Các bn làm nhanh dùm m nhé :) Tks các bn <3
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TC
27 tháng 5 2022
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 \(Max_A=1-\dfrac{6}{-2+1}=1-\left(-6\right)=1+6=7\)
27 tháng 5 2022
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
21 tháng 2 2022
\(a)(-3/5)*x=-1/20+1/2=9/20=>x=9/20:(-3/5)=-3/4\)
Các câu kia làm tương tự nhé, chúc em học giỏi
21 tháng 2 2022
a: =>-3/5x=-1/20+1/2=-1/20+10/20=-9/20
=>x=3/4
b: =>-1/15x-2/15=3/5
=>-1/15x=6/15+2/15=8/15
=>x=-8
c: \(\Leftrightarrow\left(2x-1\right)\left(x-3\right)\left(x+3\right)=0\)
hay \(x\in\left\{\dfrac{1}{2};3;-3\right\}\)
TL
1
30 tháng 7 2023
\(=\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{50}+\dfrac{1}{51}-\dfrac{102}{51\cdot52}\)
\(=\dfrac{1}{2}+\dfrac{1}{51}-\dfrac{102}{51\cdot52}\)
\(=\dfrac{1}{2}+\dfrac{52-102}{51\cdot52}=\dfrac{1}{2}+\dfrac{-50}{51\cdot52}=\dfrac{319}{663}\)
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!
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: \(\left(\dfrac{15}{21}:\dfrac{5}{7}\right):\left(\dfrac{6}{5}:2\right)\)
\(=\left(\dfrac{5}{7}\cdot\dfrac{7}{5}\right):\left(\dfrac{6}{5\cdot2}\right)\)
\(=1:\dfrac{6}{10}=\dfrac{10}{6}=\dfrac{5}{3}\)
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
các bn giải dùm mik nhé
20 tháng 10 2023
IV
1. D
2. D
3. C
4. A
5. B
6. B
7. A
8. A
9. D
10. A
11. C
12. A
13. B
14. B
a)\(\dfrac{2}{3}+\dfrac{1}{-4}-\dfrac{5}{18}=\dfrac{8}{12}-\dfrac{3}{12}-\dfrac{5}{18}=\dfrac{5}{18}-\dfrac{5}{18}=\dfrac{5}{36}\)
b)\(=\dfrac{5}{8}+\dfrac{3}{4}-\dfrac{9}{8}=\left(\dfrac{5}{8}-\dfrac{9}{8}\right)+\dfrac{3}{4}=--\dfrac{2}{4}+\dfrac{3}{4}=\dfrac{1}{4}\)
c)\(=\dfrac{6}{7}+\dfrac{1}{7}:\left(-\dfrac{21}{28}+\dfrac{24}{28}\right)=\dfrac{6}{7}+\dfrac{1}{7}\times\dfrac{28}{3}=\dfrac{6}{7}+\dfrac{4}{3}=\dfrac{46}{21}\)
e)\(=\dfrac{9}{17}-\dfrac{5}{13}-\dfrac{8}{13}+\dfrac{1}{17}=\left(\dfrac{9}{17}+\dfrac{1}{17}\right)+\left(-\dfrac{5}{13}-\dfrac{8}{13}\right)\)
\(=\dfrac{8}{17}-1=-\dfrac{9}{17}\)