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a)(x-1).(x-2)>0
\(\Leftrightarrow\hept{\begin{cases}\left(x-1\right)>0\\\left(x-2\right)>0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x>1\\x>2\end{cases}}\)
Vậy x>2
b)(x-2)2.(x+1).(x-4)<0
\(\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^2< 0\\\left(x+1\right)< 0\\\left(x-4\right)< 0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x< 2\\x< -1\\x< 4\end{cases}}\)
Vậy x<(-1)
c)Từ đề bài, ta suy ra:
\(\left(x-9\right)< 0\Leftrightarrow x< 9\)
d)\(\frac{5}{x}< 1\Leftrightarrow x< 5\)
\(\left(x-1\right)\left(x-2\right)>0\)
TH1: \(\hept{\begin{cases}x-1>0\\x-2>0\end{cases}}\Rightarrow x>2\)
TH2: \(\hept{\begin{cases}x-1< 0\\x-2< 0\end{cases}}\Rightarrow x< 1\)
tìm x sao cho :
a, 1-2x<7
b, (x-1)(x-2)>0
c, (x-2)(x+1)(x-4)<0
d, \(\frac{x^2\left(x-3\right)}{x-9}< 0\)
\(-2x< 7\Leftrightarrow x>-3,5\)
\(\left(x-1\right)\left(x-2\right)>0\Leftrightarrow x^2-3x+2>0\Leftrightarrow x^2-3x+\frac{9}{4}>\frac{1}{4}\)
\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2>\frac{1}{4}\Leftrightarrow\orbr{\begin{cases}x-\frac{3}{2}>\frac{1}{2}\\x-\frac{3}{2}< -\frac{1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>2\\x< 1\end{cases}}\)
a)\(1-2x< 1\)
\(\Leftrightarrow2x>0\)
\(\Leftrightarrow x>0\)
b)\(\left(x-2\right)^2\left(x+1\right)\left(x-4\right)< 0\)
\(\Leftrightarrow\hept{\begin{cases}x\ne2\\\left(x+1\right)\left(x-4\right)< 0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ne2\\x+1< 0\\x-4>0\end{cases}}\)hoặc \(\hept{\begin{cases}x\ne2\\x+1>0\\x-4< 0\end{cases}}\)
mà \(x+1>x-4\forall x\)
nên \(\hept{\begin{cases}x\ne2\\x+1>0\\x-4< 0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x\ne2\\x>-1\\x< 4\end{cases}}\)
hay \(\hept{\begin{cases}x\ne2\\-1< x< 4\end{cases}}\)
c)\(x-2< 0\)
\(\Leftrightarrow x< 2\)
d)\(\frac{x^2\left(x-3\right)}{x-9}< 0\left(x\ne9\right)\)
\(\Leftrightarrow\hept{\begin{cases}x\ne0\\\frac{x-3}{x-9}< 0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ne0\\x-3< 0\\x-9>0\end{cases}}\)hoặc \(\hept{\begin{cases}x\ne0\\x-3>0\\x-9< 0\end{cases}}\)
mà \(x-3>x-9\forall x\)
\(\Leftrightarrow\hept{\begin{cases}x\ne0\\x-3>0\\x-9< 0\end{cases}}\)\(\Leftrightarrow3< x< 9\)
e)\(\frac{5}{x}< 1\left(x\ne0\right)\)
\(\Leftrightarrow x>5\)
f)\(8x>2x\)
\(\Leftrightarrow6x>0\)
\(\Leftrightarrow x>0\)
g)\(x+a< a\)
\(\Leftrightarrow x< 0\)
h)\(x^3< x^2\)
\(\Leftrightarrow x^2\left(x-1\right)< 0\)
\(\Leftrightarrow\hept{\begin{cases}x\ne0\\x-1< 0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ne0\\x< 1\end{cases}}\)
câu d) nè
Ta có:5/x<1
=>\(\frac{5}{x}-1<0\Rightarrow\frac{5-x}{x}<0\Rightarrow x<0;x>5\)
Vậy x có 2 gtri
nhớ ****
mấy cái này đơn dãng vô cùng nhưng có đều bn ra đề dài quá nha
a) \(3x+4\ge7\Leftrightarrow3x\ge7-4\Leftrightarrow3x\ge3\Leftrightarrow x\ge1\) vậy \(x\ge1\)
b) \(-5x+1< 11\Leftrightarrow-5x< 11-1\Leftrightarrow-5x< 10\Leftrightarrow x>\dfrac{10}{-5}\)
\(\Leftrightarrow x>-2\) vậy \(x>-2\)
c) \(\dfrac{5}{x-3}< 0\Leftrightarrow x-3< 0\Leftrightarrow x< 3\) vậy \(x< 3\)
d) \(\dfrac{-7}{2-x}\ge0\Leftrightarrow2-x\le0\Leftrightarrow x\ge2\) vậy \(x\ge2\)
e) \(x^2+4x>0\Leftrightarrow x\left(x+4\right)>0\) \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>0\\x+4>0\end{matrix}\right.\\\left[{}\begin{matrix}x< 0\\x+4< 0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>0\\x>-4\end{matrix}\right.\\\left[{}\begin{matrix}x< 0\\x< -4\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x>0\\x< -4\end{matrix}\right.\) vậy \(x>0\) hoặc \(x< -4\)
f) \(\dfrac{x-2}{x-6}< 0\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x-2>0\\x-6>0\end{matrix}\right.\\\left[{}\begin{matrix}x-2< 0\\x-6< 0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>2\\x>6\end{matrix}\right.\\\left[{}\begin{matrix}x< 2\\x< 6\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>6\\x< 2\end{matrix}\right.\)
vậy \(x>6\) hoặc \(x< 2\)
g) \(\left(x-1\right)\left(x+2\right)\left(3-x\right)< 0\Leftrightarrow-\left[\left(x-1\right)\left(x+2\right)\left(x-3\right)\right]< 0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x-3\right)>0\)
th1: 3 số hạng đều dương : \(\Leftrightarrow\left[{}\begin{matrix}x-1>0\\x+2>0\\x-3>0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x>1\\x>-2\\x>3\end{matrix}\right.\) \(\Rightarrow x>3\)
th2: 2 âm 1 dương : (vì trong 3 số hạng ta có : \(\left(x+2\right)\) lớn nhất \(\Rightarrow\left(x+2\right)\) dương)
\(\Leftrightarrow\left[{}\begin{matrix}x-1< 0\\x+2>0\\x-3< 0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x< 1\\x>-2\\x< 3\end{matrix}\right.\) \(\Rightarrow-2< x< 1\)
vậy \(x>3\) hoặc \(-2< x< 1\)
h) \(\dfrac{x^2-1}{x}>0\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x^2-1>0\\x>0\end{matrix}\right.\\\left[{}\begin{matrix}x^2-1< 0\\x< 0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x^2>1\\x>0\end{matrix}\right.\\\left[{}\begin{matrix}x^2< 1\\x< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\\x>0\end{matrix}\right.\\\left[{}\begin{matrix}-1< x< 1\\x< 0\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>1\\-1< x< 0\end{matrix}\right.\) vậy \(x>1\) hoặc \(-1< x< 0\)
i) \(x^2+x-2< 0\Leftrightarrow x^2+x+\dfrac{1}{4}-\dfrac{9}{4}< 0\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2-\dfrac{9}{4}< 0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2< \dfrac{9}{4}\Leftrightarrow\dfrac{-3}{2}< \left(x+\dfrac{1}{2}\right)< \dfrac{3}{2}\Leftrightarrow-2< x< 1\)
vậy \(-2< x< 1\)
Mysterious Person, Đoàn Đức Hiếu, Nguyễn Đình Dũng , ... giúp mình!
\(b.\) \(\left(x-1\right).\left(x-2\right)>0\)
\(\Leftrightarrow x-1\) và \(x-2\) cùng dấu
\(\Leftrightarrow\hept{\begin{cases}x-1>0\\x-2>0\end{cases}}\) Hoặc: \(\Leftrightarrow\hept{\begin{cases}x-1< 0\\x-2< 0\end{cases}}\)
T/hợp 1: \(\hept{\begin{cases}x-1>0\\x-2>0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x>1\\x>2\end{cases}}\)
T/hợp 2: \(\hept{\begin{cases}x-1< 0\\x-2< 0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x< 1\\x< 2\end{cases}}\)
Vậy: ..................................
\(e.\)\(\frac{5}{x}< 1\)
\(\Leftrightarrow x>5\)
Vậy: .............................