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b
\(\left|6+x\right|\ge0;\left(3+y\right)^2\ge0\Rightarrow\left|6+x\right|+\left(3+y\right)^2\ge0\)
Suy ra \(\left|6+x\right|+\left(3+y\right)^2=0\)\(\Leftrightarrow\hept{\begin{cases}6+x=0\\3+y=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-6\\y=-3\end{cases}}\)
a
Ta có:\(\left|3x-12\right|=3x-12\Leftrightarrow3x-12\ge0\Leftrightarrow3x\ge12\Leftrightarrow x\ge4\)
\(\left|3x-12\right|=12-3x\Leftrightarrow3x-12< 0\Leftrightarrow3x< 12\Leftrightarrow x< 4\)
Với \(x\ge4\) ta có:
\(3x-12+4x=2x-2\)
\(\Rightarrow5x=10\)
\(\Rightarrow x=2\left(KTMĐK\right)\)
Với \(x< 4\) ta có:
\(12-3x+4x=2x-2\)
\(\Rightarrow10=x\left(KTMĐK\right)\)
1)Tìm x
a) (x+1)(x-2)<0
=>Có 2TH:
TH1:
x+1<0=>x< -1
x-2>0=>x>2
=>Vô lí
TH2:
x+1>0=>x> -1
x-2<0=>x<2
=> -1<x<2
Vậy x thuộc {0;1}
b) Tương tự a thôi ạ.
c) (x-2)(3x+2)
=> Có hai TH:
TH1:
x-2<0=>x<2
3x+2<0=>3x< -2=>x< -2/3
=>x< -2/3
TH2:
x-2>0=>x>2
3x+2>0=>3x> -2=>x> -2/3
=>x>2
Vậy x< -2/3 hoặc x>2
2)Tìm x
x.x=x
<=>x²-x=0
<=>x(x-1)=0
<=>x=0 hoặc x=1
\(2+\left(x-1\right)^2=0\)
\(\left(x-1\right)^2=-2\left(loại\right)\)
P/s : làm từng phần một
( x - 1 ) ( x - 5 ) > 0
TH1: cả x - 1 và x - 5 lớn hơn 0
+) x - 1 > 0 => x > 1
+) x - 5 > 0 => x > 5
=> x > 5
TH2 : cả x - 1 và x - 5 đều bé hơn 0
+) x - 1 < 0 => x < 1
+) x - 5 < 0 => x < 5
=> x < 1
Vậy,..........
\(a,\left(x+2\right)^{10}+\left(x+2\right)^8=0\\ \Leftrightarrow\left(x+2\right)^8\left[\left(x+2\right)^2+1\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x+2\right)^8=0\\\left(x+2\right)^2+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+2=0\\\left(x+2\right)^2=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\end{matrix}\right.\\ b,\left(x+3\right)^{10}-\left(x+3\right)^8=0\\ \Leftrightarrow\left(x+3\right)^8\left[\left(x+3\right)^2-1\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x+3\right)^8=0\\\left(x+3\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\\left(x+3\right)^2=1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x+3=1\\x+3=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\\x=-4\end{matrix}\right.\)
(x^2+1)(x-2)(3-x)<0
=>(x-2)(3-x)<0
=>(x-2)(x-3)>0
TH1: \(\left\{{}\begin{matrix}x-2>0\\x-3>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>2\\x>3\end{matrix}\right.\)
=>x>3
TH2: \(\left\{{}\begin{matrix}x-2< 0\\x-3< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< 2\\x< 3\end{matrix}\right.\)
=>x<2
Hướng làm: Áp dụng công thức \(A.B>0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}A>0\\B>0\end{matrix}\right.\\\left\{{}\begin{matrix}A< 0\\B< 0\end{matrix}\right.\end{matrix}\right.\)
\(\left(x-2\right)\left(x+\dfrac{2}{3}\right)>0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\x+\dfrac{2}{3}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\x+\dfrac{2}{3}< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x>-\dfrac{2}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x< -\dfrac{2}{3}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x< -\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(x>2\) hoặc \(x< -\dfrac{2}{3}\)
Ta có: \(\left(x-2\right)\left(x+\dfrac{2}{3}\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\x+\dfrac{2}{3}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\x+\dfrac{2}{3}< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x< -\dfrac{2}{3}\end{matrix}\right.\)