a: \(A=\dfrac{-35}{63}=\dfrac{-5}{9}\)
\(B=-\dfrac{15}{54}=\dfrac{-5}{18}\)
mà 9<18
nên A<B
b: \(C=\dfrac{2\cdot4+3}{12}\cdot\dfrac{3}{5}=\dfrac{11}{12}\cdot\dfrac{3}{5}=\dfrac{33}{60}=\dfrac{11}{20}\)
\(D=\dfrac{5-8}{20}\cdot\dfrac{-11}{3}=\dfrac{-3}{20}\cdot\dfrac{-11}{3}=\dfrac{11}{20}\)
Do đó: C=D
lllllll
a: \(=\dfrac{-7}{15}-\dfrac{12}{15}+\dfrac{10}{15}=\dfrac{\left(-7\right)-12+10}{15}=\dfrac{-19+10}{15}=\dfrac{-4}{15}\)
b: \(\dfrac{-3}{4}:\dfrac{7}{6}=\dfrac{\left(-3\right)\cdot6}{4\cdot7}=\dfrac{-9}{14}\)
c: \(=\dfrac{-12}{13}\cdot\dfrac{26}{-15}\cdot\dfrac{9}{-5}=\dfrac{\left(-12\right)\cdot26\cdot9}{13\cdot\left(-15\right)\cdot\left(-5\right)}=\dfrac{-2808}{975}\)
cho S = 2 mũ 1 + 2 mũ 2 + 2 mũ 3 + 2 mũ 4 + 2 mũ 5 + 2 mũ 6 +... + 2 mũ 28 + 2 mũ 29 + 2 mũ 30 . Chứng minh rằng S chia hết cho 7
\(S=2^1+2^2+2^3+2^4+2^5+2^6+..+2^{28}+2^{29}+2^{30}\)
\(S=2.\left(1+2+2^2\right)+2^4.\left(1+2+2^2\right)+...+2^{28}.\left(1+2+2^2\right)\)
\(S=\left(1+2+2^2\right).\left(2+2^4+...+2^{28}\right)\)
\(S=7.\left(2+2^4+...+2^{28}\right)\)
⇒ \(S⋮7\) ( điều phải chứng minh )
S=21+22+23+...+230
S=(21+22+23)+(24+25+26)+...+(228+229+230)
S=7.2+7.24+...+7.228
S=7.(2+24+...+228)
⇒S⋮7
Ta có: \(S=2^1+2^2+2^3+...+2^{28}+2^{29}+2^{30}\)
\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{28}\left(1+2+2^2\right)\)
\(=7\cdot\left(2+2^4+...+2^{28}\right)⋮7\)
Thực hiện phép tính:
\(\left(6-2\dfrac{4}{5}\right).3\dfrac{1}{8}-1\dfrac{3}{5}:\dfrac{1}{4}\)
Cảm ơn ạ!
`(6-2 4/5)*3 1/8-1 3/5:1/4`
`=(6-14/5)*25/8-8/5*4`
`=16/5*25/8-32/5`
`=10-32/5=18/5`
=(30/5-14/5).25/8.4
=16/5.25/2
=8.5/1.1
=40
Tìm x biết:
a, \(\left(4,5-2.x\right).1\dfrac{4}{7}=\dfrac{11}{14}\)
b, \(\left(2,8.x-32\right):\dfrac{2}{3}=-90\)
Cảm ơn ạ!
`a)(4,5-2x)*1 4/7=11/14`
`=>(4,5-2x)*11/7=11/14`
`=>4,5-2x=1/2`
`=>2x=4,5-0,5=4`
`=>x=2`
Vậy `x=2`
`b)(2,8x-32):2/3=-90`
`=>2,8x-32=-90*2/3=-60`
`=>2,8x=-28`
`=>x=-10`
Vậy `x=-10`
hellu
Tham khảo bài mình nhé : Tính A/BA=\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}\)B=\(\dfrac{1}{2020}+\dfrac{2}{2019}+\dfrac{3}{20... - Hoc24
\(\dfrac{2}{3}\) - \(\dfrac{5}{7}\) . \(\dfrac{14}{25}\)
`2/3-5/7xx14/25`
`=2/3-5/14`
`=28/42-15/42`
`=13/42`
\(\dfrac{2}{3}-\dfrac{5}{7}.\dfrac{14}{25}\)
\(=\dfrac{2}{3}-\dfrac{2}{5}\)
\(=\dfrac{4}{15}\)
Chúc bạn học tốt!
A= -2/4 +2/7 -5/28
`A-2/4+2/7-5/28`
`=-14/28+8/28-5/28`
`=(-14+8-5)/28`
`=-17/28`
A=\(\dfrac{-2}{4}+\dfrac{2}{7}-\dfrac{5}{28}=\dfrac{-14}{28}+\dfrac{8}{28}-\dfrac{5}{28}=\dfrac{6}{28}-\dfrac{5}{28}=\dfrac{1}{28}\)
`A-2/4+2/7-5/28` `=-14/28+8/28-5/28` `=(-14+8-5)/28` `=-11/28`
= 1/2*3 +1/3*4+1/4*5.......+ 1/99*100
\(A=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\\ =\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100.}\)
A=1/1*2+1/2*3+1/3*4+...+1/2019*2020
\(\Rightarrow A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2019}-\dfrac{1}{2020}\)
\(=1-\dfrac{1}{2020}=\dfrac{2019}{2020}\)
Vậy \(A=\dfrac{2019}{2020}\)