Ta có: \(A=\dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{2n}>\dfrac{1}{2n}+\dfrac{1}{2n}+...+\dfrac{1}{2n}=\dfrac{n}{2n}=\dfrac{1}{2}\)

Vậy \(A>B\)

@Trần Thị Tâm Phúc

rIhAmI oTaKu

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.

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

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

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

\(u_n=\dfrac{1}{2^2-1}+\dfrac{1}{3^2-1}+...+\dfrac{1}{n^2-1}\)

\(=\dfrac{1}{\left(2-1\right)\left(2+1\right)}+\dfrac{1}{\left(3-1\right)\left(3+1\right)}+...+\dfrac{1}{\left(n-1\right)\left(n+1\right)}\)

\(=\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+...+\dfrac{1}{\left(n-1\right)\cdot\left(n+1\right)}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{2\cdot4}+...+\dfrac{2}{\left(n-1\right)\left(n+1\right)}\right)\)

\(=\dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+...+\dfrac{1}{\left(n-1\right)}-\dfrac{1}{\left(n+1\right)}\right)\)

\(=\dfrac{1}{2}\left(1+\dfrac{1}{2}-\dfrac{1}{n+1}\right)=\dfrac{1}{2}\cdot\left(\dfrac{3}{2}-\dfrac{1}{n+1}\right)\)

\(=\dfrac{3}{4}-\dfrac{1}{2n+2}\)

\(\lim\limits u_n=\lim\limits\left(\dfrac{3}{4}-\dfrac{1}{2n+2}\right)\)

\(=\lim\limits\dfrac{3}{4}-\lim\limits\dfrac{1}{2n+2}\)

\(=\dfrac{3}{4}-\lim\limits\dfrac{\dfrac{1}{n}}{2+\dfrac{1}{n}}\)

=3/4

=>Chọn A

\(\dfrac{n+1}{2n+3}\) < \(\dfrac{n+1}{2n+2}\) < \(\dfrac{n+2}{2n+2}\)

1) \(ĐK:3-2a>0\Leftrightarrow a< \dfrac{3}{2}\)

2) \(ĐK:2x-5< 0\Leftrightarrow x< \dfrac{5}{2}\)

3) \(ĐK:3-5a< 0\Leftrightarrow a>\dfrac{3}{5}\)

4) \(ĐK:a< 0\)

5) \(ĐK:-a\ge0\Leftrightarrow a\le0\)

\(a,\dfrac{a}{b}>1\Leftrightarrow a>1\cdot b=b\\ \dfrac{a}{b}< 1\Leftrightarrow a< 1\cdot b=b\\ b,\dfrac{a}{b}=\dfrac{a\left(b+1\right)}{b\left(b+1\right)}=\dfrac{ab+a}{b^2+b}\\ \dfrac{a+1}{b+1}=\dfrac{b\left(a+1\right)}{b\left(b+1\right)}=\dfrac{ab+b}{b^2+b}\\ \forall a=b\Leftrightarrow\dfrac{a}{b}=\dfrac{a+1}{b+1}\\ \forall a>b\Leftrightarrow\dfrac{a}{b}>\dfrac{a+1}{b+1}\\ \forall a< b\Leftrightarrow\dfrac{a}{b}< \dfrac{a+1}{b+1}\)

\(c,\forall a>b\Leftrightarrow\dfrac{a}{b}-1=\dfrac{a-b}{b}>\dfrac{a-b}{b+n}\left(b< b+n;a-b>0\right)=\dfrac{a+n}{b+n}-1\\ \Leftrightarrow\dfrac{a}{b}>\dfrac{a+n}{b+n}\\ \forall a< b\Leftrightarrow1-\dfrac{a}{b}=\dfrac{b-a}{b}>\dfrac{b-a}{b+n}\left(b< b+n;b-a>0\right)=1-\dfrac{a+n}{b+n}\\ \Leftrightarrow1-\dfrac{a}{b}>1-\dfrac{a+n}{b+n}\Leftrightarrow\dfrac{a}{b}>\dfrac{a+n}{b+n}\\ \forall a=b\Leftrightarrow\dfrac{a+n}{b+n}=\dfrac{a}{b}\left(=1\right)\)

Chọn B. Thay \(\dfrac{1}{3}\)vào x và \(\dfrac{1}{2}\)vào y 

giải để ra được m

a, \(\sqrt{25}-3\sqrt{\dfrac{4}{9}}=5-3.\dfrac{2}{3}=3\)

b, \(\left(2-\dfrac{5}{3}\right):\left(\dfrac{2}{7}+\dfrac{5}{21}-1\right)\)

\(=\dfrac{1}{3}:\dfrac{6+5-21}{21}\)

\(=-\dfrac{1}{3}.\dfrac{21}{10}\)

\(=-\dfrac{7}{10}\)