Nếu đề bài là: (\(\dfrac{2}{3}\))⁵ x \(\dfrac{1}{2^5}\) thì làm như này em nhé.

(\(\dfrac{2}{3}\))⁵ x \(\dfrac{1}{2^5}\) = (\(\dfrac{2}{3}\) x \(\dfrac{1}{2}\))⁵ = (\(\dfrac{1}{3}\))⁵  

\(\dfrac{11}{8}\left[\left(\dfrac{-5}{11}:\dfrac{13}{8}-\dfrac{5}{11}:\dfrac{13}{5}\right)+\dfrac{-6}{33}\right]+\dfrac{3}{4}\\ =\dfrac{11}{8}\left[\left(\dfrac{-5}{11}\cdot\dfrac{8}{13}+\dfrac{-5}{11}\cdot\dfrac{5}{13}\right)+\dfrac{-2}{11}\right]+\dfrac{3}{4}\\ =\dfrac{11}{8}\left[\dfrac{-5}{11}\cdot\left(\dfrac{8}{13}+\dfrac{5}{13}\right)+\dfrac{-2}{11}\right]+\dfrac{3}{4}\\ =\dfrac{11}{8}\left(\dfrac{-5}{11}+\dfrac{-2}{11}\right)+\dfrac{3}{4}\\ =\dfrac{11}{8}\cdot\dfrac{-7}{11}+\dfrac{3}{4}\\ =\dfrac{-7}{8}+\dfrac{6}{8}\\ =\dfrac{-1}{8}\)

Bài 3:

\(a.\dfrac{3x+2}{x-3}=\dfrac{3x-9+11}{x-3}=\dfrac{3\left(x-3\right)+11}{x-3}=3+\dfrac{11}{x-3}\)

Để bt nguyên thì 11 ⋮ x - 3

=> x - 3 ∈ Ư(11) = {1; -1; 11; -11} 

=> x ∈ {4; 2; 14; -8} 

\(b.\dfrac{x-2}{3x-2}\) nguyên \(\Rightarrow\dfrac{3\left(x-2\right)}{3x-2}=\dfrac{3x-6}{3x-2}=\dfrac{3x-2-4}{3x-2}=1-\dfrac{4}{3x-2}\)

Để bt nguyên thì: 4 ⋮ 3x - 2

=> 3x - 2 ∈ Ư(4) = {1; -1; 2; -2; 4; -4}

=> 3x ∈ {3; 1; 4; 0; 5; -2} 

=> x ∈ {1; `1/3`; `4/3`; 0;`5/3`;`-2/3`} 

Mà: x nguyên => x ∈ {1;0}

Bài 4:

\(A=3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3+1\)

=>\(3A=3^{101}-3^{100}+3^{99}-3^{98}+...+3^3-3^2+3\)

=>\(3A+A=3^{101}-3^{100}+3^{99}-3^{98}+...+3^3-3^2+3+3^{100}-3^{99}+...+3^2-3+1\)

=>\(4A=3^{101}+1\)

=>\(A=\dfrac{3^{101}+1}{4}\)

a) 

\(\dfrac{24\cdot47-23}{24+47\cdot23}\cdot\dfrac{3+\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{101}-\dfrac{3}{13}}{\dfrac{6}{101}-\dfrac{6}{13}+\dfrac{6}{7}-\dfrac{6}{11}+6}\\ =\dfrac{\left(23+1\right)\cdot47-23}{24+47\cdot23}\cdot\dfrac{3\left(1+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{101}-\dfrac{1}{13}\right)}{6\left(1+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{101}-\dfrac{1}{13}\right)}\\ =\dfrac{23\cdot47+47-23}{23\cdot47+24}\cdot\dfrac{3}{6}\\ =\dfrac{23\cdot47+24}{23\cdot47+24}\cdot\dfrac{1}{2}\\ =1\cdot\dfrac{1}{2}=\dfrac{1}{2}\)

b) 

\(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{19\cdot21}\\ =\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{3}{19\cdot21}\right)\\ =\dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}\cdot\left(1-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}\cdot\dfrac{20}{21}\\ =\dfrac{10}{21}\)

Dễ dàng so sánh được:

\(-\dfrac{5}{3}< -\dfrac{3}{3}=-1=-\dfrac{5}{5}< -\dfrac{4}{5}< 0< \dfrac{1}{3}< \dfrac{5}{3}< \dfrac{6}{3}=2=\dfrac{4}{2}< \dfrac{7}{2}\)

Vẽ trục số:

Xét ΔADE vuông tại D có \(cosA=\dfrac{AD}{AE}\)

=>\(\dfrac{AD}{4}=cos30=\dfrac{\sqrt{3}}{2}\)

=>\(AD=4\cdot\dfrac{\sqrt{3}}{2}=2\sqrt{3}\left(cm\right)\)

D là trung điểm của AB

=>\(AB=2\cdot AD=4\sqrt{3}\left(cm\right)\)

Xét ΔABC vuông tại C có \(sinA=\dfrac{BC}{AB}\)

=>\(\dfrac{BC}{4\sqrt{3}}=sin30=\dfrac{1}{2}\)

=>\(BC=\dfrac{4\sqrt{3}}{2}=2\sqrt{3}\left(cm\right)\)

Xét ΔABD và ΔAED có

AB=AE

\(\widehat{BAD}=\widehat{EAD}\) 

AD chung

Do đó: ΔABD=ΔAED

=>DB=DE

=>D nằm trên đường trung trực của BE(1)

Ta có: AB=AE
=>A nằm trên đường trung trực của BE(2)

Từ (1),(2) suy ra AD là đường trung trực của BE

=>AD\(\perp\)BE

Ta có: I là trung điểm AB =>IC là đường trung tuyến

          K là trung điểm của AC =>KB là đường trung tuyến

Vì IC và KB cắt nhau tại H nên H là trực tâm

=> AG là đường trung trực đi qua H

=> A,H,G thẳng hàng

\(\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\\ =\dfrac{1}{3\cdot6}+\dfrac{1}{6\cdot9}+...+\dfrac{1}{30\cdot33}\\ =\dfrac{1}{3}\cdot\left(\dfrac{3}{3\cdot6}+\dfrac{3}{6\cdot9}+...+\dfrac{3}{30\cdot33}\right)\\ =\dfrac{1}{3}\cdot\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\\ =\dfrac{1}{3}\cdot\left(\dfrac{1}{3}-\dfrac{1}{33}\right)\\ =\dfrac{1}{3}\cdot\dfrac{11-1}{33}\\ =\dfrac{10}{99}\)