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Ta có:
\(\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)-x=\dfrac{-19}{24}\)
\(\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)-x=\dfrac{-19}{24}\)\(\left(\dfrac{4-3}{3.4}+\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+\dfrac{7-6}{6.7}+\dfrac{8-7}{7.8}+\dfrac{9-8}{8.9}+\dfrac{10-9}{9.10}\right)-x=\dfrac{-19}{24}\)
\(\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)-x=\dfrac{-19}{24}\)\(\left(\dfrac{1}{3}-\dfrac{1}{10}\right)-x=\dfrac{-19}{24}\)
\(\dfrac{7}{30}-x=\dfrac{-19}{24}\)
\(x=\dfrac{7}{30}-\dfrac{-19}{24}\)
\(x=\dfrac{41}{40}\)
\(\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)-x=\dfrac{-19}{24}\)
\(\Leftrightarrow\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)-x=\dfrac{-19}{24}\)
\(\Leftrightarrow\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)-x=\dfrac{-19}{24}\)
\(\Leftrightarrow\left(\dfrac{1}{3}-\dfrac{1}{10}\right)-x=\dfrac{-19}{24}\)
\(\Leftrightarrow\dfrac{7}{30}-x=\dfrac{-19}{24}\)
\(\Rightarrow x=\dfrac{7}{30}-\dfrac{-19}{24}\)
\(\Rightarrow x=\dfrac{41}{40}\)
a, 18.(-24) + 35.42 - 18.76 + 35.(-142)
= [ 18.(-24) - 18.76] + (35.42 + 35(-142)]
= -18.[ 24 + 76] - 35.( 142 - 42)
= -18.100 - 35. 100
= - 100.(18 + 35)
= - 100. 53
= - 5300
h, 104 : (-13) - [56 - 220 : (-4)]
= 104 : (-13) - [ 56 + 55]
= -8 - 111
= - 119
UCLN(54,42,48)=6;UCLN(24,36,72)=12
UCLN(180,168)=12;UCLN(48,72)=24
xrminhmk...
\(ƯC\left(8,12\right)=\left\{\pm1;\pm2;\pm4\right\}\)
\(ƯC\left(12;15;30\right)=\left\{\pm1;\pm3\right\}\)
\(ƯC\left(60;72\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
\(ƯC\left(24;42\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
a) { 1; 2; 4 }
b) { 1; 3 }
c) { 1; 2; 3; 4; 6; 12 }
d) { 1; 2; 3; 6 }
Ý bn là tìm phần tử à:
a, ƯC(8;12)= ƯCLN (8;12)
Ta có: 8= 23 và 12 = 22.3
\(\Rightarrow\)ƯCLN(8;12)= 22= 4
\(\Rightarrow\)ƯC (8;12)= Ư(4)= {1;2;4}
b, ƯC (12;15;30)= ƯCLN (12;15;30)
Ta có: 12= 22.3
15= 3.5
30= 3.2.5
\(\Rightarrow\)ƯCLN (12;15;30)= 2.3= 6
\(\Rightarrow\)ƯC (12;15;30)= Ư(6)= {1;2;3;6}
c, ƯC (60;72)= ƯCLN (60;72)
Ta có: 60= 22.3.5 và 72= 23.32
\(\Rightarrow\)ƯCLN (60;72)= 22= 4
\(\Rightarrow\)ƯC(60;72)= Ư(4)= {1;2;4}
d, ƯC (24;42)= ƯCLN (24;42)
Ta có: 24= 23.3 và 42= 2.3.7
\(\Rightarrow\)ƯCLN (24;42)= 3
\(\Rightarrow\)ƯC (24;42)= Ư(3)= {1;3}
Chúc bn học tốt
`@Bao`
`42−24+x=72`
`18+x=72`
`x=72−18`
`x=54`
Vậy `x=54`
\(\text{42−24+x=72}\)
\(\text{18 + x = 72}\)
\(\text{x = 72 - 18}\)
\(\text{x = 54}\)