TH1: \(-\dfrac{1}{2}\le x\le\dfrac{5}{3}\)

\(\left(2x+1\right)+\left(5-3x\right)=6\\ =>2x+1+5-3x=6\\ =>\left(2x-3x\right)+6=6\\ =>x=0\left(tm\right)\)

TH2: \(x>\dfrac{5}{3}\)

\(\left(2x+1\right)-\left(5-3x\right)=6\\ =>2x+1-5+3x=6\\ =>2x+3x=6-1+5\\ =>5x=10\\ =>x=\dfrac{10}{5}=2\left(tm\right)\)

TH3: \(x< -\dfrac{1}{2}\)

\(-\left(2x+1\right)+\left(5-3x\right)=6\\ =>-2x-1+5-3x=6\\ =>-2x-3x+4=6\\ =>-5x=6-4=2\\ =>x=-\dfrac{2}{5}\left(ktm\right)\)

\(A=\dfrac{1}{4}:\left(1,5-1\dfrac{5}{6}\right)+1\\ =\dfrac{1}{4}:\left(\dfrac{3}{2}-\dfrac{11}{6}\right)+1\\ =\dfrac{1}{4}:\dfrac{9-11}{6}+1\\ =\dfrac{1}{4}:\dfrac{-2}{6}+1\\ =\dfrac{1}{4}\cdot\left(-3\right)+1\\ =\dfrac{-3}{4}+1\\ =\dfrac{1}{4}\)

Ta có: 

\(y^3+2y^2+y+4=\left(y+2\right)^3-\left(4y^2+11y+4\right)\)

Do y là số tự nhiên \(\Rightarrow4y^2+11y+4>0\Rightarrow\left(y+2\right)^3>y^3+2y^2+y+4\)

Đồng thời cũng do y tự nhiên \(\Rightarrow2y^2+y+4>0\Rightarrow y^3+2y^2+y+4>y^3\)

\(\Rightarrow y^3< y^3+2y^2+y+4< \left(y+2\right)^3\)

\(\Rightarrow y^3+2y^2+y+4\) là lập phương của 1 số tự nhiên khi và chỉ khi:

\(y^3+2y^2+y+4=\left(y+1\right)^3\)

\(\Leftrightarrow y^2+2y-3=0\Rightarrow y=1\)

\(\Rightarrow x^3=8\Rightarrow x=2\)

Ta có:

`(2x-5)^2022>=0` với mọi x 

`(3y-4)^2024>=0` với mọi y

`=>(2x-5)^2022+(3y-4)^2024>=0` với mọi x,y 

Mặt khác: `(2x-5)^2022+(3y-4)^2024<=0` 

`=>2x-5=0` và `3y-4=0`

`=>x=5/2` và `y=4/3` 

\(P+\left(5\cdot\dfrac{5}{2}-2\cdot\dfrac{4}{3}\right)=6\cdot\left(\dfrac{5}{2}\right)^2+9\cdot\dfrac{5}{2}\cdot\dfrac{4}{3}-\left(\dfrac{4}{3}\right)^2\\ =>P+\dfrac{59}{6}=\dfrac{1183}{18}\\ =>P=\dfrac{1183}{18}-\dfrac{59}{6}\\ =>P=\dfrac{503}{9}\)

\(\dfrac{1}{99\cdot97}-\dfrac{1}{97\cdot95}-...-\dfrac{1}{5\cdot3}-\dfrac{1}{3\cdot1}\\ =\dfrac{1}{99\cdot97}-\left(\dfrac{1}{97\cdot95}+...+\dfrac{1}{5\cdot3}+\dfrac{1}{3\cdot1}\right)\\ =\dfrac{1}{99\cdot97}-\left(\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{93}-\dfrac{1}{95}+...+\dfrac{1}{3}-\dfrac{1}{5}+1-\dfrac{1}{3}\right)\\ =\dfrac{1}{99\cdot97}-\left(-\dfrac{1}{97}+1\right)\\ =\dfrac{1}{99}-\dfrac{1}{97}+\dfrac{1}{97}-1\\ =\dfrac{1}{99}-1\\ =-\dfrac{98}{99}\)

\(\dfrac{11}{48}=\dfrac{11.3}{48.3}=\dfrac{33}{144}\)

\(\dfrac{17}{36}=\dfrac{17.4}{36.4}=\dfrac{68}{144}\)

Do \(68>33\Rightarrow\dfrac{68}{144}>\dfrac{33}{144}\Rightarrow\dfrac{17}{36}>\dfrac{11}{48}\)

\(\left(x^2y^2+2x^2y+x^2\right)\left(x^2+2x-2\right)=\left(xy+x\right)^2\left(x^2+2x-2\right)\)

Do \(\left(xy+x\right)^2\) chính phương với mọi x;y nguyên nên biểu thức đã cho chính phương khi \(x^2+2x-2\) là SCP

\(\Rightarrow x^2+2x-2=k^2\)

\(\Rightarrow\left(x+1\right)^2-k^2=3\)

\(\Rightarrow\left(x+1-k\right)\left(x+1+k\right)=3\)

Pt ước số cơ bản, dễ dàng giải ra \(x=\left\{-3;1\right\}\)

Vậy \(\left(x;y\right)=\left(-3;k\right);\left(1;k\right)\) với k là số nguyên bất kì

\(2\left|\dfrac{1}{2}-\dfrac{3}{4}\right|+\sqrt{\dfrac{4}{9}}\\ =2\left|\dfrac{2}{4}-\dfrac{3}{4}\right|+\sqrt{\left(\dfrac{2}{3}\right)^2}\\ =2\left|\dfrac{-1}{4}\right|+\dfrac{2}{3}\\ =2\cdot\dfrac{1}{4}+\dfrac{2}{3}\\ =\dfrac{1}{2}+\dfrac{2}{3}\\ =\dfrac{7}{6}\)

\(2\left|\dfrac{1}{2}-\dfrac{3}{4}\right|+\sqrt{\dfrac{4}{9}}\)

\(=2\left|\dfrac{2}{4}-\dfrac{3}{4}\right|+\sqrt{\left(\dfrac{2}{3}\right)^2}\)

\(=2\left|-\dfrac{1}{4}\right|+\dfrac{2}{3}\)

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

\(=\dfrac{1}{2}+\dfrac{2}{3}\)

\(=\dfrac{3}{6}+\dfrac{4}{6}\)

\(=\dfrac{7}{6}\)

a) Ta có:

\(13>12=>\dfrac{13}{40}>\dfrac{12}{40}=>\dfrac{-13}{40}< \dfrac{-12}{40}\)

b) Ta có: 

\(\dfrac{-91}{104}=\dfrac{-13}{14}=\dfrac{1}{14}-1< \dfrac{1}{6}-1=\dfrac{-5}{6}\) 

c) Ta có: 

\(\dfrac{-15}{21}=\dfrac{-5}{7}=1-\dfrac{2}{7}\\ \dfrac{-36}{44}=\dfrac{-9}{11}=1-\dfrac{2}{11}\)

Mà: \(\dfrac{2}{7}>\dfrac{2}{11}=>\dfrac{-2}{7}< \dfrac{-2}{11}=>1-\dfrac{2}{7}< 1-\dfrac{2}{11}=>-\dfrac{15}{21}< \dfrac{-36}{44}\) 

d) Ta có:

\(\dfrac{-16}{30}=\dfrac{-8}{15}=\dfrac{7}{15}-1\\ \dfrac{-35}{84}=\dfrac{-5}{12}=\dfrac{7}{12}-1\)

Mà: \(\dfrac{7}{15}< \dfrac{7}{12}=>\dfrac{7}{15}-1< \dfrac{7}{12}-1=>-\dfrac{16}{30}< \dfrac{-35}{84}\)

e) Ta có:

\(\dfrac{-5}{91}=\dfrac{-5\cdot101}{91\cdot101}=\dfrac{-505}{9191}< \dfrac{-501}{9191}\)

f) Ta có: 

\(\dfrac{-11}{3^7\cdot7^3}=\dfrac{-11\cdot7}{3^7\cdot7^3\cdot7}=\dfrac{-77}{3^7\cdot7^4}>\dfrac{-78}{3^7\cdot7^4}\)