`(2x+3)^2 =` \(\dfrac{9}{121}\)

`=> (2x + 3)^2 =` \(\left(\dfrac{3}{11}\right)^2\)

=> \(\left[{}\begin{matrix}2x+3=\dfrac{3}{11}\\2x+3=-\dfrac{3}{11}\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}2x=\dfrac{3}{11}-3\\2x=-\dfrac{3}{11}-3\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}2x=\dfrac{3}{11}-\dfrac{33}{11}\\2x=-\dfrac{3}{11}-\dfrac{33}{11}\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}2x=-\dfrac{30}{11}\\2x=-\dfrac{36}{11}\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-\dfrac{15}{11}\\x=-\dfrac{18}{11}\end{matrix}\right.\)

Vậy ...

\(\left(2x+3\right)^2=\dfrac{9}{121}\)

\(\left(2x+3\right)^2=\left(\dfrac{3}{11}\right)^2\)

\(2x+3=\dfrac{2}{11}\) hoặc \(2x+3=-\dfrac{2}{11}\)

\(2x=-\dfrac{31}{11}\) hoặc \(2x=-\dfrac{35}{11}\)

\(x=-\dfrac{31}{22}\) hoặc \(x=-\dfrac{35}{22}\)

3: Ta có: Om là phân giác của góc xOz

=>\(\widehat{mOz}=\dfrac{\widehat{xOz}}{2}\)

On là phân giác của góc yOz

=>\(\widehat{nOz}=\dfrac{\widehat{yOz}}{2}\)

\(\widehat{mOn}=\widehat{mOz}+\widehat{nOz}=\dfrac{1}{2}\left(\widehat{xOz}+\widehat{yOz}\right)\)

\(=\dfrac{1}{2}\cdot180^0=90^0\)

4: Om là phân giác của góc xOz

=>\(\widehat{zOm}=\dfrac{\widehat{xOz}}{2}\)

On là phân giác của góc yOz

=>\(\widehat{zOn}=\dfrac{\widehat{yOz}}{2}\)

\(\widehat{mOn}=\widehat{mOz}+\widehat{nOz}=\dfrac{1}{2}\left(\widehat{xOz}+\widehat{yOz}\right)\)

\(=\dfrac{1}{2}\cdot180^0=90^0\)

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

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

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

a: Ta có: Ox\(\perp\)Oy

Ox\(\perp\)Az

Do đó: Oy//Az

b: Om không song song với Am nha bạn

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

\(\sqrt{2}+x=\sqrt{3}\)

=>\(x=\sqrt{3}-\sqrt{2}\simeq0,32\)

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

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

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

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}