1,

a,\(2020-\left(249+2020\right)+\left(249-573\right)\)

\(=2020-249-2020+249-573\)

\(=-573\)

b,\(\left|-257\right|+\left(-3\right)^0-\left(18+257\right)\)

\(=257+1-18-257\)

\(=1-18=-17\)

\(c,25.\left(85-47\right)-85.\left(47+25\right)\)

\(=25.85-25.47-47.85+85.25\)

\(=85.\left(25-47+25\right)-25.47\)

\(=85.3-25.47\)

\(=-920\)

2,

\(a,15-5.\left(x+2\right)=-30\)

\(=>5.\left(x+2\right)=15+30=45\)

\(=>x+2=\frac{45}{5}=9\)

\(=>x=7\)

\(b,\left(x+2\right)^2+5=105\)

\(=>\left(x+2\right)^2=100\)

\(=>\left(x+2\right)^2=10^2\)

\(=>x+2=10\)

\(=>x=8\)

\(c,\left|2x-5\right|-\left(-6\right)=11\)

\(=>\left|2x-5\right|=11-6=5\)

\(=>\orbr{\begin{cases}2x-5=5\\2x-5=-5\end{cases}}\)

\(=>\orbr{\begin{cases}2x=5-5=0\\2x=-5+5=0\end{cases}=>x=0}\)

a: \(\Leftrightarrow3x+7\in\left\{1;-1;3;-3;11;-11;33;-33\right\}\)

hay \(x=-6\)

b: \(\Leftrightarrow3x-1\in\left\{1;-1;7;-7\right\}\)

hay \(x\in\left\{0;-2\right\}\)

a) \(10⋮\left(x-1\right)\) (đkxđ \(x\ne1\))

\(\Rightarrow x-1\in\left\{-1;1;-2;2;-5;5;-10;10\right\}\)

\(\Rightarrow x\in\left\{0;2;-1;3;-4;6;-9;11\right\}\)

b) \(\left(x+5\right)⋮\left(x-2\right)\left(đkxđ,x\ne2\right)\)

\(\Rightarrow\left(x+5\right)-\left(x-2\right)⋮\left(x-2\right)\)

\(\Rightarrow x+5-x+2⋮\left(x-2\right)\)

\(\Rightarrow7⋮\left(x-2\right)\)

\(\Rightarrow x-2\in\left\{-1;1;-7;7\right\}\)

\(\Rightarrow x\in\left\{1;3;-5;9\right\}\)

c) \(\left(3x+8\right)⋮\left(x-1\right)\left(đkxd,x\ne1\right)\)

\(\Rightarrow\left(3x+8\right)-3\left(x-1\right)⋮\left(x-1\right)\)

\(\Rightarrow3x+8-3x+3⋮\left(x-1\right)\)

\(\Rightarrow11⋮\left(x-1\right)\)

\(\Rightarrow x-1\in\left\{-1;1;-11;11\right\}\)

\(\Rightarrow x\in\left\{0;2;-10;12\right\}\)

a) x∈{0;2;−1;3;−4;6;−9;11}

b) x∈{1;3;−5;9}

c) x ∈ {0;2;−10;12}

2 câu dưới là b, c

a) <=> 10 - 2x + 5 = 1 - 3x

   <=>    -2x + 3x    = 1 - 10 - 5

   <=>      x             = -14

b) \(\Leftrightarrow\frac{x-1}{x+1}-\frac{8}{9}=0\)

    \(\Leftrightarrow\frac{9\left(x-1\right)-8\left(x+1\right)}{9\left(x+1\right)}=0\)

     \(\Leftrightarrow\frac{9x-9-8x-8}{9x+9}=0\)

     \(\Leftrightarrow\frac{x-17}{9x+9}=0\)

ĐK: 9x + 9 \(\ne\)0 => x \(\ne\)-1

     \(\Leftrightarrow x-17=0\)

     \(\Leftrightarrow x=17\)

Ps: Câu b không chắc lắm. 

phan b Vu Nhu Mai sai roi phai the nay moi dung

x-1/x+1=8/9

(x-1).9=(x+1).8

9x-9=8x+8

9x-8x=8+9

1x=17

x=17;1

x=17 

a) x + 7 = 19

x = 19 - 7

x = 12

b) 42 - 3x = 27

3x = 42 - 27

3x = 15

x = 15 : 3

x = 5

c) 45 : (3 - x) = 5

3 - x = 45 : 5

3 - x = 9

x = 3 - 9

x = (-6)

a) x = 29

b) x = 1 và x = -5

c) x = 9 và x = -1