Các bn làm nhanh lên nhé m camon mn nhiều :)
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20 tháng 9 2021
b: Ta có: \(\left(2+4+6+...+100\right)\cdot\left(36\cdot333-108\cdot111\right)\)
\(=\left(2+4+6+...+100\right)\cdot36\cdot111\cdot\left(3-3\right)\)
=0
5 tháng 3 2022
Đặt A=\(\dfrac{1}{11}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{13}\)+...\(\dfrac{1}{69}\)+\(\dfrac{1}{70}\)
=(\(\dfrac{1}{11}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{13}\)+...\(\dfrac{1}{19}\)+\(\dfrac{1}{20}\))+(\(\dfrac{1}{21}\)+\(\dfrac{1}{22}\)+\(\dfrac{1}{23}\)+...+\(\dfrac{1}{29}\)+\(\dfrac{1}{30}\))+(\(\dfrac{1}{31}\)+\(\dfrac{1}{32}\)+\(\dfrac{1}{33}\)+...+\(\dfrac{1}{59}\)+\(\dfrac{1}{60}\))+...+\(\dfrac{1}{70}\)
Nhận xét:
\(\dfrac{1}{11}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{13}\)+...+\(\dfrac{1}{19}\)+\(\dfrac{1}{20}\)>\(\dfrac{1}{20}\)+\(\dfrac{1}{20}\)+...+\(\dfrac{1}{20}\)=\(\dfrac{10}{20}\)=\(\dfrac{1}{2}\)
\(\dfrac{1}{21}\)+\(\dfrac{1}{22}\)+\(\dfrac{1}{23}\)+...+\(\dfrac{1}{29}\)+\(\dfrac{1}{30}\)>\(\dfrac{1}{30}\)+\(\dfrac{1}{30}\)+...+\(\dfrac{1}{30}\)=\(\dfrac{10}{30}\)=\(\dfrac{1}{3}\)
\(\dfrac{1}{31}\)+\(\dfrac{1}{32}\)+\(\dfrac{1}{33}\)+...+\(\dfrac{1}{59}\)+\(\dfrac{1}{60}\)>\(\dfrac{1}{60}\)+\(\dfrac{1}{60}\)+...+\(\dfrac{1}{60}\)=\(\dfrac{30}{60}\)=\(\dfrac{1}{2}\)
=>A>\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{2}\)+\(\dfrac{1}{61}\)+...+\(\dfrac{1}{69}\)+\(\dfrac{1}{70}\)>\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{2}\)=\(\dfrac{4}{3}\)
=>A>\(\dfrac{4}{3}\)
Vậy: \(\dfrac{1}{11}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{13}\)+...+\(\dfrac{1}{69}\)+\(\dfrac{1}{70}\)>\(\dfrac{4}{3}\) (ĐPCM)
Thấy đúng cho 1 tick và 1 follow nha!
Chúc bạn học tốt!
2
27 tháng 5 2022
`-1 5/27-(3x-7/9)^3=-24/27`
`-32/27-(3x-7/9)^3=-24/27`
`(3x-7/9)^3=-8/27`
`(3x-7/9)^3=(-2/3)^3`
`3x-7/9=-2/3`
`3x=1/9`
`x=1/27`
NA
27 tháng 5 2022
`=>(3x-7/9)^3=-32/27+24/27`
`(3x-7/9)^3=-8/27`
`(3x-7/9)^3=(-2/3)^3`
`=>3x-7/9=-2/3`
`=>3x=-2/3+7/9`
`=>3x=-1/9`
`=>x=-1/27`
TL
1
30 tháng 7 2023
\(=\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{50}+\dfrac{1}{51}-\dfrac{102}{51\cdot52}\)
\(=\dfrac{1}{2}+\dfrac{1}{51}-\dfrac{102}{51\cdot52}\)
\(=\dfrac{1}{2}+\dfrac{52-102}{51\cdot52}=\dfrac{1}{2}+\dfrac{-50}{51\cdot52}=\dfrac{319}{663}\)
TL
1
9 tháng 1
Bài 2:
a: \(x-\dfrac{1}{2}=\dfrac{7}{13}\cdot\dfrac{13}{28}\)
=>\(x-\dfrac{1}{2}=\dfrac{7}{28}=\dfrac{1}{4}\)
=>\(x=\dfrac{1}{4}+\dfrac{1}{2}=\dfrac{3}{4}\)
b: \(\dfrac{x}{15}=\dfrac{-3}{11}\cdot\dfrac{77}{36}\)
=>\(\dfrac{x}{15}=\dfrac{-3}{36}\cdot\dfrac{77}{11}=7\cdot\dfrac{-1}{12}=-\dfrac{7}{12}\)
=>\(x=-\dfrac{7}{12}\cdot15=-\dfrac{105}{12}=-\dfrac{35}{4}\)
c: \(x:\dfrac{15}{11}=\dfrac{-3}{12}:8\)
=>\(x:\dfrac{15}{11}=-\dfrac{1}{4}:8=-\dfrac{1}{32}\)
=>\(x=-\dfrac{1}{32}\cdot\dfrac{15}{11}=\dfrac{-15}{352}\)
Bài 1:
a: \(\dfrac{-12}{25}\cdot\dfrac{10}{9}=\dfrac{-12}{9}\cdot\dfrac{10}{25}=\dfrac{-4}{3}\cdot\dfrac{2}{5}=\dfrac{-8}{15}\)
b: \(\dfrac{10}{21}-\dfrac{3}{8}\cdot\dfrac{4}{5}\)
\(=\dfrac{10}{21}-\dfrac{12}{40}\)
\(=\dfrac{10}{21}-\dfrac{3}{10}=\dfrac{100-63}{210}=\dfrac{37}{210}\)
c: \(\dfrac{28}{11}:\dfrac{21}{22}\cdot9=\dfrac{28}{11}\cdot\dfrac{22}{21}\cdot9\)
\(=\dfrac{28}{21}\cdot\dfrac{22}{11}\cdot9=\dfrac{4}{3}\cdot2\cdot9=\dfrac{4}{3}\cdot18=24\)
d: \(-\dfrac{10}{21}\cdot\left[\dfrac{9}{15}+\left(\dfrac{3}{5}\right)^2\right]\)
\(=\dfrac{-10}{21}\cdot\left[\dfrac{3}{5}+\dfrac{9}{25}\right]\)
\(=\dfrac{-10}{21}\cdot\dfrac{15+9}{25}\)
\(=\dfrac{-10}{25}\cdot\dfrac{24}{21}=\dfrac{-2}{5}\cdot\dfrac{8}{7}=\dfrac{-16}{35}\)
e: \(\left(\dfrac{2}{3}-\dfrac{1}{2}-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}-\dfrac{1}{7}\right)\)
\(=\left(\dfrac{1}{3}-\dfrac{1}{2}\right)\cdot\dfrac{28-7-4}{28}\)
\(=\dfrac{-1}{6}\cdot\dfrac{17}{28}=\dfrac{-17}{168}\)
f: \(\left(\dfrac{15}{21}:\dfrac{5}{7}\right):\left(\dfrac{6}{5}:2\right)\)
\(=\left(\dfrac{5}{7}\cdot\dfrac{7}{5}\right):\left(\dfrac{6}{5\cdot2}\right)\)
\(=1:\dfrac{6}{10}=\dfrac{10}{6}=\dfrac{5}{3}\)
20 tháng 9 2021
a) \(1+2+3+...+48=\dfrac{\left(48+1\right)\left(\dfrac{48-1}{1}+1\right)}{2}=1176\)
b) \(2+4+6+...+212=\dfrac{\left(212+2\right)\left(\dfrac{212-2}{2}+1\right)}{2}=11342\)
30 tháng 7 2023
a: =>4y+15/16=1
=>4y=1/16
=>y=1/64
b: =>10y+1/2+1/4+...+1/1024=1
=>10y+1023/1024=1
=>10y=1/1024
=>y=1/10240
b)Để A đạt GTNN : \(=>\dfrac{6}{n+1}\) phải lớn nhất
\(=>n+1=1\Leftrightarrow n=0\)
Vậy \(Min_A=1-\dfrac{6}{0+1}=1-6=-5\left(khi\right)n=0\)
Để A đạt GTLN : \(n+1\) phải là số âm lớn nhất
\(=>n+1=-1\Leftrightarrow n=-2\)
Vậy \(Max_A=1-\dfrac{6}{-2+1}=1-\left(-6\right)=1+6=7\)
a, để A là số âm, thì n-5 và n+1 khác dấu, mà n-5<n+1
=> n-5<0 và n+1>0
=> n<5 và n> -1
=> n thuộc {0;1;2;3;4}
b,để A có GTNN thì n+1 có giá trị dương nhỏ nhất có thể
=> n+1=1
=>n=0
c,gọi UCLN(n-5,n+1)=d(d thuộc N*)
=> n-5 chia hết cho d
=> n+1 chia hết cho d
=> (n+1)-(n-5)chia hết cho d
=> 6 chia hết cho d
=> d là ước của 6
nếu d=2
thì n-5 chia hết cho 2
n-5+6 chia hết cho 2
n+1 chia hết cho 2
=> n=2k+1(k thuộc N)
để A là p/s tối giản, thì n khác 2k+1