a) ta co:

1/18<x/12<y/9<1/4

=>2/36<x.3/36<y.4/36<9/36

=>x.3thuộc{3;6};y.4thuộc{4;8}

=>x thuộc{1;2};y thuộc{1:2}

b) ta co

7/8<x/40<9/10

=>70/80<x.2/40<72/80

=>x.2 =71

=>x=71/2

a: =6x5/4=30/4=15/2

b: =8x7/12=56/12=14/3

c: =5x8/7=40/7

d: =15x3/5=9

a: =6x5/4=30/4=15/2

b: =8x7/12=56/12=14/3

c: =5x8/7=40/7

d: =15x3/5=9

bạn cứ tính 2 vế là xong mà:

a) x\(\in\){1;2;3;4;5;6;7}

b) x=0

Sửa đề: Tìm các giá trị nguyên X thỏa mãn

3: \(\frac{-5}{11}<\frac{9}{x}<\frac{-5}{12}\)

=>\(\frac{-45}{99}<\frac{-45}{-5x}<\frac{-45}{108}\)

=>\(\frac{45}{99}>\frac{45}{-5x}>\frac{45}{108}\)

=>99<-5x<108

=>\(-\frac{99}{5}>x>-\frac{108}{5}\)

mà x nguyên

nên x∈{-20;-21}

4: Ta có: \(\frac{-11}{13}<\frac{9}{x}<\frac{-11}{15}\)

=>\(\frac{-99}{117}<\frac{-99}{-11x}<\frac{-99}{135}\)

=>\(\frac{99}{117}>\frac{99}{-11x}>\frac{99}{135}\)

=>117<-11x<135

=>\(-\frac{117}{11}>x>-\frac{135}{11}\)

mà x nguyên

nên x∈{-11;-12}

5: Ta có: \(\frac{-4}{5}<\frac{9}{x}<\frac{-4}{7}\)

=>\(\frac{-36}{45}<\frac{-36}{-4x}<\frac{-36}{28}\)

=>\(\frac{36}{45}>\frac{36}{-4x}>\frac{36}{28}\)

=>45<-4x<28

=>\(-\frac{45}{4}>x>-\frac{28}{4}\)

mà x nguyên

nên x∈∅

Ta có: \(\frac12\times\frac{45}{33}\times\frac19\times\frac{11}{6}\)

\(=\frac12\times\frac19\times\frac{15}{11}\times\frac{11}{6}\)

\(=\frac{1}{18}\times\frac{15}{6}=\frac16\times\frac56=\frac{5}{36}\)

\(x=\dfrac{3}{5}-\dfrac{7}{8}=-\dfrac{11}{40}\)

x = 3/5 - 7/8

x =  -11/40

a: \(\dfrac{x+2}{27}=\dfrac{x}{-9}\)

=>x+2=-3x

=>4x=-2

hay x=-1/2

b: \(\dfrac{-7}{x}=\dfrac{21}{34-x}\)

=>-7(34-x)=21x

=>34-x=-3x

=>2x=-34

hay x=-17

c: \(\dfrac{-8}{15}< \dfrac{x}{40}< \dfrac{-7}{15}\)

\(\Leftrightarrow-64< 3x< -56\)

hay \(x\in\left\{-21;-20;-19\right\}\)

d: \(\dfrac{-1}{2}< \dfrac{x}{18}< \dfrac{-1}{3}\)

=>-9<x<-6

hay \(x\in\left\{-8;-7\right\}\)

a)

Ta có \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{20^2}< \dfrac{1}{19.20}\)

Cộng vế với vế ta được 

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{20^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}-\dfrac{1}{20}\)

\(\Rightarrow T< 2-\dfrac{1}{20}=\dfrac{39}{20}\)

mà 39/20 < 8/7 => T < 8/7 

Lời giải:
a.

\(\frac{n+1}{n+2}=\frac{n+1}{n+2}+1-1=\frac{2n+3}{n+2}-1\)

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

b.

\(10A=\frac{10^{12}-10}{10^{12}-1}=\frac{(10^{12}-1)-9}{10^{12}-1}=1-\frac{9}{10^{12}-1}<1\)

\(10B=\frac{10^{11}+10}{10^{11}+1}=\frac{(10^{11}+1)+9}{10^{11}+1}=1+\frac{9}{10^{11}+1}>1\)

$\Rightarrow 10A< 10B\Rightarrow A< B$