dung phương pháp làm trội làm giảm

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

a) \(A=\dfrac{n-5}{n+1}=\dfrac{n-\left(6-1\right)}{n+1}=\dfrac{n+1}{n+1}-\dfrac{6}{n+1}=1-\dfrac{6}{n+1}\)

để A là số nguyên âm 

=> \(\dfrac{6}{n+1}>1\Rightarrow6>n+1\Rightarrow n< 5\)

vậy nếu n < 5 thì A là số nguyên âm

\(a)(-3/5)*x=-1/20+1/2=9/20=>x=9/20:(-3/5)=-3/4\)

Các câu kia làm tương tự nhé, chúc em học giỏi

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 \(x\in\left\{\dfrac{1}{2};3;-3\right\}\)

a, \(-\dfrac{3}{5}x=-\dfrac{1}{20}+\dfrac{1}{2}=\dfrac{9}{20}\Leftrightarrow x=-\dfrac{3}{4}\)

b, \(-\dfrac{2}{5}x+\dfrac{1}{3}x-\dfrac{28}{105}=\dfrac{3}{5}\Leftrightarrow-\dfrac{1}{15}x=\dfrac{13}{15}\Leftrightarrow x=-13\)

c, \(\left(\dfrac{1}{2}x-\dfrac{1}{4}\right)\left(x-3\right)\left(x+3\right)=0\Leftrightarrow x=\dfrac{3}{4};x=3;x=-3\)

\(=\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{50}+\dfrac{1}{51}-\dfrac{102}{51\cdot52}\)

\(=\dfrac{1}{2}+\dfrac{1}{51}-\dfrac{102}{51\cdot52}\)

\(=\dfrac{1}{2}+\dfrac{52-102}{51\cdot52}=\dfrac{1}{2}+\dfrac{-50}{51\cdot52}=\dfrac{319}{663}\)

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!

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: \(\left(\dfrac{15}{21}:\dfrac{5}{7}\right):\left(\dfrac{6}{5}:2\right)\)

\(=\left(\dfrac{5}{7}\cdot\dfrac{7}{5}\right):\left(\dfrac{6}{5\cdot2}\right)\)

\(=1:\dfrac{6}{10}=\dfrac{10}{6}=\dfrac{5}{3}\)

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

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