\(A=2^1+2^2+2^3+...+2^{10}\)

\(\Rightarrow2A=2\cdot\left(2+2^2+2^3+...+2^{10}\right)\)

\(\Rightarrow2A=2^2+2^3+...+2^{11}\)

\(\Rightarrow2A-A=\left(2^2+2^3+...+2^{11}\right)-\left(2+2^2+...2^{10}\right)\)

\(\Rightarrow A=2^{11}-2\) 

\(B=3^1+3^2+...+3^{100}\)

\(\Rightarrow3B=3\cdot\left(3+3^2+...+3^{100}\right)\)

\(\Rightarrow3B=3^2+3^3+...+3^{101}\)

\(\Rightarrow3B-B=\left(3^2+3^3+...+3^{101}\right)-\left(3+3^2+3^3+...+3^{100}\right)\)

\(\Rightarrow2B=3^{101}-3\)

\(\Rightarrow B=\dfrac{3^{101}-3}{2}\)

phần B thiếu 3 mũ 3 ak

Ta có:

\(\dfrac{20}{45}=\dfrac{5}{9}\)

\(\dfrac{14}{35}=\dfrac{2}{5}\)

\(\dfrac{32}{44}=\dfrac{8}{11}\) 

Mẫu số chung là:  495

\(\dfrac{5}{9}=\dfrac{5\cdot55}{9\cdot55}=\dfrac{275}{495}\)

\(\dfrac{2}{5}=\dfrac{2\cdot99}{5\cdot99}=\dfrac{198}{495}\)

\(\dfrac{8}{11}=\dfrac{8\cdot45}{11\cdot45}=\dfrac{360}{495}\)

 \(\dfrac{20}{45}\) = \(\dfrac{4}{9}\) = \(\dfrac{4.5.11}{9.5.11}\) = \(\dfrac{220}{495}\)

 \(\dfrac{14}{35}\) = \(\dfrac{2}{5}\) = \(\dfrac{2.9.11}{9.5.11}\) = \(\dfrac{198}{495}\)

  \(\dfrac{32}{44}\) = \(\dfrac{8}{11}\) = \(\dfrac{8.9.5}{9.5.11}\)= \(\dfrac{360}{495}\)

Ta có:

\(\dfrac{54}{90}=\dfrac{3}{5}\)

\(\dfrac{180}{288}=\dfrac{5}{8}\)

\(\dfrac{60}{135}=\dfrac{4}{9}\) 

Có mẫu số chung là 360

\(\dfrac{3}{5}=\dfrac{3\cdot72}{5\cdot72}=\dfrac{216}{360}\)

\(\dfrac{5}{8}=\dfrac{5\cdot45}{8\cdot45}=\dfrac{225}{360}\)

\(\dfrac{4}{9}=\dfrac{4\cdot40}{9\cdot40}=\dfrac{160}{360}\)

a) \(\widehat{AOB}=\widehat{AOM}+\widehat{MOB}\)

\(\Rightarrow\widehat{MOB}=\widehat{AOB}-\widehat{AOM}\)

mà \(\widehat{AOM}=90^o\) (OM vuông góc OA)

\(\Rightarrow\widehat{MOB}=100^o-90^o\)

\(\Rightarrow\widehat{MOB}=10^o\)

Câu b bạn xem lại đề

b) \(\widehat{B'OA}+\widehat{AOB}=180^o\) (2 góc kề bù)

\(\Rightarrow\widehat{B'OA}=180^o-\widehat{AOB}\)

\(\Rightarrow\widehat{B'OA}=180^o-100^o\)

\(\Rightarrow\widehat{B'OA}=80^o\)

\(\widehat{MOB'}=\widehat{B'OA}+\widehat{AOM}\)

\(\Rightarrow\widehat{MOB'}=80^o+90^o\)

\(\Rightarrow\widehat{MOB'}=170^o\)

a) x ⋮ 2 và x < 19 nên

x ∈ {0; 2; 4; ...; 16; 18}

b) x ⋮ 2 và 28 < x < 40 nên

x ∈ {30; 32; 34; 36; 38}

c) x ⋮ 3 và 45 < x ≤ 63 nên

x ∈ {48; 51; 54; 57; 60; 63}

d) x ⋮ 5 và 152 < x < 175 nên

x ∈ {155; 160; 164; 170}

1. Gọi số cần tìm là: abcdef

Ta có: abcde4

         4abcde = abcde4 . 4

         4 . 100000 + abcde = (abcde . 10 + 4)

Đạt abcde = x, ta có:

         400000 + x = (x . 10 + 4) . 4

         400000 + x = x . 10 . 4 + 4 . 4

         400000 + x = x . 40 + 16

         400000 - 16 = x . 40 - x 

               399984 = x . 39

                x = 399984 : 39

                x =          102564

Vậy số cần tìm là: 102564

có thể lập được 27 số

\(MSC\left(150;50;75\right)=150\)

\(\dfrac{17}{150}\)

\(\dfrac{7}{50}=\dfrac{14}{150}\)

\(\dfrac{4}{75}=\dfrac{16}{150}\)

Đính chính \(\dfrac{7}{50}=\dfrac{21}{150}\)

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

\(\Rightarrow C=-2\left|\dfrac{1}{3}x+4\right|+\dfrac{5}{3}\)

mà \(-2\left|\dfrac{1}{3}x+4\right|\le0,\forall x\)

\(\Rightarrow C=-2\left|\dfrac{1}{3}x+4\right|+\dfrac{5}{3}\le\dfrac{5}{3}\)

\(\Rightarrow GTLN\left(C\right)=\dfrac{5}{3}\left(tạix=-12\right)\)

Lời giải:

a.

$3\leq x< 10$

$\Rightarrow x\in\left\{3; 4; 5;6 ;7; 8;9\right\}$

b.

$24< x< 27$

$\Rightarrow x\in\left\{25; 26\right\}$

c.

$x\vdots 2; x<10$

$\Rightarrow x\in\left\{0; 2; 4; 6; 8\right\}$

d.

$x\in\left\{1;2;4;5;7;8;;10;11;13;14\right\}$