tìm x
|10-x|+|8-x|=0
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,\left(x+2\right)^{10}+\left(x+2\right)^8=0\\ \Leftrightarrow\left(x+2\right)^8\left[\left(x+2\right)^2+1\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x+2\right)^8=0\\\left(x+2\right)^2+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+2=0\\\left(x+2\right)^2=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\end{matrix}\right.\\ b,\left(x+3\right)^{10}-\left(x+3\right)^8=0\\ \Leftrightarrow\left(x+3\right)^8\left[\left(x+3\right)^2-1\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x+3\right)^8=0\\\left(x+3\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\\left(x+3\right)^2=1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x+3=1\\x+3=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\\x=-4\end{matrix}\right.\)
\(\dfrac{x+8}{12}+\dfrac{x+9}{11}+\dfrac{x+10}{10}+3=0\\ \Leftrightarrow\dfrac{x+8}{12}+1+\dfrac{x+9}{11}+1+\dfrac{x+10}{10}+1=0\\ \Leftrightarrow\dfrac{x+20}{12}+\dfrac{x+20}{11}+\dfrac{x+20}{10}=0\\ \Leftrightarrow\left(x+20\right)\left(\dfrac{1}{12}+\dfrac{1}{11}+\dfrac{1}{10}\right)=0\\ \Leftrightarrow x+20=0\Leftrightarrow x=-20\\ KL:...\)
`<=>((x+8)/12+1)+((x+9)/11+1)+((x+10)/10+1)=0`
`<=>(x+20)/12+(x+20)/11+(x+20)/10=0`
`<=>(x+20)(1/12+1/11+1/10)=0`
Vì `1/12+1/11+1/10 ≠ 0`
`=>x+20=0`
`=>x=0-20`
`=>x=-20`
Ta có :\(\frac{x+8}{12}+\frac{x+9}{11}+\frac{x+10}{10}+3=0\)
=> \(\left(\frac{x+8}{12}+1\right)+\left(\frac{x+9}{11}+1\right)+\left(\frac{x+10}{10}+1\right)=0\)
=> \(\frac{x+20}{12}+\frac{x+20}{11}+\frac{x+20}{10}=0\)
=> \(\left(x+20\right)\left(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}\right)=0\)
Vì \(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}\ne0\)
=> x + 20 = 0
=> x = -20
Vậy x = -20
\(\frac{x+8}{12}+\frac{x+9}{11}+\frac{x+10}{10}+3=0\)
\(\Leftrightarrow\left(\frac{x+8}{12}+1\right)+\left(\frac{x+9}{11}+1\right)+\left(\frac{x+10}{10}+1\right)=0\)
\(\Leftrightarrow\frac{x+20}{12}+\frac{x+20}{11}+\frac{x+20}{10}=0\)
\(\Leftrightarrow\left(x+20\right)\left(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}\right)=0\)
Vì \(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}\ne0\)
\(\Rightarrow x+20=0\Rightarrow x=-20\)
a) => x - 8 = 0 hoặc x3 + 8 = 0
+) x - 8 = 0 => x = 8
+) x3 + 8 = 0 => x3 = - 8 = (-2)3 => x = -2
Vậy x = 8; -2
b) => 4x - 3 - x - 5 = 30 - 3x
=> 3x - 8 = 30 - 3x
=> 3x + 3x = 30 + 8
=> 6x = 38 => x = 38/6 = 19/3
Vậy x = 19/3
a) => x - 8 = 0 hoặc x3 + 8 = 0
+) x - 8 = 0 => x = 8
+) x3 + 8 = 0 => x3 = - 8 = (-2)3 => x = -2
Vậy x = 8; -2
b) => 4x - 3 - x - 5 = 30 - 3x
=> 3x - 8 = 30 - 3x
=> 3x + 3x = 30 + 8
=> 6x = 38 => x = 38/6 = 19/3
Vậy x = 19/3
a, x + 8 = 15 => x = 15 – 8 = 7. Tập hợp: A = {7}
b, 19 – x = 7 => x = 19 – 7 =12. Tập hợp: B = {12}
c. x : 10 = 0 => x = 0. Tập hợp: C = {0}
d, 0 : x = 0 => x ∈ {1;2;3;...}. Tập hợp: D = N*
Do \(\left|10-x\right|,\left|8-x\right|\ge0\forall x\)
\(\Rightarrow\left\{{}\begin{matrix}10-x=0\\8-x=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=10\\x=8\end{matrix}\right.\)(Vô lý)
Vậy \(S=\varnothing\)