2​9.19 -29.13-19.29+19.13=(29.19-19.29)+13.(19-29)=0+13.(-10)=-130. Tìm x +2 =1 => x=1-2=-1

\(210-\left(x+31\right)=-29\)

\(\Rightarrow\left(x+31\right)=210-\left(-29\right)\)

\(\Rightarrow x+31=239\)

\(\Rightarrow x=239-31=208\)

\(-\frac{5}{2}< \frac{3}{x}< \frac{2}{3}\)

Các số nguyên x thỏa mãn là: x \(\in\left\{5;6;......\right\}\)

ta có

\(-\frac{5}{6}+\frac{8}{3}+\frac{29}{-6}=\frac{-5+16-29}{6}=-\frac{18}{6}=-3\)

\(-\frac{1}{2}+2+\frac{5}{2}=2+2=4\)

vì vậy \(-3< x< 4\Rightarrow x\in\left\{-2,-1,0,1,2,3\right\}\)

\(1,-\frac{3}{29}+\frac{-7}{29}\le\frac{x}{29}\le-\frac{3}{29}-\frac{5}{29}\)

\(\Rightarrow-\frac{10}{29}\le\frac{x}{29}\le-\frac{8}{29}\Rightarrow-10\le x\le-8\)

\(\Rightarrow x=\left\{-8;-9;-10\right\}\)

\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\)

\(\Rightarrow2S=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)

             \(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\)

\(\Rightarrow2S-S=S=1-\frac{1}{2^{100}}\)

\(\left(3-x\right)+\left(x-5\right)=29-x\)

\(3-x+x-5+x=29\)

\(x-2=29\)

\(x=31\)

vậy \(x=31\)

\(15+x-\left(3+x\right)=x-14\)

\(15+x-3-x-x=-14\)

\(-x+12=-14\)

\(-x=-26\)

\(x=26\)

vậy \(x=26\)

a, => 3-x+x-5 = 29-x

=> -2 = 29-x

=> x=29-(-2) = 31

b, => 15+x-3-x=x-14

=> 12=x-14

=> x=12+14=26

Tk mk nha

1, 

     (3-x)+(x-5) =29-x

=> (x-x) + (3-5) = 29 - x

=>            -2     = 29 -x

=>            -31  = -x

Vậy x=31

2,

    15+x-(3+x)=x-14

=> (x-x) + (15-3) = x-14

=>            12       = x-14

=>            12+14 =x

Vậy x= 26

biết j mới dk bn??

(x^2-1).(x^2-3).(x^2-5).(x^2-7)\(\le\)0

Biết j vậy phải có điều kiện chứ ...

\(a,3x-75=45+3.\left(-25\right)\)

\(3x-75=-30\)

\(3x=45\)

\(x=15\)

\(b,2.\left(-7\right)-11=6x+17\)

\(-14-11-17=6x\)

\(-42=6x\)

\(-7=x\Leftrightarrow x=-7\)

\(d,-4.|x-3|=-48\)

\(|x-3|=12\)

\(\Rightarrow\orbr{\begin{cases}x-3=12\\x-3=-12\end{cases}\Rightarrow\orbr{\begin{cases}x=15\\x=-9\end{cases}}}\)