(x - 10)x+1- (x - 10)x+11= 0
GIÚP VS MN ƠI!!! 
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Những câu hỏi liên quan
NT
1
21 tháng 12 2021
\(\Leftrightarrow x-\left[3-x+3+x-2\right]=0\)
=>x=4
KC
1
YN
2 tháng 10 2021
\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Rightarrow\left(x^3+27\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x^2-2x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x\left(x-2\right)=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=0\\x=2\end{matrix}\right.\)
Vậy \(x\in\left\{-3;0;2\right\}\)
AM
1
12 tháng 4 2023
\(2\left(x^2-x\right)-x\left(x+2\right)+4=0\)
\(\Leftrightarrow2x^2-2x-x^2-2x+4=0\)
\(\Leftrightarrow x^2-4x+4=0\)
\(\Leftrightarrow\left(x-2\right)^2=0\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
Vậy \(S=\left\{2\right\}\)
A
1
14 tháng 8 2019
Ta có: \(\hept{\begin{cases}|x-40|\ge0;\forall x,y\\|x-y+10|\ge0;\forall x,y\end{cases}}\)
\(\Rightarrow|x-40|+|x-y+10|\ge0;\forall x,y\)
Do đó: \(|x-40|+|x-y+10|=0\)
\(\Leftrightarrow\hept{\begin{cases}|x-40|=0\\|x-y+10|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=40\\y=50\end{cases}}\)
19 tháng 10 2021
\(a,\Rightarrow\left(x-2000\right)\left(5x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=2000\\x=\dfrac{1}{5}\end{matrix}\right.\\ b,\Rightarrow x\left(x^2-13\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{13}\\x=-\sqrt{13}\end{matrix}\right.\\ c,\Rightarrow3x\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ d,\Rightarrow\left(x-5\right)\left(x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\\ e,\Rightarrow\left(3x-2\right)\left(3x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
HT
2
9 tháng 10 2021
1)
a) \(=15x^3-20x^2+10x\)
b) \(=3x^4-x^3+4x^2-9x^3+3x-12x=3x^4-10x^3+4x^2-9x\)
2)
a) \(\Rightarrow x\left(x^2-6x+12\right)=0\)
\(\Rightarrow x=0\)(do \(x^2-6x+12=\left(x^2-6x+\dfrac{36}{4}\right)+3=\left(x-\dfrac{6}{2}\right)^2+3\ge3>0\))
b) \(\Rightarrow\left(x+3\right)^3=0\Rightarrow x=-3\)
23 tháng 3 2022
(3x²-5x+2)+(3x²+5x)= bao nhiêu ạ
Giúp em vs ạ . Em cảm ơn
Ta có :
\(\left(x-10\right)^{x+1}-\left(x-10\right)^{x+11}=0\)
\(\Leftrightarrow\left(x-10\right)^{x+1}\left[1-\left(x-10\right)^{10}\right]=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-10=0\\1-\left(x-10\right)^{10}=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=10\\\left[\begin{array}{nghiempt}x-10=1\\x-10=-1\end{array}\right.\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=10\\\left[\begin{array}{nghiempt}x=11\\x=9\end{array}\right.\end{array}\right.\)
Vậy x = 10 ; x = 11 ; x = 9
\(\left(x-10\right)^{x+1}-\left(x-10\right)^{x+11}=0\)
\(\Rightarrow\left(x-10\right)^{x+1}.\left[1-\left(x-10\right)^{10}\right]=0\)
\(\Rightarrow\left(x-10\right)^{x+1}=0\) hoặc \(1-\left(x-10\right)^{10}=0\)
+) \(\left(x-10\right)^{x+1}=0\)
\(\Rightarrow x-10=0\)
\(\Rightarrow x=10\)
+) \(1-\left(x-10\right)^{10}=0\)
\(\Rightarrow\left(x-10\right)^{10}=1\)
\(\Rightarrow x-10=\pm1\)
+ \(x-10=1\Rightarrow x=11\)
+ \(x-10=-1\Rightarrow x=9\)
Vậy \(x\in\left\{10;11;9\right\}\)