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62 . 10 : { 780 : [ 103 - ( 2 . 53 + 35 . 14 ) ] }
= 360 : { 780 : [ 1000 - ( 250 + 490 ) ] }
= 360 : { 780 : [ 1000 - 740 ] }
= 360 : { 780 : 260 }
= 360 : 3
= 120.
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\(=\frac{5^3\cdot2^3+2+5^3}{55}=\frac{5^3\left(8+2+1\right)}{55}=\frac{125\cdot11}{11\cdot5}=\frac{25\cdot1}{1\cdot1}=25\)
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`[x-3]/5=[2x-6]/10`
`[2(x-3)]/10=[2x-6]/10`
`2x-6=2x-6`
`2x-2x=-6+6`
`0x=0` (LĐ)
Vậy `x in RR`
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\(\dfrac{2}{3}+\dfrac{1}{5}.\dfrac{10}{7}=\dfrac{2}{3}+\dfrac{10}{35}=\dfrac{70}{105}+\dfrac{30}{105}=\dfrac{100}{105}=\dfrac{50}{21}\)
a) Ta có: \(\dfrac{2}{3}+\dfrac{1}{5}\cdot\dfrac{10}{7}\)
\(=\dfrac{2}{3}+\dfrac{2}{7}\)
\(=\dfrac{14}{21}+\dfrac{6}{21}\)
\(=\dfrac{20}{21}\)
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a: \(\left[504-\left(25\cdot8+70\right)\right]:9-15+19^0\)
\(=\left(504-270\right):9-15+1\)
\(=29-14\)
=15
a) [504 - (25 . 8 + 70)] : 9 - 15 + 190
[504 - (25 . 8 + 70)] : 9 - 15 + 1
[504 - (200 + 70)] : 9 - 15 + 1
(504 - 270) : 9 - 15 + 1
234 : 9 - 15 + 1
= 26 - 15 + 1
= 11 + 1
= 12
b) 5 . {26 - [3 . (5 + 2 . 5) + 15] : 15}
5 . {26 - [3 . (5 + 10) + 15] : 15}
5 . [26 - (3 . 15 + 15) : 15]
5 . [26 - (45 + 15) : 15]
5 . (26 - 60 : 15)
5 . (26 - 4)
= 5 . 22
= 110
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\(\dfrac{x-3}{10}-\dfrac{x-3}{15}=\dfrac{x-3}{12}-\dfrac{x-3}{18}\)
\(\Rightarrow\dfrac{x-3}{1}-\dfrac{x-3}{15}-\dfrac{x-3}{12}+\dfrac{x-3}{18}=0\)
\(\Rightarrow\left(x-3\right)\left(1-\dfrac{1}{15}-\dfrac{1}{12}+\dfrac{1}{18}\right)=0\)
Mà \(1-\dfrac{1}{15}-\dfrac{1}{12}+\dfrac{1}{18}\ne0\)
\(\Rightarrow x-3=0\Leftrightarrow x=3\)
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\(\dfrac{3}{7}.\dfrac{5}{9}+\dfrac{4}{9}.\dfrac{-3}{7}+\left(-2023^0\right)\)
\(=\left(\dfrac{3}{7}.\dfrac{5}{9}\right)+\left(\dfrac{4}{9}.\dfrac{-3}{7}\right)+\left(-1\right)\)
\(=\dfrac{5}{21}+\dfrac{-4}{21}+\left(-1\right)\)
\(=\dfrac{5}{21}+\dfrac{-4}{21}+\dfrac{-21}{21}\)
\(=\dfrac{5-4-21}{21}=\dfrac{-20}{21}\)
\(=\dfrac{5^3\cdot2^3+2\cdot5^3+5^3}{55}=\dfrac{5^3\left(2^3+2+1\right)}{55}=5^2=25\)