a: 3x=81

nên x=27

b: \(5\cdot4^x=80\)

\(\Leftrightarrow4^x=16\)

hay x=2

c: \(2^x=4^5:4^3\)

\(\Leftrightarrow2^x=2^4\)

hay x=4

-3x + 5 = 41                                    52 - |x| = 80                                     7x + 1 = -20

   -3x     = 41 - 5                                     |x| = 52 - 80                                  7x    = -20 - 1

     -3x   =  36                                         |x| =  - 28                                      7x    = -21

=> 3x    = -36                                  Vì |x| > hoặc = 0                                     x    = -3

        x   =  -12                                 nên không tìm được x                         Vậy x = - 3

Vậy x = - 12                                    Vậy không tìm được x             

a: \(\Leftrightarrow\dfrac{x}{-4}=\dfrac{21}{y}=\dfrac{z}{-80}=\dfrac{3}{4}\)

=>x=-3; y=28; z=-60

b: 5/12=x/-72

=>x=-72*5/12=-6*5=-30

c: =>x+3=-5

=>x=-8

-29-9(2x-1)\(^2\)= -110

(=) 9(2x-1)² = (-29) +110

(=) 9(2x-1)² = 81

(=) (2x-1)² =81: 9

(=) (2x-1)² =9

(=) (2x-1)² = 3² =(-3)²

\(\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\)

\(\orbr{\begin{cases}2x=4\\2x=-2\end{cases}}\)

\(\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)

vậy : ........

a,\(-29-9\left(2x-1\right)^2=-110\)

\(=>-29+110=9.\left(2x-1\right)^2\)

\(=>81=9.\left(2x-1\right)^2\)

\(=>\left(2x-1\right)^2=9\)

\(=>\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}=>\orbr{\begin{cases}x=\frac{4}{2}=2\\x=\frac{-2}{2}=-1\end{cases}}}\)

b,\(-2.\left(-x-5\right)+28=20-3.4\)

\(=>2x+10+28=20-12=8\)

\(=>2x=8-28-10=-30\)

\(=>x=\frac{-30}{2}=-15\)

\(x=-\dfrac{1}{2}=-0.5,y=\dfrac{5}{4}=1.25\\x=2,y=\dfrac{7}{2}=3.5\)

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

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

Từ đó bạn tìm những giá trị của x nha!

a) \(5\times x-123=12\)

\(\Rightarrow5\times x=135\)

\(\Rightarrow x=27\)

b) \(x+3x+5x+7x=96\)

\(\Rightarrow16x=96\)

\(\Rightarrow x=6\)

a) \(5\times x-123=12\)

\(5x=12+123\)

\(5x=135\)

\(x=135:5\)

\(x=27\)

________

b) \(x+3x+5x+7x=96\)

\(x\left(1+3+5+7\right)=96\)

\(x.16=96\)

\(x=96:16\)

\(x=6\)

a)

\(5\times x-123=12\\\Rightarrow 5\times x=12+123\\\Rightarrow 5\times x=135\\\Rightarrow x=135:5\\\Rightarrow x=27\\---\)

b) 

\(x+3x+5x+7x=96\\\Rightarrow16x=96\\\Rightarrow x=96:16\\\Rightarrow x=6\)

Cần bổ sung điều kiện \(x;y\inℤ\)

a) \(\left(x-7\right)\left(xy+1\right)=9=1\cdot9=3\cdot3=\left(-1\right)\cdot\left(-9\right)=\left(-3\right)\cdot\left(-3\right)\)

Xét bảng :

Vì x,y thuộc Z nên ta có (x;y)={(8;1),(16;0),(-2;1),(4;-1)

b) \(\left(x+5\right)\left(3x-12\right)>0\)

TH1 : \(\hept{\begin{cases}x+5>0\\3x-12>0\end{cases}\Leftrightarrow\hept{\begin{cases}x>-5\\x>4\end{cases}\Leftrightarrow x>4}}\)

TH2 : \(\hept{\begin{cases}x+5< 0\\3x-12< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x< -5\\x< 4\end{cases}\Leftrightarrow x< -5}}\)

Vậy....

c) \(\left|x+3\right|< 3\)

Mặt khác : \(0< \left|x+3\right|< 3\)

\(\Rightarrow\left|x+3\right|\in\left\{1;2\right\}\)

\(\Rightarrow\left(x+3\right)\in\left\{\pm1;\pm2\right\}\)

\(\Rightarrow x\in\left\{-2;-4;-1;-5\right\}\)

Vậy....