\(\frac{25.9-25.17}{-8.80-8.10}=\frac{25\left(9-17\right)}{-8\left(80+10\right)}=\frac{25.\left(-8\right)}{\left(-8\right).90}=\frac{25}{90}=\frac{5}{18}\)

Ta có: \(\left|2x+3\right|\ge0\)

\(\Rightarrow50-\left|2x+3\right|\le50\)

\(\Rightarrow A\le50\)

Dấu "=" xảy ra <=> 2x + 3 = 0 <=> x = \(\frac{-3}{2}\)

Vậy A_Max = 50 khi x = \(\frac{-3}{2}\)

1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 +...+ 1998 - 1999 - 2000 + 2001

= 1 + (2 - 3 - 4 + 5) + (6 - 7 - 8 + 9) +...+ (1998 - 1999 - 2000 + 2001)

= 1 + 0 + 0 +...+ 0

= 1

a, Vì \(\widehat{xOt}\) và \(\widehat{yOt}\) là hai góc kề bù nên:

\(\widehat{yOt}=180^o-\widehat{xOt}=180^o-30^o=150^o\)

Vậy \(\widehat{yOt}=150^o\)

b, Vì \(\widehat{yOt}>\widehat{yOt'}\left(150^o>60^o\right)\) nên tia Ot' nằm giữa 2 tia Oy và Ot

Ta có: \(\widehat{yOt'}+\widehat{tOt'}=\widehat{yOt}\)

\(60^o+\widehat{tOt'}=150^o\)

\(\widehat{tOt'}=150^o-60^o\)

\(\widehat{tOt'}=90^o\)

Vậy \(\widehat{tOt'}=90^o\)

Bài 1:

a, n - 18 chia hết cho 17

=> n - 18 thuộc B(17)

=> n - 18 = 17k (k thuộc N)

=> n = 17k + 18 (k thuộc N)

Vậy n có dạng 17k + 18 (k thuộc N)

Bài 2:

a, 2x - 5 chia hết cho x - 1

=> 2x - 2 - 3 chia hết cho x - 1

=> 2(x - 1) - 3 chia hết cho x - 1

Vì 2(x - 1) chia hết cho x - 1 nên để 2(x - 1) - 3 chia hết cho x - 1 thì 3 chia hết cho x - 1

=> x - 1 thuộc Ư(3) = {1;-1;3;-3}

Vậy x = {2;0;4;-2}

b, x + 1 thuộc Ư(x² + 1)

=> x² + 1 chia hết cho x + 1

=> x² + x - x + 1 chia hết cho x + 1

=> x(x + 1) - x - 1 + 2 chia hết cho x + 1

=> x(x + 1) - (x + 1) + 2 chia hết cho x + 1

Vì x(x + 1) và x + 1 chia hết cho x + 1 nên để x(x + 1) - (x + 1) + 2 chia hết cho x + 1 thì 2 chia hết cho x + 1

=> x + 1 thuộc Ư(2) = {1;-1;2;-2}

Vậy x = {0;-2;1;-3}

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2010}\)

\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2010}\)

\(\Rightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2009}{2010}\)

\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2009}{2010}\)

\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2009}{2010}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{2010}:2\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4020}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2009}{4020}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{4020}\)

\(\Rightarrow x+1=4020\)

=> x = 4020 - 1

=> x = 4019

Khi tia Oy nằm giữa 2 tia Ox và Oz thì \(\widehat{xOy}+\widehat{yOz}=\widehat{xOz}\)

Để \(\frac{5}{x-2}\) là số nguyên <=> x - 2 \(\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Vậy để \(\frac{5}{x-2}\) nguyên thì x = {3;1;7;-3}

Ư(-3) = {1;-1;3;-3}

Vì n - 2 ước của -3 nên ta có:

n - 2 = 1 => n = 3

n - 2 = -1 => n = 1

n - 2 = 3 => n = 5

n - 2 = -3 => n = -1

Vậy n = {3;1;5;-1}

a, x² + 11x = 0

x(x + 11) = 0

\(\Rightarrow\left[\begin{matrix}x=0\\x+11=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=0\\x=-11\end{matrix}\right.\)

b, (x² - 1)(x² - 9) = 0

\(\Rightarrow\left[\begin{matrix}x^2-1=0\\x^2-9=0\end{matrix}\right.\)\(\Rightarrow\left[\begin{matrix}x^2=1\\x^2=9\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=1\\x=-1\\x=3\\x=-3\end{matrix}\right.\)

Vậy \(x\in\left\{1;-1;3;-3\right\}\)

c, ( |x + 1| - 5)(x² - 9) = 0

\(\Rightarrow\left[\begin{matrix}\left|x+1\right|-5=0\\x^2-9=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}\left|x+1\right|=5\\x^2=9\end{matrix}\right.\)

\(\Rightarrow\left[\begin{matrix}x+1=5\\x+1=-5\\x=3\\x=-3\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=4\\x=-6\\x=3\\x=-3\end{matrix}\right.\)

Vậy \(x\in\left\{4;-6;3;-3\right\}\)

d, \(3x-16⋮x+2\)

\(\Rightarrow3x+6-22⋮x+2\)

\(\Rightarrow3\left(x+2\right)-22⋮x+2\)

Vì \(3\left(x+2\right)⋮x+2\) nên để \(3\left(x+2\right)-22⋮x+2\) thì \(22⋮x+2\)

\(\Rightarrow x+2\inƯ\left(22\right)=\left\{\pm1;\pm2;\pm11;\pm22\right\}\)

Vậy x = {-1;-3;0;-4;9;-13;20;-24}