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

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

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

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

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

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

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

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

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: 

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

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

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

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

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

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

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

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

Vậy \(x=\dfrac{1}{9}\)

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

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} 

\(a,1,7-\dfrac{5}{9}+\dfrac{3}{10}\\ =\left(\dfrac{17}{10}+\dfrac{3}{10}\right)-\dfrac{5}{9}\\ =2-\dfrac{5}{9}\\ =\dfrac{13}{9}\\ b,-5,2+\left(-\dfrac{2}{9}\right)-\left(-4,2\right)\\ =\left(-5,2+4,2\right)+\dfrac{-2}{9}\\ =-1+\dfrac{-2}{9}\\ =-\dfrac{11}{9}\\ c,2,53-\dfrac{4}{11}+2,47-\dfrac{7}{11}\\ =\left(2,53+2,47\right)+\left(\dfrac{-4}{11}+\dfrac{-7}{11}\right)\\ =5-\dfrac{11}{11}\\ =5-1\\ =4\\ d,-\dfrac{-19}{13}+0,7+\dfrac{7}{13}-\dfrac{17}{10}\\ =\left(\dfrac{7}{10}-\dfrac{17}{10}\right)+\left(\dfrac{19}{13}+\dfrac{7}{13}\right)\\ =-\dfrac{10}{10}+\dfrac{13}{13}\\ =-1+1\\ =0\)

\(e,-\left(-3,498\right)+\dfrac{5}{13}+1,502-\dfrac{18}{13}\\ =3,498+\left(\dfrac{5}{13}-\dfrac{18}{13}\right)+1,502\\ =\left(3,498+1,502\right)+\dfrac{-13}{13}\\ =5-1\\ =4\\ f,4,2+\left(\dfrac{-9}{15}\right)-\left(-5,8\right)\\ =\left(4,2+5,8\right)+\left(\dfrac{-3}{5}\right)\\ =10+\left(\dfrac{-3}{5}\right)\\ =\dfrac{47}{5}\\ g,-\dfrac{5}{9}-0,385+\left(\dfrac{-4}{9}\right)+1,385\\ =\left(-0,385+1,385\right)+\left(\dfrac{-5}{9}+\dfrac{-4}{9}\right)\\ =1-1\\ =0\\ h,3,75-\dfrac{5}{9}+\left(\dfrac{-3}{4}\right)\\ =\left(3,75-0,75\right)-\dfrac{5}{9}\\ =3-\dfrac{5}{9}\\ =\dfrac{22}{9}\)

Ta có: \(\widehat{xOm}=\dfrac{\widehat{xOb}}{2}\)

\(\widehat{yOn}=\dfrac{\widehat{yOa}}{2}\)

mà \(\widehat{xOb}=\widehat{yOa}\)(hai góc đối đỉnh)

nên \(\widehat{xOm}=\widehat{yOn}\)

mà \(\widehat{xOm}+\widehat{mOy}=180^0\)(hai góc kề bù)

nên \(\widehat{yOm}+\widehat{yOn}=180^0\)

=>Om và On là hai tia đối nhau

a: Ta có: Ox\(\perp\)Oy

Ox\(\perp\)Az

Do đó: Oy//Az

b: Om không song song với Am nha bạn

\(BC=\sqrt{AB^2+AC^2}=\sqrt{3^2+4^2}=5\left(cm\right)\)

\(=>S_{xq}=\left(AB+AC+BC\right)\cdot CF=\left(3+4+5\right)\cdot7=84\left(cm^2\right)\) 

Diện tích hai mặt đáy là: \(2S_đ=2\cdot\dfrac{1}{2}\cdot AB\cdot AC=2\cdot\dfrac{1}{2}\cdot3\cdot4=12\left(cm^2\right)\)

Diện tích toàn phần: \(S_{tp}=2S_đ+S_{xq}=84+12=96\left(cm^2\right)\) 

Thể tích của hình lăng trụ là:

\(V=CF\cdot S_đ=7\cdot\dfrac{12}{2}=42\left(cm^3\right)\)

ΔABC vuông tại A

=>\(BC^2=AB^2+AC^2\)

=>\(BC=\sqrt{3^2+4^2}=5\left(cm\right)\)

Chu vi đáy là 3+4+5=12(cm)

Diện tích xung quanh là:

\(S_{xq}=12\cdot7=84\left(cm^2\right)\)

Diện tích đáy là: \(S_{đáy}=\dfrac{1}{2}\cdot3\cdot4=6\left(cm^2\right)\)

Diện tích toàn phần là: \(S_{tp}=S_{xq}+2\cdot S_{đáy}=84+2\cdot6=96\left(cm^2\right)\)

Thể tích là: \(V=S_{đáy}\cdot7=6\cdot7=42\left(cm^3\right)\)