cho anh hỏi cái ba chấm là cái chi mô??? hc nhiều nên hơi trốc tru xíu

Ta có: x + 7 ⋮ x + 2

⇒ ( x + 2 ) + 5 ⋮ x + 2

⇒ 5 ⋮ x + 2

⇒ x + 2 U ( 5 ) = { − 1 ; 1 ; − 5 ; 5 }

\(\Rightarrow\left\{{}\begin{matrix}x+2=-1\Rightarrow x=-3\\x+2=1\Rightarrow x=-1\\x+2=-5\Rightarrow x=-7\\x+2=5\Rightarrow x=3\end{matrix}\right.\)

vậy \(x\in\left\{\text{ − 3 ; − 1 ; − 7 ; 3 }\right\}\)

b) Ta có: x + 5 \(⋮\) x + 1

(x + 1) + 4 \(⋮\) x + 1

\(\Rightarrow\left\{{}\begin{matrix}x+1⋮x+1\\4⋮x+1\end{matrix}\right.\)

4 \(⋮\) x + 1 \(\Rightarrow\) (x + 1) \(\in\) Ư (4) = \(\left\{-4;-2;-1;1;2;4\right\}\)

\(\Rightarrow x\in\left\{-5;-3;-2;0;1;3\right\}\)

=>700+10a+5+800+10b+4 chia hết cho 9 và a-b=6

=>10(a+b)+9+1500 chia hết cho 9 và a-b=6

=>10(a+b) chia 9 dư 3 và a-b=6

=>10(a+b)=120 và a-b=6

=>a+b=12 và a-b=6

=>a=9; b=3

\(3y-30\%y=-16\\ 3y-\dfrac{3}{10}y=-16\\ \dfrac{27}{10}y=-16\\ y=-16:\dfrac{-27}{10}\\ y=\dfrac{-160}{27}\)

3y−30%y=−163y−310y=−162710y=−16y=−16:−2710y=−16027

a) (-64) + 32 -54=  -32-54= -86

b) 125 + 45 - (-23) + (-124)=193-124=69  bước này mình nhóm 3 cái đầu với nhau tính ra kết quả

c) [43 + (-16)] - (18 - 27)=27+9=36

d) - 52 - (34 - 52) + (-65 + 34)= -52+18-31=-65

e) 237 x 28 - 28 x 137 =28(237-137)=28x100=2800

 f) 3 x 117 - 3 x16 - 3=3(117-16-1)=3x100=300

Chúc bạn học tốt

A=108−1108+2​=108−1108−1+3​=108−1108−1​+108−13​=1+108−13​
 

I don't know if it's true

\(2008-\dfrac{2009}{3}-\dfrac{2009}{6}-\dfrac{2009}{10}-\dfrac{2009}{15}-...-\dfrac{2009}{45}\)

\(=2008-2009\left(\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{45}\right)\)

\(=2008-2009\left(\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{90}\right)\)

\(=2008-2009\cdot2\left(\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{90}\right)\)

\(=2008-4018\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

\(=2008-4018\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=2008-4018\cdot\dfrac{4}{10}=400,8\)

\(\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{101}{303}-\dfrac{3}{303}=\dfrac{98}{303}\)

Ta có: \(\dfrac{3}{10}\cdot\left(-\dfrac{4}{9}+\dfrac{2}{5}\right)-\dfrac{3}{10}\left(\dfrac{5}{9}+\dfrac{-3}{5}\right)\)

\(=\dfrac{3}{10}\left(-\dfrac{4}{9}+\dfrac{2}{5}-\dfrac{5}{9}+\dfrac{3}{5}\right)\)

\(=\dfrac{3}{10}\left(-\dfrac{9}{9}+\dfrac{5}{5}\right)\)

\(=\dfrac{3}{10}\cdot\left(-1+1\right)=0\)

\(50+\dfrac{50}{3}+\dfrac{25}{3}+...+\dfrac{100}{98\cdot99}+\dfrac{1}{99}\)

\(=\dfrac{100}{2}+\dfrac{100}{6}+\dfrac{100}{12}+...+\dfrac{100}{99\cdot100}\)

\(=100\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)\)

\(=100\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=100\left(1-\dfrac{1}{100}\right)=100\cdot\dfrac{99}{100}\)

=99

:( sao ko ai rep ht dzạy

Ta có: \(\dfrac{2}{5}+x=-5^2\cdot\dfrac{3}{5}\)

=>\(x+\dfrac{2}{5}=-15\)

=>\(x=-15-\dfrac{2}{5}=-\dfrac{77}{5}\)

`2/5 + x = -5^2 xx 3/5`

`=> 2/5 + x = -25 xx 3/5`

`=> 2/5 + x = -75/5`

`=> 2/5 + x = -15`

`=> x = -15 - 2/5`

`=> x = -15 - 0,4`

`=> x = -15,4`

Vậy ...