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\(A=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)
\(A=\frac{2^{10}.3^8.\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}\)
\(A=\frac{1-3}{1+5}\)
\(A=\frac{-1}{3}\)
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a) 32 - 6 . (8 - 23) + 18 = 32 - 6 . (8 - 8) + 18
= 32 - 6 . 0 + 18 = 32 + 18 = 50
b) (3 . 5 - 9)3 . (1 + 2 . 3)2 + 42
= (15 - 9)3 . (1 + 6)2 + 42
= 63 . 72 + 42 = 216 . 49 + 16 = 10 584 + 16 = 10 600
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Sửa đề : `P=3/1.2+3/2.3+3/3.4+....+3/11.12`
`P=3/1.2+3/2.3+3/3.4+....+3/11.12`
`=3(1/1.2+1/2.3+1/3.4+...+1/11.12)`
`=3(1/1-1/2+1/2-1/3+1/3-1/4+...+1/11-1/12)`
`=3(1/1-1/12)`
`=3(12/12-1/12)`
`=3 . 11/12`
`=33/12`
`=11/4`
Vậy `P=11/4`
`#`𝐷𝑎𝑖𝑙𝑧𝑖𝑒𝑙
hình đề bị sai thì phải
\(\dfrac{3}{1\cdot2}+\dfrac{3}{2\cdot3}+\dfrac{3}{3\cdot4}+...+\dfrac{3}{11\cdot12}\) đề phải ntn chứ nhỉ?
\(=3\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{11\cdot12}\right)\)
\(=3\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{11}-\dfrac{1}{12}\right)\)
\(=3\left(\dfrac{1}{1}-\dfrac{1}{12}\right)\)
\(=3\left(\dfrac{12}{12}-\dfrac{1}{12}\right)\\ =3\cdot\dfrac{11}{12}\\ =\dfrac{33}{12}\\ =\dfrac{11}{4}\)
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a)
\(\begin{array}{l}\frac{2}{3} + \frac{{ - 2}}{5} + \frac{{ - 5}}{6} - \frac{{13}}{{10}}\\ = \frac{2}{3} + \frac{{ - 5}}{6} + \frac{{ - 2}}{5} - \frac{{13}}{{10}}\\ = \left( {\frac{2}{3} + \frac{{ - 5}}{6}} \right) + \left( {\frac{{ - 2}}{5} - \frac{{13}}{{10}}} \right)\\ = \left( {\frac{4}{6} + \frac{{ - 5}}{6}} \right) + \left( {\frac{{ - 4}}{{10}} - \frac{{13}}{{10}}} \right)\\ = \frac{{ - 1}}{6} + \frac{{ - 17}}{{10}}\\ = \frac{{ - 5}}{{30}} + \frac{{ - 51}}{{30}}\\ = \frac{{ - 56}}{{30}}\\ = \frac{{ - 28}}{{15}}\end{array}\)
b)
\(\begin{array}{l}\frac{{ - 3}}{7}.\frac{{ - 1}}{9} + \frac{7}{{ - 18}}.\frac{{ - 3}}{7} + \frac{5}{6}.\frac{{ - 3}}{7}\\ = \frac{{ - 3}}{7}.\left( {\frac{{ - 1}}{9} + \frac{7}{{ - 18}} + \frac{5}{6}} \right)\\ = \frac{{ - 3}}{7}.\left( {\frac{{ - 2}}{{18}} + \frac{{ - 7}}{{18}} + \frac{{15}}{{18}}} \right)\\ = \frac{{ - 3}}{7}.\frac{{ 6}}{{18}}\\ = \frac{-1}{7}\end{array}\).
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\(\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}...\frac{98.100}{99^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{98.100}{99.99}\)
\(=\frac{1.2.3...98}{2.3.4...99}.\frac{3.4.5...100}{2.3.4...99}\)
\(=\frac{1}{99}.\frac{100}{2}\)
\(=\frac{1}{99}.50=\frac{50}{99}\)
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a) \(\frac{{ - 3}}{7}.\frac{2}{5} + \frac{2}{5}.\left( { - \frac{5}{{14}}} \right) - \frac{{18}}{{35}}\)
\(\begin{array}{l} = \frac{2}{5}.\left( {\frac{{ - 3}}{7} + \frac{{ - 5}}{{14}}} \right) - \frac{{18}}{{35}}\\ = \frac{2}{5}.\left( {\frac{{ - 6}}{{14}} + \frac{{ - 5}}{{14}}} \right) - \frac{{18}}{{35}}\\ = \frac{2}{5}.\frac{{ - 11}}{{14}} - \frac{{18}}{{35}} = \frac{{ - 11}}{{35}} - \frac{{18}}{{35}} = \frac{{ -29}}{{35}}\end{array}\)
b) \(\left( {\frac{2}{3} - \frac{5}{{11}} + \frac{1}{4}} \right):\left( {1 + \frac{5}{{12}} - \frac{7}{{11}}} \right)\)
\(\begin{array}{l} = \left( {\frac{{2.11.4}}{{3.11.4}} - \frac{{5.3.4}}{{11.3.4}} + \frac{{1.3.11}}{{4.3.11}}} \right):\left( {\frac{11.12}{11.12} + \frac{{5.11}}{{12.11}} - \frac{{7.12}}{{11.12}}} \right)\\ = \left( {\frac{{88 - 60 + 33}}{{121}}} \right):\left( { \frac{{121+55 - 84}}{{121}}} \right)\\ = \frac{{61}}{{121}}:\frac{{92}}{{121}} = \frac{{61}}{{121}}.\frac{{121}}{{92}}= \frac{{61}}{{92}}\end{array}\)
c) \(\left( {13,6 - 37,8} \right).\left( { - 3,2} \right)\)
\( = \left( { - 24,2} \right).\left( { - 3,2} \right) = 77,44\)
d) \(\left( { - 25,4} \right).\left( {18,5 + 43,6 - 16,8} \right):12,7\)
\(\begin{array}{l} = \left( { - 25,4} \right).\left( {62,1 - 16,8} \right):12,7\\ = \left( { - 25,4} \right).45,3:12,7\\ = \left( { - 25,4} \right):12,7.45,3\\ = (- 2).45,3 = - 90,6\end{array}\)
a: \(=\dfrac{2}{5}\cdot\left(-\dfrac{3}{7}-\dfrac{5}{14}\right)-\dfrac{18}{35}\)
\(=\dfrac{2}{5}\cdot\dfrac{-6-5}{14}-\dfrac{18}{35}\)
\(=\dfrac{2}{5}\cdot\dfrac{-11}{14}-\dfrac{18}{35}=-\dfrac{22}{70}-\dfrac{18}{35}=\dfrac{-58}{70}=-\dfrac{29}{35}\)
b: \(=\dfrac{88-60+33}{132}:\dfrac{132+55-84}{132}\)
\(=\dfrac{61}{132}\cdot\dfrac{132}{103}=\dfrac{61}{103}\)
c: \(=-24.2\cdot\left(-3.2\right)=24.2\cdot3.2=77.44\)
d: \(=\dfrac{-25.4}{12.7}\cdot45.3=-2\cdot45.3=-90.6\)