Question

a) Tìm số : (\(\frac{1}{24.25}\)+ \(\frac{1}{25.26}\)+ ... + \(\frac{1}{29.30}\)) . 120 + x : \(\frac{1}{3}\)=  -4
b) Cho A= \(\frac{3n-5}{n-4}\) (n\(\in\)Z)  . Tìm n nguyên để A có giá trị nguyên

Answer 1

 Câu a) :

x=-5/3

Câu b) :

GỢI Ý : 3n-5 phải chia hết cho n-4 để A là số nguyên ( đk : n khác 4)

Answer 2

\(a,\left(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\right).120+x:\frac{1}{3}=-4\)

\(\left(\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\right).120+3x=-4\)

\(\left(\frac{1}{24}-\frac{1}{30}\right).120+3x=-4\)

\(\frac{1}{120}.120+3x=-4\)

\(1+3x=-4\)

\(\Rightarrow3x=-5\)

\(\Rightarrow x=-\frac{5}{3}\)

\(b,A=\frac{3n-5}{n-4}=\frac{3n-12+7}{n-4}=3+\frac{7}{n-4}\)

Để \(A\in Z\Rightarrow7⋮n-4\Leftrightarrow n-4\in\left(1;-1;7;-7\right)\)

\(\Rightarrow n\in\left(5;3;11;-3\right)\)

Answer 3

           \(\)\(\left(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\right).120+x:\frac{1}{3}=-4\)  \(\)

\(\Rightarrow\)\(\left(\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\right).120+x:\frac{1}{3}=-4\)

\(\Rightarrow\)\(\left(\frac{1}{24}-\frac{1}{30}\right).120+x:\frac{1}{3}=-4\)

\(\Rightarrow\)\(\frac{1}{120}.120+x:\frac{1}{3}=-4\)

\(\Rightarrow\)\(1+x:\frac{1}{3}=-4\)

\(\Rightarrow\)\(3x=-3\)

\(\Rightarrow\)\(x=-1\)

Answer 4

\(\left(\frac{1}{24\cdot25}+\frac{1}{25\cdot26}+...+\frac{1}{29\cdot30}\right)\cdot120+x:\frac{1}{3}=-4\)

\(\Leftrightarrow\left(\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\right)\cdot120+3x=-4\)

\(\Leftrightarrow\left(\frac{1}{24}-\frac{1}{30}\right)\cdot120+3x=-4\)

\(\Leftrightarrow\frac{1}{120}\cdot120+3x=-4\)

\(\Leftrightarrow1+3x=-4\)

\(\Leftrightarrow3x=-5\)

\(\Leftrightarrow x=-\frac{5}{3}\)

\(A=\frac{3n-5}{n-4}=\frac{3\left(n-4\right)+7}{n-4}=3+\frac{7}{n-4}\)

Để A nguyên => \(\frac{7}{n-4}\)nguyên

=> \(7⋮n-4\)

=> \(n-4\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

n-4	1	-1	7	-7
n	5	3	11	-3

Vậy n = { 5 ;  3 ; 11 ; -3 }