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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: \(x\in\left\{24;-6\right\}\)
a) \(2x-8=5\)
\(\Leftrightarrow2x=13\)
\(\Leftrightarrow x=\frac{13}{2}\)
b) \(\left(6x-12\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}6x-12=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}6x=12\\x=-5\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-5\end{cases}}}\)
c) \(|2x-1|+3=6^2\)
\(\Leftrightarrow|2x-1|+3=36\)
\(\Leftrightarrow|2x-1|=33\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=33\\2x-1=-33\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=34\\2x=-32\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=17\\x=-16\end{cases}}}\)
a,2x-8=5
2x =5+8
2x =13
x =13:2 k đi rồi tui làm o
x =13/2=6,5
Bài giải
a, \(2x-\left(-17\right)=15\)
\(2x=15-17\)
\(2x=-2\)
\(x=-1\)
b, \(\left|x-2\right|=8\)
\(\Rightarrow\text{ }\orbr{\begin{cases}x-2=-8\\x-2=8\end{cases}}\Rightarrow\orbr{\begin{cases}x=-8+2\\x=8+2\end{cases}}\Rightarrow\orbr{\begin{cases}x=-6\\x=10\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{-6\text{ ; }10\right\}\)
c, \(\left|x+9\right|\cdot2-9=1\)
\(2\left|x+9\right|=1+9\)
\(2\left|x+9\right|=10\)
\(\left|x+9\right|=10\text{ : }2\)
\(\left|x+9\right|=5\)
\(\Rightarrow\orbr{\begin{cases}x+9=-5\\x+9=5\end{cases}}\Rightarrow\orbr{\begin{cases}x=-5-9\\x=5-9\end{cases}}\Rightarrow\orbr{\begin{cases}x=-14\\x=-4\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{-14\text{ ; }-4\right\}\)
\(2x\cdot\dfrac{-4}{9}+2x\cdot\dfrac{-5}{9}=\dfrac{8}{11}\)
\(2x\cdot\left(\dfrac{-4}{9}+\dfrac{-5}{9}\right)=\dfrac{8}{11}\)
\(2x\cdot\left(-1\right)=\dfrac{8}{11}\)
2x = \(\dfrac{8}{11}:\left(-1\right)=\dfrac{-8}{11}\)
x = \(\dfrac{-8}{11}:2=\dfrac{-4}{11}\)
a) (3x - 1)³ + 17 = 710 : 5
(3x - 1)³ + 17 = 142
(3x - 1)³ = 142 - 17
(3x - 1)³ = 125
(3x - 1)³ = 5³
3x - 1 = 5
3x = 5 + 1
3x = 6
x = 6 : 3
x = 2
a,x-12=-9-(-15)+2x
x - 12 = -9 + 15 + 2x
x - 12 = 6 + 2x
x -2x = 6 + 12
-1x = 18
x = 18 : (-1)
x = -18
b,17-{x-[-x-(-x)]}=-16
17 - {x-[-x+x]} = -16
17 - {x+x+-x} = -16
17 - 2x - x = - 16
17 - x = - 16
x = 17 - (-16)
x = 33
c,x+{(x+3)-[(x+3)-(-x-2)]} = x
x+{x+3-[x+3+x+2]} = x
x + {x+3-x-3-x-2} = x
x + x + 3 - x - 3 - x - 2 = x
(x+x)-(x-x) + (3 - 3) = x
2x - 2x + 0 = x
0 + 0 = x
x = 0
\(2x-13=5-x\)
\(\Rightarrow2x+x=5+13\)
\(\Rightarrow3x=18\)
\(\Rightarrow x=6\)
\(-10⋮n-5\Rightarrow n-5\inƯ\left(-10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
\(\Rightarrow n\in\left\{6;4;7;3;10;0;15;-5\right\}\)
Vậy.............................
x=-8
tk minh nha
2x - 9 = - 17 + x
2x - x = - 17 + 9
1x = - 8
x = - 8 : 1
x = - 8