Lời giải:

\(B=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+....+\frac{2021}{4^{2021}}\)

\(4B=1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2021}{4^{2020}}\)

\(4B-B=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2020}}-\frac{2021}{4^{2021}}\)

\(3B=1+\frac{1}{4}+\frac{1}{4^2}+....+\frac{1}{4^{2020}}-\frac{2021}{4^{2021}}\)

\(12B=4+1+\frac{1}{4}+...+\frac{1}{4^{2019}}-\frac{2021}{4^{2020}}\)

\(9B=4-\frac{6067}{4^{2021}}<4\Rightarrow B< \frac{4}{9}< \frac{1}{2}\)

Đặt \(\dfrac{a}{b^2}=\dfrac{b^2}{c^3}=\dfrac{c^3}{a^4}=k\)

\(\Rightarrow\left\{{}\begin{matrix}a=k.b^2\\b^2=k.c^3\\c^3=k.a^4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=k.k.c^3=k^2c^3\\c^3=k.a^4\end{matrix}\right.\)

\(\Rightarrow a=k^2.k.a^4\)

\(\Rightarrow a=k^3a^4\)

\(\Rightarrow\left(ka\right)^3=1\)

\(\Rightarrow ka=1\)

\(\Rightarrow a=\dfrac{1}{k}\) (1)

Thế vào \(c^3=k.a^4\Rightarrow c^3=k.\dfrac{1}{k^4}=\dfrac{1}{k^3}\)

\(\Rightarrow c=\dfrac{1}{k}\) (2)

Thế vào \(b^2=kc^3\Rightarrow b^2=k.\dfrac{1}{k^3}=\dfrac{1}{k^2}\)

\(\Rightarrow b=\dfrac{1}{k}\) hoặc \(b=-\dfrac{1}{k}\) (3)

(1);(2);(3) \(\Rightarrow\left[{}\begin{matrix}a=b=c\\a=c=-b\end{matrix}\right.\)

TH1: \(a=b=c\)

\(\Rightarrow P=\left(1+\dfrac{a}{a}\right)\left(1+\dfrac{a}{a}\right)\left(1+\dfrac{a}{a}\right)=2.2.2=8\)

Th2: \(a=c=-b\)

\(\Rightarrow P=\left(1+\dfrac{-b}{b}\right)\left(1+\dfrac{b}{-b}\right)\left(1+\dfrac{-b}{-b}\right)=0.0.2=0\)

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.

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

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

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

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 

a)

`2/3+5/2-3/4`

`=10/4-3/4+2/3`

`=7/4+2/3`

`=21/12+8/12`

`=29/12`

b)

`2/5xx1/2:1/3`

`=2/10xx3/1`

`=6/10=3/5`

c)

`2/9:2/9xx1/3`

`=2/9xx9/2xx1/3`

`=1xx1/3`

`=1/3`

a, \(\dfrac{2}{3}\) + \(\dfrac{5}{2}\) - \(\dfrac{3}{4}\)

= \(\dfrac{8}{12}\) + \(\dfrac{30}{12}\) - \(\dfrac{9}{12}\)

= \(\dfrac{38-9}{12}\)

= \(\dfrac{29}{12}\)

b, \(\dfrac{2}{5}\) x \(\dfrac{1}{2}\) : \(\dfrac{1}{3}\)

= \(\dfrac{1}{5}\) x \(\dfrac{3}{1}\)

= \(\dfrac{3}{5}\)

c, \(\dfrac{2}{9}\) : \(\dfrac{2}{9}\) x \(\dfrac{1}{3}\)

= 1 x \(\dfrac{1}{3}\)

= \(\dfrac{1}{3}\)

a) Ta có: \(\dfrac{3a^2-10a+3}{2\left(a-3\right)}\)

\(=\dfrac{3a^2-9a-a+3}{2\left(a-3\right)}\)

\(=\dfrac{3a\left(a-3\right)-\left(a-3\right)}{2\left(a-3\right)}\)

\(=\dfrac{\left(a-3\right)\left(3a-1\right)}{2\left(a-3\right)}\)

\(=\dfrac{3a-1}{2}\)

\(=\dfrac{3}{2}a-\dfrac{1}{2}\)(đpcm)

b) Ta có: \(\dfrac{b^2+3b+9}{b^3-27}\)\(=\dfrac{b^2+3b+9}{\left(b-3\right)\left(b^2+3b+9\right)}\)

\(=\dfrac{1}{b-3}\)

\(=\dfrac{b-2}{\left(b-3\right)\left(b-2\right)}\)

\(=\dfrac{b-2}{b^2-5b+6}\)(đpcm)

Đề bài yêu cầu gì?