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1) 5 + (-4) = 1
2) (-8) + 2 = -6
3) 8 + (-2) = 6
4) 11 + (-3) = 8
5) (-11) + 2 = -9
6) (-7) + 3 = -4
7) (-5) + 5 = 0
8) 11 + (-12) = -1
9) (-18) + 20 = 2
10) (15) + (-12) = 3
11) (-17) + 17 = 0
12) 16 + (-2) = 14
13) (30) + (-14) = 16
14) (-19) + 20 = 1
15) (-18) + 15 = -3
16) (10) + (-6) = 4
17) (-28) + 14 = -14
18) 15 + (-30) = -15
19) (15) + (-4) = 11
20) (-21) + 11 = -10
21) 8 + (-22) = -14
22) (-15) + 4 = -11
23) (-3) + 2 = -1
24) 17 + (-14) = 3
25) 17 + (-14) = 3
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c) Ta có: \(\dfrac{3}{5}+\dfrac{-5}{20}+\dfrac{30}{75}+\dfrac{-7}{4}\)
\(=\dfrac{3}{5}+\dfrac{2}{5}+\dfrac{-1}{4}+\dfrac{-7}{4}\)
\(=1-2=-1\)
Giải:
a)-1/12+4/3=-1/12+16/12=15/12=5/4
b)(-4/14-3/15)-(1/5-20/35-(-1)).7
=-17/35-22/35.7
=-17/35-22/5
=-171/35
c)3/5+-5/20+30/75+-7/4
=3/5+-1/4+2/5+-7/4
=(3/5+2/5)+(-1/4+-7/4)
=1+-2
=-1
d)5/6.-12/14+7/13
=-5/7+7/13
=-16/91
e)2/-9-5/-36-1/4
=-1/12-1/4
=-1/3
f)2/23+-5/12+7/18+21/23+-7/12
=(2/23+21/23)+(-5/12+-7/12)+7/18
=1+-1+7/18
=7/18
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a)\(12-\left[25+\left(6+3\right)^2\right]\)
\(=12-\left[25+9^2\right]\)
\(=12-\left[25+81\right]\)
\(=12-106\)
\(=\left(-94\right)\)
b) \(315-\left(8^6\div8^4-12.3+7+7^3\right)\)
\(=316-\left(8^2-36+7+343\right)\)
\(=316-\left(64-43+343\right)\)
\(=316-\left(21+343\right)\)
\(=316-364\)
\(=\left(-48\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(12^7:6^7=\left(12:6\right)^7=2^7\)
b) \(27^5:81^3=\left(3^3\right)^5:\left(3^4\right)^3=3^{15}:3^{12}=3^3=27\)
c) \(18^3:9^3=\left(18:9\right)^3=2^3=8\)
d) \(125^3:25^4=\left(5^3\right)^3:\left(5^2\right)^4=5^9:5^8=5\)
a, 127 : 67 = (12 : 6)7 = 27 = 128.
b,275 : 813 = (33)5 : (34)3 = 315 : 312 = 315 - 12 = 33 = 27
c,183 : 93 = (18 : 9)3 = 23 = 8.
d, 1253 : 254 = (53)3 : (52)4 = 59 : 58 = 59 - 8 = 5.
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a) \(-12:\left(\frac{3}{4}-\frac{5}{6}\right)^2=-\frac{12}{\left(\frac{3}{4}-\frac{5}{6}\right)^2}=-\frac{12}{\frac{1}{12^2}}\)
\(=-\frac{12^2.12}{1}=12^2.12\)
\(=-12^3=--1728\)
b) \(\frac{7}{23}.\left[\left(-\frac{8}{6}\right)-\frac{45}{18}\right]=\frac{7}{23}.\left(-\frac{4}{3}-\frac{5}{2}\right)\)
\(=\frac{7}{23}.\left(-\frac{23}{6}\right)=-\frac{7}{23}.\frac{23}{6}\)
\(=-\frac{7}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(A=\frac{2^{12}.3^4-4^5.9^2}{\left(2^2.3\right)^6+8^4.3^5}\)
\(A=\frac{2^{12}.3^4-2^{10}.3^4}{2^{12}.3^6+2^{12}.3^5}\)
\(A=\frac{2^{10}.3^4\left(2^2-1\right)}{2^{10}.3^4\left(2^2.3^2+2^2.3\right)}\)
\(A=\frac{2^2-1}{2^2.3^2+2^2.3}\)
\(A=\frac{4-1}{36+12}\)
\(A=\frac{3}{48}=\frac{1}{16}\)
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(-12) + (-18)
= -30
------------------
(-90) + 27 + (-10) + 73
= [(-90) + (-10) + (27 + 73)
= -100 + 100
= 0
---------------------------------
25 . (-5) . (-4)
= 25 . (-4) . (-5)
= -100 . (-5)
= 100 . 5
= 500
----------------------------------------
(-15) . 40 : 6
= [(-15) . 40] : 6
= (-600) : 6
= -100
-----------------------------------
(-27) . 58 + (-27) . 42
= (-27) . (58 + 42)
= (-27) . 100
= -2700
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Lời giải:
a.
$-18: \frac{3}{5}=-18.\frac{5}{3}=-30$
b.
$\frac{3}{4}:(-9)=\frac{3}{4}.\frac{-1}{9}=\frac{-1}{12}$
c.
$\frac{13}{20}-\frac{6}{7}: \frac{10}{21}=\frac{13}{20}-\frac{6}{7}.\frac{21}{10}$
$=\frac{13}{20}-\frac{9}{5}=\frac{13}{20}-\frac{36}{20}=\frac{-23}{20}$
d.
$\frac{-21}{5}: (\frac{7}{3}.\frac{7}{5})=\frac{-21}{5}: \frac{49}{15}$
$=\frac{-21}{5}.\frac{15}{49}=\frac{-9}{7}$
e.
$(\frac{-2}{5}+\frac{1}{4}): (1-\frac{2}{5})$
$=\frac{-3}{20}: \frac{3}{5}=\frac{-1}{4}$