\(a,A=\dfrac{x+1}{x-2}=\dfrac{x-2+3}{x-2}=1+\dfrac{3}{x-2}\)

Để A nguyên thì: 3 ⋮ x - 2

=> x - 2 ∈ Ư(3) = {1; -1; 3; -3} 

=> x ∈ {3; 1; 5; -1} 

\(b,B=\dfrac{2x-1}{x+5}=\dfrac{\left(2x+10\right)-11}{x+5}=\dfrac{2\left(x+5\right)-11}{x+5}=2-\dfrac{11}{x+5}\)

Để B nguyên thì 11 ⋮ x + 5

=> x + 5 ∈ Ư(11) = {1; -1; 11; -11}

=> x ∈ {-4; -6; 6; -16} 

\(c,C=\dfrac{10x-9}{2x-3}=\dfrac{\left(10x-15\right)+6}{2x-3}=\dfrac{5\left(2x-3\right)+6}{2x-3}=5+\dfrac{6}{2x-3}\)

Để C nguyên thì 6 ⋮ 2x - 3

=> 2x - 3 ∈ Ư(6) = {1; -1; 2; -2; 3; -3; 6; -6}

Mà: 2x - 3 luôn lẻ

=> 2x - 3 ∈ {1; -1; 3; -3}

=> 2x ∈ {4; 2; 6; 0}

=> x ∈ {2; 1; 3; 0} 

\(\left(3x-1\right)^3=\dfrac{-8}{27}\\ =>\left(3x-1\right)^3=\dfrac{\left(-2\right)^3}{3^3}\\ =>\left(3x-1\right)^3=\left(-\dfrac{2}{3}\right)^3\\ =>3x-1=-\dfrac{2}{3}\\ =>3x=-\dfrac{2}{3}+1\\ =>3x=\dfrac{1}{3}\\ =>x=\dfrac{1}{3}:3\\ =>x=\dfrac{1}{9}\)

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

\(\Rightarrow\) \(3x-1=\dfrac{-2}{3}\)

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

\(3x=\dfrac{1}{3}\)

\(x=\dfrac{1}{3}\div3\)

\(x=\dfrac{1}{9}\)

Vậy \(x=\dfrac{1}{9}\)

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: 

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

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

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

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

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

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

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

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

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

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` 

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

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: \(\sqrt{5}-x=\sqrt{7}\)

=>\(x=\sqrt{5}-\sqrt{7}\simeq-0,41\)

b: \(\sqrt{7}+2x=\sqrt{11}\)

=>\(2x=\sqrt{11}-\sqrt{7}\)

=>\(x=\dfrac{1}{2}\left(\sqrt{11}-\sqrt{7}\right)\simeq0,34\)

a: ĐKXĐ: x>=1

\(\sqrt{x-1}=3\)

=>\(x-1=3^2=9\)

=>x=9+1=10(nhận)

b: ĐKXĐ: x>=-1

\(\sqrt{x+1}=\sqrt{4}+\sqrt{9}\)

=>\(\sqrt{x+1}=2+3=5\)

=>x+1=25

=>x=24(nhận)

c: ĐKXĐ: x>=3

\(\sqrt{x-3}=\sqrt{25}-\sqrt{16}\)

=>\(\sqrt{x-3}=5-4=1\)

=>x-3=1

=>x=4(nhận)

sai !

a: \(\sqrt{25}-2\cdot\sqrt{36}+5\cdot\sqrt{49}\)

\(=5-2\cdot6+5\cdot7\)

=5-12+35

=40-12=28

b: \(5\sqrt{100}-\sqrt{9+16}-\sqrt{121}:11\)

\(=5\cdot10-5-11:11\)

=50-5-1=44