\(D=\left\lbrace m\vert m=n^2\in\mathbb{N}^{\ast},n\le5\right\rbrace\)

7(3x - 15) = 42

3x - 15 = 42 : 7

3x - 15 = 6

3x = 6 + 15

3x = 21

x = 21 : 3

x = 7

Vậy x = 7

\(4x^2(2x-3)=1-6x\)

\(\rArr8x^3-12x^2=1-6x\)

\(\rArr8x^3-12x^2-1+6x=0\)

\(\rArr(2x-1)^3=0\)

\(\rArr2x-1=0\)

\(\rArr x=\dfrac12\)

Vậy \(x=\dfrac12\)

(x - 5)(2x - 12) = 0

Xét 2 TH:

TH1: x - 5 = 0

x = 0 + 5

x = 5

TH2: 2x - 12 = 0

2x = 0 + 12

2x = 12

x = 12 : 2

x = 6

Vậy x ∈ {5 ; 6}

Ta có:

\(\underbrace{10+20+30+40+...+180+190}_\text{có (190 - 10) : 10 + 1 = 19 số hạng}\)

\(=\dfrac{\left(190+10\right)\times19}{2}\)

\(=1900\)

\(\frac12-\left(-\frac25\right)+\frac13+\frac57-\frac16+\left(-\frac{4}{35}\right)+\frac{1}{41}\)

\(=\frac12+\frac25+\frac13+\frac57-\frac16-\frac{4}{35}+\frac{1}{41}\)

\(=\left(\frac12+\frac13-\frac16\right)+\left(\frac25+\frac57-\frac{4}{35}\right)+\frac{1}{41}\)

\(=\frac23+1+\frac{1}{41}\)

\(=\frac{208}{123}\)

\(A=\left(1-\frac12\right)\left(1-\frac13\right)\left(1-\frac14\right)\left(1-\frac15\right)\left(1-\frac16\right)\left(1-\frac17\right)\left(1-\frac18\right)\left(1-\frac19\right)\ldots\left(1-\frac{1}{2023}\right)\)

\(A=\frac12\cdot\frac23\cdot\frac34\cdot\frac45\cdot\frac56\cdot\frac67\cdot\frac78\cdot\frac89\cdot\cdot\cdot\frac{2022}{2023}\)

\(A=\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot\cdot\cdot2022}{2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot\cdot\cdot2023}\)

\(A=\frac{1}{2023}\)

Vậy \(A=\frac{1}{2023}\)

Đặt \(A=\frac{2}{1\times2}+\frac{2}{2\times3}+\cdots+\frac{2}{18\times19}+\frac{2}{19\times20}\)

Ta có:

\(A=2\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\cdots+\frac{1}{18\times19}+\frac{1}{19\times20}\right)\)

\(A=2\times\left(\frac{2-1}{1\times2}+\frac{3-2}{2\times3}+\cdots+\frac{19-18}{18\times19}+\frac{20-19}{19\times20}\right)\)

\(A=2\times\left(\frac{2}{1\times2}-\frac{1}{1\times2}+\frac{3}{2\times3}-\frac{2}{2\times3}+\cdots+\frac{20}{19\times20}-\frac{19}{19\times20}\right)\)

\(A=2\times\left(1-\frac12+\frac12-\frac13+\ldots+\frac{1}{19}-\frac{1}{20}\right)\)

\(A=2\times\left(1-\frac{1}{20}\right)\)

\(A=2\times\frac{19}{20}\)

\(A=\frac{19}{10}\)

Vậy \(A=\frac{19}{10}\)

Vì C nằm giữa A và B nên BC = AB - AC = 7 - 3 = 4 (cm)

Vì M là trung điểm của BC nên BM = BC/2 = 4/2 = 2 (cm)

Vậy BM = 2cm

\(a,\) Để \(\dfrac{6}{x-2}\in\Z\) thì \(6\) \(\vdots\) \(x-2\)

\(\lrArr x-2\in\left\lbrace-1;1;-2;2;-3;3;-6;6\right\rbrace\)

\(\lrArr x\in\left\lbrace1;3;0;4;-1;5;-4;8\right\rbrace\)

Vậy \(x\in\left\lbrace1;3;0;4;-1;5;-4;8\right\rbrace\)

\(b,\) Ta có: \(\dfrac{4x-3}{x+2}=\dfrac{\left(4x+8\right)-8-3}{x+2}=\dfrac{4\left(x+2\right)-11}{x+2}=4-\dfrac{11}{x+2}\in\Z\)

\(\lrArr x+2\in\left\lbrace-1;1;-11;11\right\rbrace\)

\(\lrArr x\in\left\lbrace-3;-1;-13;9\right\rbrace\)

Vậy \(x\in\left\lbrace-3;-1;-13;9\right\rbrace\)