a) Ta có: \(3-\left(17-x\right)=-12\)

\(\Leftrightarrow3-17+x+12=0\)

\(\Leftrightarrow x-2=0\)

hay x=2

Vậy: x=2

b) Ta có: \(\left(2x+4\right)\left(10-2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+4=0\\10-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-4\\2x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)

Vậy: \(x\in\left\{-2;5\right\}\)c) Ta có: \(\left|x-9\right|=-2+17\)

\(\Leftrightarrow\left|x-9\right|=15\)

\(\Leftrightarrow\left[{}\begin{matrix}x-9=15\\x-9=-15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=24\\x=-6\end{matrix}\right.\)

Vậy: \(x\in\left\{24;-6\right\}\)

x + x : 1/5 + x : 1/8 - x × 4 = 1250

x + x × 5 + x × 8 - 4 × x = 1250

x × (1 + 5 + 8 - 4) = 1250

x × 10 = 1250

x = 1250 : 10

x = 125

\(xy+2y-x=5\)

\(\Leftrightarrow xy+2y-x-2=5-2\)

\(\Leftrightarrow y\left(x+2\right)-\left(x+2\right)=3\)

\(\Leftrightarrow\left(x+2\right)\left(y-1\right)=3\)

\(\Leftrightarrow x+2;y-1\inƯ\left(3\right)\)

Suy ra :

+) \(\hept{\begin{cases}x+2=1\\y-1=3\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=-1\\y=4\end{cases}}\)

+) \(\hept{\begin{cases}x+2=-1\\y-1=-3\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=-3\\y=-2\end{cases}}\)

+) \(\hept{\begin{cases}x+2=-3\\y-1=-1\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=-5\\y=0\end{cases}}\)

+) \(\hept{\begin{cases}x+2=3\\y-1=1\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=1\\y=2\end{cases}}\)

Vậy ..

cảm ơn bạn nhiều nhé

\(\left(1+x\right)+\left(5+x-4\right)+\left(9+x-8\right)+...=501501\)

\(\left(1+x\right)+\left(x+1\right)+\left(x+1\right)+...=501501\) có x :2 dấu ngoặc

\(\left(1+x\right).x:2=501501\)

\(\left(1+x\right).x=1003002=1002.1001\Rightarrow x=1001\)

Chúc bạn học tốt! 

a) Ta có: \(x+\left(-5\right)=-\left(-7\right)\)

\(\Leftrightarrow x-5=7\)

\(\Leftrightarrow x=7+5=12\)

Vậy: x=12

b) Ta có: \(-5\cdot\left|x\right|=-30\)

\(\Leftrightarrow\left|x\right|=-30:\left(-5\right)=6\)

hay \(x\in\left\{6;-6\right\}\)

Vậy: \(x\in\left\{6;-6\right\}\)

c) Ta có: \(2x+20=-22\)

\(\Leftrightarrow2x=-22-20=-42\)

hay x=-21

Vậy: x=-21

d) Ta có: \(\left|x+2\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)

Vậy: \(x\in\left\{2;-6\right\}\)

=>-1+(3-5)+(7-9)+...+[(x-4)-(x-2)]+x=600

=>-1+(-2)+(-2)+...+(-2)+x=600                                      (\(\frac{x-1}{4}\) số hạng -2)

=>-1+(-2)\(\frac{x-1}{4}\)+x=600

=>\(\frac{-4+\left(-2\right)\left(x-1\right)+4x}{4}=600\)

=>-4+(-2)(x-1)+4(x-1)+4=600.4

=>(-2+4)(x-1)=2400+4-4

=>2(x-1)=2400

=>x-1=2400:2

=>x-1=1200

=>x=1201

a) 3^{2 x + 2} = 9^{x+ 3} 

     x         = \(\phi\)

b) ( x - 5 )⁴ = ( x - 5 ) ⁶

 =>x - 5 = 0 hoặc 1.

x-5=0 thì x=0+5=5

x-5=1 thì x=1+5=6

=>x=5;6

c) x¹⁵ = x²

     x    = 0 ;1

Tick đúng cho mình nha.Cảm ơn nhìu

5-[-|-4|+(-3+x-|-6|)]=15-94+|-8|-9

5-[-4+(-3+x-6)] = 15-94+8-9

5-[-4+(-3)+x-6] = (-79)+8-9

5+4-(-3) - x + 6 = -71 - 9

9-(-3) - x + 6 = -80

12 - x + 6 = -80

12-x = -80 - 6

12-x = -86

x = 12 - (-86)

x = 98

mk ko chắc lắm

5 - [ - l -4 l + ( - 3 + x - l - 6 l ) ] = 15 - 94 + l -8 l - 9

           5 - [ - 4 + ( - 3 + x - 6 ) ] = - 78 + 8 - 9

                  5 - [ - 4 - 3 + x - 6 ]  = - 70 - 9

                   5 + 4 + 3 - x + 6      = - 79

                          9 + 3 - x + 6      = - 79

                              12 - x + 6      = - 79

                           12 + ( - x ) + 6  = - 79

                                 12 + ( - x )  = - 79 - 6

                                   12 + ( - x ) = - 85

                                              - x  = - 85 - 12

                                               - x  = - 97 => x = 97

Vậy x = 97

Xin lỗi mk sai , mk tính nhầm 1 bước