`a)`

`x-15 = -21`

`<=> x = -21 + 15`

`<=> x = -6`

`b)`

`42 - x = -7`

`<=> x = 42 - (-7)`

`<=> x = 42 + 7`

`<=> x = 49`

`c)`

`12 - (30-x) = -23`

`30-x = 12 - (-23)`

`30-x = 35`

`x = 30-35`

`x = -5`

`d)`

`31 - (17+x)=18`

`<=> 17+x = 31-18`

`<=> 17+x =13`

`<=> x = 13-17`

`<=> x = -4`

Sai thông cảm nha :)

a) \(x-15=-21\)

\(\Rightarrow x=-21+15=-6\)

b) \(42-x=-7\)

\(\Rightarrow x=42+7=49\)

c) \(12-\left(30-x\right)=-23\)

\(\Rightarrow30-x=12+23=35\)

\(\Rightarrow x=30+35=65\)

d) \(31-\left(17+x\right)=18\)

\(\Rightarrow17+x=31-18=13\)

\(\Rightarrow x=13-17=-4\)

a) $x-15=-21$

$\Rightarrow x=-21+15=-6$

b) $42-x=-7$

$x=42+7=49$

c) $12-\left(30-x\right)=-23$

$30-x=12+23=35$

$x=30+35=65$

d) $31-\left(17+x\right)=18$

$17+x=31-18=13$

$x=13-17=-4$

− 12 27 + 2 3 + − 2 9 ≤ x ≤ 11 7 + 2 5 + − 7 5 + 3 7 ⇒ − 4 9 + − 2 9 + 2 3 ≤ x ≤ 11 7 + 2 5 + − 7 5 + 3 7 ⇒ − 2 3 + 2 3 ≤ x ≤ 11 7 + 3 7 + 2 5 + − 7 5 ⇒ 0 ≤ x ≤ 2 + − 1 ⇒ 0 ≤ x ≤ 1 ⇒ x = 1

a: =>19/23>19/x>19/29

=>\(x\in\left\{24;25;26;27;28\right\}\)

b: =>88/132<88/x<88/128

=>132>x>128

=>\(x\in\left\{131;130;129\right\}\)

c: =>\(\left\{{}\begin{matrix}\dfrac{4}{x}-\dfrac{x}{8}< 0\\\dfrac{x}{8}-\dfrac{5}{x}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{32-x^2}{8x}< 0\\\dfrac{x^2-40}{8x}< 0\end{matrix}\right.\)

=>32<x^2<40

=>x=6

a, 117 - \(x\) = 28 - (-7)

   117 - \(x\) = 28 + 7

    117 - \(x\) = 35

            \(x\) =  117 - 35

             \(x\) = 82

b, \(x\) - (-38 - 2\(x\)) = (-3) - 8 + 2\(x\) 

    \(x\) + 38 + 2\(x\) = - 11 + 2\(x\)

   3\(x\)  + 38       = - 11 + 2\(x\)

   3\(x\) - 2\(x\)        = - 11 - 38

   \(x\)                = - 49

Giấy hơi nhàu, thông cảm :<

a) Đặt: \(A=1+2^2+2^3+...+2^{10}\)

\(\Rightarrow2A=2\left(1+2^2+2^3+...+2^9+2^{10}\right)\)

\(\Rightarrow2A=2+2^3+2^4+...+2^{10}+2^{11}\)

\(\Rightarrow2A-A=\left(2+2^3+2^4+...+2^{10}+2^{11}\right)-\left(1+2^2+2^3+...+2^{10}\right)\)

\(\Rightarrow A=\left(2^3-2^3\right)+\left(2^4-2^4\right)+...+\left(2-1\right)+\left(2^{11}-2^2\right)\)

\(\Rightarrow A=0+0+...+1+\left(2^{11}-2^2\right)\)

\(\Rightarrow A=1+2^{11}-2^2=1+2048-4=2045\)

Vậy: \(1+2^2+2^3+...+2^{10}=2045\)

b) 

a] \(60-3\left(x-1\right)=2^3\cdot3\)

\(\Rightarrow60-3\left(x-1\right)=24\)

\(\Rightarrow3\left(x-1\right)=36\)

\(\Rightarrow x-1=12\)

\(\Rightarrow x=13\)

b] \(\left(3x-2\right)^3=2\cdot2^5\)

\(\Rightarrow\left(3x-2\right)^3=2^6\)

\(\Rightarrow\left(3x-2\right)^3=\left(2^2\right)^3\)

\(\Rightarrow3x-2=2^2\)

\(\Rightarrow3x=6\)

\(x=2\)

c] \(5^{x+1}-5^x=500\)

\(\Rightarrow5^x\left(5-1\right)=500\)

\(\Rightarrow5^x\cdot4=500\)

\(\Rightarrow5^x=125\)

\(\Rightarrow5^x=5^3\)

\(\Rightarrow x=3\)

d] \(x^2=x^4\)

\(\Rightarrow x=x^2\)

\(\Rightarrow x-x^2=0\)

\(\Rightarrow x\left(1-x\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

giúp mình đi các bạn

\(a,\left(x+3\right)\left(5-x\right)=0\\ \Rightarrow\left\{{}\begin{matrix}x+3=0\\5-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)

\(c,x+17⋮x+3\\ x+3+14⋮x+3\\ 14⋮x+3\\ x+3\inƯ\left(14\right)=\left\{\pm14;\pm7\pm2;\pm1\right\}\)

Từ đó bạn tìm những giá trị của x nha!