Bài 1:

A = \(\frac{x+4}{x-7}\) (7 ≠ \(x\) ∈ Z)

A ∈ Z ⇔ (\(x+4\)) ⋮ (\(x\) - 7)

[(\(x-7\)) + 11] ⋮ (\(x\) - 7)

11 ⋮ (\(x\) - 7)

(\(x-7\)) ∈ Ư(11) = {-11; - 1; 1 ; 11}

Lập bảng ta có:

Theo bảng trên ta có: \(x\in\) {-4; 6; 8; 18}

Vậy \(x\) ∈ {-4; 6; 8; 18}

Bài 2:

B = \(\frac{2x+3}{2x-1}\) (\(x\) ∈ Z; B ∈ Z)

B ∈ Z ⇔ (2\(x+3\)) ⋮ (2\(x\) - 1)

[(2\(x-1\)) + 4] ⋮ (2\(x\) - 1)

4 ⋮ (2\(x-1\))

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

Lập bảng ta có:

Theo bảng trên ta có:

\(x\) ∈ {0; 1}

Vậy \(x\) ∈ {0; 1}

\(\frac{11}{30}\) + \(\frac{18}{35}\) x (\(\frac{35}{54}\) - \(\frac{49}{18}\) - \(\frac{28}{48}\))

= \(\frac{11}{30}\) + \(\frac{18}{35}\) x \(\frac{35}{54}\) - \(\frac{18}{35}\) x \(\frac{49}{18}\) - \(\frac{18}{35}\) x \(\frac{28}{48}\)

= \(\frac{11}{30}\) + \(\frac13\) - \(\frac75\) - \(\frac{3}{10}\)

= \(\frac{11}{30}\) + \(\frac{1\times10}{3\times10}\) - \(\frac{7\times6}{5\times6}\) - \(\frac{3\times3}{10\times3}\)

= \(\frac{11}{30}\) + \(\frac{10}{30}\) - \(\frac{42}{30}\) - \(\frac{9}{30}\)

= \(\frac{21}{30}\) - \(\frac{42}{30}\) - \(\frac{9}{30}\)

= \(\frac{-21}{30}\) - \(\frac{9}{30}\)

= \(\frac{-30}{30}\)

= 1

=1

(2\(x\) - 1)\(^3\) = - \(\frac{8}{64}\)

(2\(x-1\))\(^3\) = (- \(\frac24\))\(^3\)

2\(x-1\) = - \(\frac24\)

2\(x\) = - \(\frac24\) + 1

2\(x\) = \(\frac12\)

\(x=\frac12:2\)

\(x\) = \(\frac14\)

Vậy \(x=\frac14\)

- \(\frac{5}{17}\) x (\(\frac35x\) - 0,75) = 0

\(\frac35x\) - 0,75 = 0

\(\frac35\)\(x\) = 0,75

\(x=0,75:\frac35\)

\(x\) = \(\frac54\) \(\)

Vậy \(x=\frac54\)

x=5/4

(\(x+2)\)\(^2\) - 9

= (\(x+2\))(\(x+2\)) -9

= \(x\left(x+2\right)\) + 2(\(x\) + 2) - 9

= \(x^2\) + 2\(x\) + 2\(x\) + 4 - 9

= \(x^2\) + (2\(x+2x\)) - (9 - 4)

= \(x^2\) + 4\(x\) - 5

(x + 2)\(^2\) - 9

= (x + 2) . (x + 2) - 9

= x (x + 2) + 2 (x + 2) - 9

= (\(x^2\) + 2x) + (2x + 4) - 9

= \(x^2\) + 2x + 2x + 4 - 9

= \(x^2\) + 4x + 4 - 9

= \(x^2\) + 4x - 5

1,8 nek bạn

= 9/5 = 1,8 nha

\(1-\frac{4}{1\cdot3}-\frac{4}{3\cdot5}-\frac{4}{5\cdot7}-\cdots-\frac{4}{99\cdot101}\)

\(=1-2\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\cdots+\frac{2}{99\cdot101}\right)\)

\(=1-2\cdot\left(\frac11-\frac13+\frac13-\frac15+\cdots+\frac{1}{99}-\frac{1}{101}\right)\)

\(=1-2\cdot\left(\frac11-\frac{1}{101}\right)=1-2\cdot\frac{100}{101}\)

\(=1-\frac{200}{101}=\frac{101}{101}-\frac{200}{101}=-\frac{99}{101}\)

vậy \(B=-\frac{99}{101}\)

Bài 1:

\(\frac{a}{b}=\frac{c}{d}\)

\(\frac{a}{b}+2=\frac{c}{d}+2\)

\(\frac{a+2b}{b}\) = \(\frac{c+2d}{d}\) (đpcm)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=>a=bk; c=dk

\(\frac{a+2b}{b}=\frac{bk+2b}{b}=\frac{b\left(k+2\right)}{b}=k+2\)

\(\frac{c+2d}{d}=\frac{dk+2d}{d}=\frac{d\left(k+2\right)}{d}=k+2\)

Do đó: \(\frac{a+2b}{b}=\frac{c+2d}{d}\)

(2x-3^2)=16

2x-9 =16

2x =16+9

2x =25

x =25.2

x =50

2\(x-3^2\) = 16

2\(x-9\) = 16

\(2x=16+9\)

2\(x\) = 25

\(x=25:2\)

\(x=\frac{25}{2}\)

Vậy \(x=\frac{25}{2}\)

C={x∈N|x=6k+1; 0<=k<=10; k∈N}

Tính chất đặc trưng của dãy số này là đều là số lẻ nha bạn.