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Bài 1 :
a) x < 0
b) x > 0
c) <=> 3 + |3x - 1| = 5
<=> |3x - 1| = 5 - 3 = 2
<=> 3x - 1 = 2 hoặc -3x + 1 = 2
<=> 3 x = 3 hoặc -3x = 1
<=> x = 1 hoặc x = -1/3
Bài 2 :
a) 27 = 33 < 3n < 243 = 35
<=> 3 < n < 5
Vì n thuộc N* nên n thuộc {4; 5}
b) 32 = 25 < 2n < 128 = 27
<=> 5 < n < 7. Vì n thuộc N* nên n = 6
c) 125 = 5 . 25 = 5 . 52 < 5.5n < 5 . 125 = 5 . 53
<=> 2 < n < 3. Vì n thuộc N* nên n = 3
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!
a)\(\sqrt{x}=4\Leftrightarrow x=4^2\Leftrightarrow x=16\)
b)\(\sqrt{x-2}=3\Leftrightarrow x-2=3^2\Leftrightarrow x=9-2=7\)
c)\(\sqrt{\dfrac{x}{3}-\dfrac{7}{6}}=\dfrac{1}{6}\Leftrightarrow\dfrac{x}{3}-\dfrac{7}{6}=\dfrac{1}{36}\Leftrightarrow\dfrac{x}{3}=-\dfrac{41}{36}\Leftrightarrow x=-\dfrac{41}{12}\)
d)\(x^2=7vớix< 0\)
\(\Leftrightarrow\left(-x\right)^2=7\Leftrightarrow-x=\sqrt{7}\Leftrightarrow x=-\sqrt{7}\)
e)\(x^2-4=0với>0\)
\(\Leftrightarrow x^2=4\Leftrightarrow x=\sqrt{4}=2\)
f)\(\left(2x+7\sqrt{7}\right)^2=7\)
\(\Leftrightarrow4x^2+\sqrt{5488}+343=7\)
\(\Leftrightarrow4x^2+\sqrt{5488}=-336\)
\(\Leftrightarrow4x^2=28\left(12-\sqrt{7}\right)\Leftrightarrow x^2=\dfrac{28\left(12-\sqrt{7}\right)}{4}=7\left(12-\sqrt{7}\right)\)
\(\Leftrightarrow x=\sqrt{7\left(12-\sqrt{7}\right)}=\sqrt{84-7\sqrt{7}}\)
Ể? \(x^2+x+1=0\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}=0\left(VL\right)\) rồi mà SP.
\(\left(x-\frac{2}{5}\right)\left(x+\frac{2}{7}\right)>0\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{2}{5}>0\\x+\frac{2}{7}>0\end{cases}\Leftrightarrow\orbr{\begin{cases}x>\frac{2}{5}\\x>-\frac{2}{7}\end{cases}\Leftrightarrow}x>\frac{2}{5}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{2}{5}< 0\\x+\frac{2}{7}< 0\end{cases}\Leftrightarrow\orbr{\begin{cases}x< \frac{2}{5}\\x< -\frac{2}{7}\end{cases}\Leftrightarrow}x< -\frac{2}{7}}\)
b) \(\left(2x-\frac{1}{2}\right)\left(3x-\frac{1}{3}\right)< 0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-\frac{1}{2}>0\\3x-\frac{1}{3}< 0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>\frac{1}{4}\\x< \frac{1}{9}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-\frac{1}{2}< 0\\3x-\frac{1}{3}>0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x< \frac{1}{4}\\x>\frac{1}{9}\end{cases}}\)
a) ( x - 2/5 )( x + 2/7 ) > 0
Xét hai trường hợp :
1. \(\hept{\begin{cases}x-\frac{2}{5}>0\\x+\frac{2}{7}>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>\frac{2}{5}\\x>-\frac{2}{7}\end{cases}\Leftrightarrow}x>\frac{2}{5}\)
2. \(\hept{\begin{cases}x-\frac{2}{5}< 0\\x+\frac{2}{7}< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< \frac{2}{5}\\x< -\frac{2}{7}\end{cases}}\Leftrightarrow x< -\frac{2}{7}\)
Vậy với x > 2/5 hoặc x < -2/7 thì ( x - 2/5 )( x + 2/7 ) > 0
b) ( 2x - 1/2 )( 3x - 1/3 ) < 0
Xét hai trường hợp :
1. \(\hept{\begin{cases}2x-\frac{1}{2}>0\\3x-\frac{1}{3}< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x>\frac{1}{2}\\3x< \frac{1}{3}\end{cases}}\Leftrightarrow\hept{\begin{cases}x>\frac{1}{4}\\x< \frac{1}{9}\end{cases}}\)( loại )
2. \(\hept{\begin{cases}2x-\frac{1}{2}< 0\\3x-\frac{1}{3}>0\end{cases}\Leftrightarrow}\hept{\begin{cases}2x< \frac{1}{2}\\3x>\frac{1}{3}\end{cases}}\Leftrightarrow\hept{\begin{cases}x< \frac{1}{4}\\x>\frac{1}{9}\end{cases}}\Leftrightarrow\frac{1}{9}< x< \frac{1}{4}\)
Vậy với 1/9 < x < 1/4 thì ( 2x - 1/2 )( 3x - 1/3 ) < 0
\(\frac{2x+1}{x+3}< 0\)khi \(2x+1< 0\)hoặc \(x+3< 0\)
\(\Leftrightarrow x< \frac{-1}{2}\)hoặc \(x< -3\)
a) nếu x-1 >= 0 hay x >=1 ta có |x-1|=x-1
nếu x-1 < 0 hay x < 1 ta có |x-1| = 1-x
với x >= 1 ta có
|x-1| = 2x - 5
x-1 = 2x - 5
x-2x = -5 + 1
-x = -4
x=4 ( thỏa mãn khoảng xét x>=1)
với x < 1 ta có
|x-1| = 2x - 5
1-x = 2x - 5
-x - 2x = -5 -1
-3x = -6
x=2 ( không thỏa mãn khoảng xét x < 1 )
