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\(c,\dfrac{12}{25}.\dfrac{-5}{24}=\dfrac{-60}{600}=-\dfrac{1}{10}\\ d,\dfrac{8}{15}:\dfrac{-4}{9}=\dfrac{8}{15}.\dfrac{9}{-4}=\dfrac{72}{-60}=-\dfrac{6}{5}\)
a) (-25)-(-17)+24-12
=(-25)+17+24-12
=(-25)+(17+24-12)
=(-25)+29
=29-25=4
b) 4.(15-18)-(3-5)3^2
=4.(-3)-2.9
=-12-18
=-(12+18)=-30
a/
(-25)-(-17)+24-12
= -8 + 12
= 4
b/
4.(15-18)-(3-5).32
= 4.(-3)-(-2).9
= -12 - (-18)
= 6
\(C=-\left[\dfrac{1}{3}\cdot\dfrac{\left(3+1\right)\cdot3}{2}+\dfrac{1}{4}\cdot\dfrac{\left(4+1\right)\cdot4}{2}+...+\dfrac{1}{50}\cdot\dfrac{\left(50+1\right)\cdot50}{2}\right]\\ C=-\left(\dfrac{1}{3}\cdot\dfrac{4\cdot3}{2}+\dfrac{1}{4}\cdot\dfrac{5\cdot4}{2}+...+\dfrac{1}{50}\cdot\dfrac{51\cdot50}{2}\right)\\ C=-\left(2+\dfrac{5}{2}+...+\dfrac{51}{2}\right)\\ C=-\dfrac{4+5+...+51}{2}=-\dfrac{\dfrac{\left(51+4\right)\left(51-4+1\right)}{2}}{2}=-\dfrac{55\cdot48}{4}=-660\)
\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}:\dfrac{1}{5}+\dfrac{3}{5}.\dfrac{1}{3}\)
=\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}.\dfrac{5}{1}+\dfrac{3}{5}.\dfrac{1}{3}\)
=\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{3}+\dfrac{3}{5}.\dfrac{1}{3}\)
=\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{1}{3}.2+\dfrac{3}{5}.\dfrac{1}{3}\)
=\(\dfrac{1}{3}.\left(\dfrac{2}{5}+\dfrac{3}{5}-2\right)\)=\(\dfrac{1}{3}.\left(-1\right)=\dfrac{-1}{3}\)
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
1. a) Ta có BCNN(12, 15) = 60 nên ta lấy mẫu chung của hai phân số là 60.
Thừa số phụ:
60:12 =5; 60:15=4
Ta được:
\(\frac{5}{{12}} = \frac{{5.5}}{{12.5}} = \frac{{25}}{{60}}\)
\(\frac{7}{{15}} = \frac{{7.4}}{{15.4}} = \frac{{28}}{{60}}\)
b) Ta có BCNN(7, 9, 12) = 252 nên ta lấy mẫu chung của ba phân số là 252.
Thừa số phụ:
252:7 = 36; 252:9 = 28; 252:12 = 21
Ta được:
\(\frac{2}{7} = \frac{{2.36}}{{7.36}} = \frac{{72}}{{252}}\)
\(\frac{4}{9} = \frac{{4.28}}{{9.28}} = \frac{{112}}{{252}}\)
\(\frac{7}{{12}} = \frac{{7.21}}{{12.21}} = \frac{{147}}{{252}}\)
2. a) Ta có BCNN(8, 24) = 24 nên:
\(\frac{3}{8} + \frac{5}{{24}} = \frac{{3.3}}{{8.3}} + \frac{5}{{24}} = \frac{9}{{24}} + \frac{5}{{24}} = \frac{{14}}{{24}} = \frac{7}{{12}}\)
b) Ta có BCNN(12, 16) = 48 nên:
\(\frac{7}{{16}} - \frac{5}{{12}} = \frac{{7.3}}{{16.3}} - \frac{{5.4}}{{12.4}} = \frac{{21}}{{48}} - \frac{{20}}{{48}} = \frac{1}{{48}}\).
\(a,=\dfrac{21}{48}+\dfrac{10}{48}=\dfrac{31}{48}\\ b,=\dfrac{44}{60}-\dfrac{25}{60}=\dfrac{19}{60}\)
a, 36:{336:[200–(12+8.20)]}
= 36:{336:[200–(12+160)]}
= 36:{336:[200–172]}
= 36:{336:28}
= 36:12 = 3
b, {145–[130–(246–236)]:2}.5
= {145–[130–10:2]}.5
= {145–130}.5
= 20.5 = 100
c, 100:{250:[450–(4. 5 3 – 2 2 .25]}
= 100:{250:[450–400]}
= 100:{250:50}
= 100:5 = 20
d, 798+100:[16–2.( 5 2 –22)]
= 798+100:10
= 798+10 = 808
e, (6954+1525:5+47.19).(29–58.2)
= (6954+1525:5+47.19).0 = 0
f, 2 4 .157– 2 4 .58+16
= 16.(157–58+1) = 1600
c) 53 48