac{2}{5}-\dfrac{11}{30}=\dfrac{-2.6-11}{30}=-\dfrac{29}{30}\)
a) 5/17 * 8/-7+8/17*-7/3+-7/3*4/17
-40/119 + 12/17 × -7/3
-40/119 + -28/17 =-236/119
b) -10/13 + 5/17 - 3/13 + 12/17 - 11/20
(5/17+12/17)-(10/13+3/13)-11/20
-11/20
a) 5/17 * 8/-7+8/17*-7/3+-7/3*4/17
-40/119 + 12/17 × -7/3
-40/119 + -28/17 =-236/119
b) -10/13 + 5/17 - 3/13 + 12/17 - 11/20
(5/17+12/17)-(10/13+3/13)-11/20
-11/20
a) \(4^8\cdot4^4=\left(2^2\right)^8\cdot\left(2^2\right)^4=2^{16}\cdot2^8=2^{16+8}=2^{24}\)
b) \(5^{12}\cdot7-5^{11}\cdot10\)
\(=5^{11}\cdot\left(5\cdot7-10\right)=5^{11}\cdot\left(35-10\right)=5^{11}\cdot25\)
\(=5^{11}\cdot5^2=5^{11+2}=5^{13}\)
d) \(27^{16}:9^{10}\)
\(=\left(3^3\right)^{16}:\left(3^2\right)^{10}=3^{48}:3^{20}=3^{48-20}=3^{28}\)
e) \(125^3:25^4=\left(5^3\right)^3:\left(5^2\right)^4=5^9:5^8=5^{9-8}=5\)
f) \(24^4:3^4-32^{12}:16^{12}\)
\(=\left(24:4\right)^4-\left(32:16\right)^{12}\)
\(=6^4-2^{12}\)
\(=2^4\cdot\left(3^4-2^8\right)=2^4\cdot-175=-2800\)
Q=(2^9.3+2^9.5):2^12
Đặt A=2^9.3+2^9.5
A=2^9.(3+5)
A=2^9.8
Mặt khác:8=2^3
=>A=2^9.2^3
A=2^12
Theo đề bài ta có Q=(2^9.3+2^9.5):2^12
=>Q=2^12:2^12
Q=1
Nhìn dài dòng thế thôi chứ đơn giản lắm.Nếu thấy đúng thì cho mình nhé!
a) \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+........+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.........+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
b) \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+..........+\frac{2}{73.75}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+.......+\frac{1}{73}-\frac{1}{75}\)
\(=\frac{1}{3}-\frac{1}{75}=\frac{8}{25}\)
c) \(\frac{4}{4.6}+\frac{4}{6.8}+\frac{4}{8.10}+..........+\frac{4}{64.66}\)
\(=2.\left(\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+..........+\frac{2}{64.66}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+.....+\frac{1}{64}-\frac{1}{66}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{66}\right)=2.\frac{31}{132}=\frac{31}{66}\)
d) \(\frac{9}{5.8}+\frac{9}{8.11}+\frac{9}{11.14}+........+\frac{9}{497.500}\)
\(=3.\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+..........+\frac{3}{497.500}\right)\)
\(=3.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+......+\frac{1}{497}-\frac{1}{500}\right)\)
\(=3.\left(\frac{1}{5}-\frac{1}{500}\right)=3.\frac{99}{500}=\frac{297}{500}\)
e) \(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+......+\frac{1}{93.95}\)
\(=\frac{1}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+........+\frac{2}{93.95}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+........+\frac{1}{93}-\frac{1}{95}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{95}\right)=\frac{1}{2}.\frac{18}{95}=\frac{9}{95}\)
g) \(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+..........+\frac{1}{200.203}\)
\(=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+........+\frac{3}{200.203}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{200}-\frac{1}{203}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{203}\right)=\frac{1}{3}.\frac{201}{406}=\frac{67}{406}\)
17+17+17=17x3=41