1.a.ta có:\(\frac{2017+2018}{2018+2019}=\frac{2017}{2018+2019}+\frac{2018}{2018+2019}\)

mà \(\frac{2017}{2018}>\frac{2017}{2018+2019};\frac{2018}{2019}>\frac{2018}{2018+2019}\)

\(\Rightarrow M>N\)

b.ta thấy:

\(\frac{n+1}{n+2}>\frac{n+1}{n+3}>\frac{n}{n+3}\Rightarrow\frac{n+1}{n+2}>\frac{n}{n+3}\)

=> A>B

Trịnh Thùy Linh ơi mk cảm ơn bạn nhìu nha =)), iu bạn nhìu

Để A có giá trị là một số nguyên thì \(3n+2⋮n\)

\(\Rightarrow3n+2⋮3n\Rightarrow2⋮n\)

\(\Rightarrow n\inƯ\left(2\right)=\left\{-1;1;2;-2\right\}\)

Vậy để A có giá trị nguyên thì \(n\in\left\{-1;1;2;-2\right\}\)

a) Ta có: \(A=\frac{3n+2}{n}=3+\frac{2}{n}\)

A là số nguyên <=> n \(\in\)Ư ( 2 ) = { -2; -1; 1; 2 }

b) Thiếu điều kiện n là số nguyên dương.

Xét hiệu: \(\frac{a+n}{b+n}-\frac{a}{b}=\frac{b\left(a+n\right)-a\left(b+n\right)}{b\left(b+n\right)}=\frac{ba+bn-ab-an}{b\left(b+n\right)}\)

\(=\frac{bn-an}{b\left(b+n\right)}=\frac{n\left(b-a\right)}{b\left(b+n\right)}\)

TH1: b > a 

=> b - a > 0

=> \(\frac{n\left(b-a\right)}{b\left(b+n\right)}>0\)

=> \(\frac{a+n}{b+n}>\frac{a}{b}\)

TH2: b <  a 

=> b - a < 0

=> \(\frac{n\left(b-a\right)}{b\left(b+n\right)}< 0\)

=> \(\frac{a+n}{b+n}< \frac{a}{b}\)

TH1: b = a 

=> b - a = 0

=> \(\frac{n\left(b-a\right)}{b\left(b+n\right)}=0\)

=> \(\frac{a+n}{b+n}=\frac{a}{b}\)

Kết luận:...

a)Để A nguyên thì (3n+2)chia hết  cho n mà 3n chia hết cho n nên 2 phải chia hết cho n =>n\(\varepsilon\){2;1;-1;-2}

b)\(\frac{a+n}{b+n}\)=\(\frac{a}{b}\)+1>\(\frac{a}{b}\)=> Điều cần chứng minh

a) Ta có:

\(\begin{array}{l}\frac{6}{{10}} = \frac{{6:2}}{{10:2}} = \frac{3}{5};\\\frac{9}{{15}} = \frac{{9:3}}{{15:3}} = \frac{3}{5}\end{array}\)

\(\begin{array}{l}\frac{{6 + 9}}{{10 + 15}} = \frac{{15}}{{25}} = \frac{{15:5}}{{25:5}} = \frac{3}{5};\\\frac{{6 - 9}}{{10 - 15}} = \frac{{ - 3}}{{ - 5}} = \frac{3}{5}\end{array}\)

Ta được: \(\frac{{6 + 9}}{{10 + 15}} = \frac{{6 - 9}}{{10 - 15}} = \frac{6}{{10}} = \frac{9}{{15}}\)

b) - Vì \(k = \frac{a}{b} \Rightarrow a = k.b\)

Vì \(k = \frac{c}{d} \Rightarrow c = k.d\)

- Ta có:

\(\begin{array}{l}\frac{{a + c}}{{b + d}} = \frac{{k.b + k.d}}{{b + d}} = \frac{{k.(b + d)}}{{b + d}} = k;\\\frac{{a - c}}{{b - d}} = \frac{{k.b - k.d}}{{b - d}} = \frac{{k.(b - d)}}{{b - d}} = k\end{array}\)

- Như vậy, \(\frac{{a + c}}{{b + d}}\) =\(\frac{{a - c}}{{b - d}}\) = \(\frac{a}{b}\) =\(\frac{c}{d}\)( = k)

a: \(\dfrac{6+9}{10+15}=\dfrac{15}{25}=\dfrac{3}{5};\dfrac{6-9}{10-15}=\dfrac{-3}{-5}=\dfrac{3}{5}\)

=>Bằng nhau

b: a/b=c/d=k

=>a=bk; c=dk

\(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=k;\dfrac{a-c}{b-d}=\dfrac{bk-dk}{b-d}=k\)

=>\(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}=\dfrac{a}{b}=\dfrac{c}{d}\)

a, Để x là số nguyên

=> a - 5 chia hét cho a

Vì a chia hết cho a

=> -5 chia hết cho a

=> a \(\in\){1; -1; 5; -5}

\(\frac{a}{b}=\frac{a\left(b+n\right)}{b\left(b+n\right)}=\frac{ab+an}{b\left(b+n\right)}\)

\(\frac{a+n}{b+n}=\frac{b\left(a+n\right)}{b\left(b+n\right)}=\frac{ab+bn}{b\left(b+n\right)}\)

TH1: a = b

=> an = bn

=> ab+an = ab+bn

=> \(\frac{a}{b}=\frac{a+n}{b+n}\)

TH2: a > b

=> an > bn

=> ab + an > ab + bn

=> \(\frac{a}{b}>\frac{a+n}{b+n}\)

TH3: a < b

=> an < bn

=> ab + an < ab + bn

=> \(\frac{a}{b}