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a) \(96 : 2^4 + 3^2 - 2018^0 \)
= 96 : 16 + 9 - 1
= 6 + 9 - 1
= 15 - 1
= 14
b) 32 . 77 + 33 . 24 - 32
= 2541 + 768 - 32
= 3309 - 32
= 3327
c) \({2 . 3^2 + [144 - 5^2 . (15 - 169 : 13)]} - 1^2019\)
1^2019 nha bạn
= 2 . 9 + [144 - 25 . (15 - 169 : 13)]}- 1
= 2 . 9 + [144 - 25 . (15 - 13)]}- 1
= 2 . 9 + [144 - 25 . 2]}- 1
= 2 . 9 + [144 - 50]}- 1
= 2 . 9 + 94 - 1
= 18 + 94 - 1
= 112 - 1
= 111
d) 5^27 . 5 : 5^25 + -125 = 0
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\(2B=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{23}-\frac{1}{25}\right)\)
\(2B=2\left(\frac{1}{3}-\frac{1}{25}\right)\)
\(2B=2\times\frac{22}{75}\)
\(B=\frac{44}{75}\)
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a) = 8 . 15 - [115- 72 ]
= 8 . 15 - [115- 49 ]
= 120 - 56
= 64
b) = 100 : {250 : [450 - (2.125 - 8.25)]}
= 100 : {250 : [450 - (250 - 200 )]}
= 100 : {250 : [450 - 50]}
= 100 : {250 :400}
= 100 : 0,265
= 0,00265
=
\(a,2^3.15-\left[115-\left(12-5\right)^2\right]\)
\(=8.15-\left[115-\left(12-5\right)^2\right]\)
\(=120-\left[115-7^2\right]\)
\(=120-\left[115-49\right]\)
\(=120-66\)
\(=54\)
\(b,100:\left\{250:\left[450-\left(2.5^3-2^3.25\right)\right]\right\}\)
\(=100:\left\{250:\left[450-\left(2.125-8.25\right)\right]\right\}\)
\(=100:\left\{250:\left[450-\left(250-200\right)\right]\right\}\)
\(=100:\left\{250:\left[450-50\right]\right\}\)
\(=100:\left\{250:400\right\}\)
\(=100:0,625\)
\(=160\)
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\(2^3.15-\left[115-\left(12-5\right)^2\right]=8.15-\left[115-7^2\right]=120-\left[115-49\right]=120-66=54\)
\(100:\left\{250:\left[450-\left(4.5^3-2^2.25\right)\right]\right\}=100:\left\{250:\left[450-\left(4.125-4.25\right)\right]\right\}=100:\left\{250:\left[450-\left(500-100\right)\right]\right\}\)
\(=100:\left[250:\left(450-400\right)\right]=100:\left(250:50\right)=100:5=20\)
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A=\(\dfrac{2}{1.3}-\dfrac{2}{3.5}-\dfrac{2}{5.7}-.....-\dfrac{2}{23.25}-\dfrac{1}{27}\)
A=\(\dfrac{2}{3}-\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+....+\dfrac{2}{23.25}\right)-\dfrac{1}{27}\)
A=\(\dfrac{2}{3}-\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+......+\dfrac{1}{23}-\dfrac{1}{25}\right)-\dfrac{1}{27}\)
A=\(\dfrac{2}{3}-\left(\dfrac{1}{3}-\dfrac{1}{25}\right)-\dfrac{1}{27}\)
A=\(\dfrac{2}{3}-\dfrac{22}{75}-\dfrac{1}{27}\)
A=\(\dfrac{227}{675}\)
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