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\((x-6)(3x-9)>0\)
TH1:
\(\orbr{\begin{cases}x-6< 0\\3x-9< 0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x< 6\\x< 3\end{cases}}\)\(\Rightarrow x< 3\)
TH2:
\(\orbr{\begin{cases}x-6>0\\3x-9>0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x>6\\x>3\end{cases}}\)\(\Rightarrow x>6\)
Vậy \(x< 3\) hoặc \(x>6\)thì \((x-6)(3x-9)>0\)
Học tốt!
20.
\((2x-1)(6-x)>0\)
TH1:
\(\orbr{\begin{cases}2x-1>0\\6-x>0\end{cases}\Rightarrow\orbr{\begin{cases}x< \frac{1}{2}\\x< 6\end{cases}}\Rightarrow x< 6}\)
TH2
\(\orbr{\begin{cases}2x-1< 0\\6-x< 0\end{cases}\Rightarrow\orbr{\begin{cases}x>\frac{1}{2}\\x>6\end{cases}}\Rightarrow x>\frac{1}{2}}\)
Vậy \(x< 6\)hoặc \(x>\frac{1}{2}\)thì \((2x-1)(6-x)>0\)
l) (x + 9) . (x2 – 25) = 0
<=> (x + 9) . (x – 5) . (x + 5) = 0
<=> \(\left[{}\begin{matrix}\text{x + 9 = 0}\\x-5=0\\x+5=0\end{matrix}\right.\left[{}\begin{matrix}x=-9\\x=5\\x=-5\end{matrix}\right.\)
Vậy S = \(\left\{-9,5,-5\right\}\)
e) |x - 4 |< 7
<=> \(\left[{}\begin{matrix}x-4=7\\x-4=-7\end{matrix}\right.< =>\left[{}\begin{matrix}x=11\\x=-3\end{matrix}\right.\)
Vậy S = \(\left\{11;-3\right\}\)
I,(x+9).(x^2-25)=0
tương đương:x+9=0
x^2-25=0
tương đương : x=-9
x=5
e,\(\left|x-4\right|\)=7
tương đương x-4=4
x-4=-4
tương đương :x=0
x=-8
a: -5x-16=x+40
=>-6x=56
hay x=-28/3
b: \(4x-10=15-x\)
=>5x=25
hay x=5
c: \(-12\left(x-5\right)+7\left(3-x\right)=5\)
=>-12x+60+21-7x=5
=>-19x+71=5
=>-19x=-76
hay x=4
d: \(\left(x-2\right)\left(x+15\right)=0\)
=>x-2=0 hoặc x+15=0
=>x=2 hoặc x=-15
1.
a, \(x-14=3x+18\)
\(\Rightarrow x-3x=18+14\)
\(\Rightarrow-2x=32\Rightarrow x=\frac{32}{-2}=-16\)
b, \(\left(x+7\right).\left(x-9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+7=0\\x-9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-7\\x=9\end{cases}}}\)
c, \(\left|2x-5\right|-7=22\)
\(\Rightarrow\left|2x-5\right|=22+7\)
\(\Rightarrow\left|2x-5\right|=29\)
\(\Rightarrow\orbr{\begin{cases}2x+5=29\\2x-5=29\end{cases}}\Rightarrow\orbr{\begin{cases}2x=24\\2x=34\end{cases}\Rightarrow}\orbr{\begin{cases}x=12\\x=17\end{cases}}\)
d, \(\left(\left|2x\right|-5\right)-7=22\)
\(\Rightarrow\left(\left|2x\right|-5\right)=29\)
\(\Rightarrow\left|2x\right|=29+5\Rightarrow\left|2x\right|=34\Rightarrow x=\pm17\)
e, \(\left|x+3\right|+\left|x+9\right|+\left|x+5\right|=4x\)
Vì \(\left|x+3\right|\ge0;\left|x+9\right|\ge0;\left|x+5\right|\ge0;4x\ge0\)
Nên \(\left|x+3\right|+\left|x+9\right|+\left|x+5\right|=4x\ge0\)
\(\Rightarrow\left|x+3\right|>0\Rightarrow\left|x+3\right|=x+3\)
\(\left|x+9\right|>0\Rightarrow\left|x+9\right|=x+9\)
\(\left|x+5\right|>0\Rightarrow\left|x+5\right|=x+5\)
Ta có :
\(x+3+x+9+x+5=4x\)
\(\Rightarrow3x+\left(3+9+5\right)=4x\)
\(\Rightarrow4x-3x=17\)
\(\Rightarrow x=17\)
2. a , b sai đề bn
c, \(\left(5x+1\right).\left(y-1\right)=4\)
\(\Rightarrow\left(5x+1\right).\left(y-1\right)\inƯ\left(4\right)\)
\(\text{ }Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
Ta có bảng sau :
| 5x+1 | 1 | -1 | 2 | -2 | 4 | -4 |
| y-1 | -4 | 4 | -2 | 2 | -1 | 1 |
| x | 0 | -2/5 | 1/5 | -3/5 | 3/5 | -1 |
| y | -3 | 5 | -1 | 3 | 0 | 2 |
d, \(5xy-5x+y=5\)
\(\Rightarrow\left(5xy-5x\right)+y=5\)
\(\Rightarrow5x.\left(y-1\right)+y=5\)
\(\Rightarrow\left(5x+1\right).\left(y-1\right)=4\)
\(\Rightarrow\left(5x+1\right).\left(y-1\right)\inƯ\left(4\right)\)
\(Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
Ta có bảng sau :
| 5x+1 | 1 | -1 | 2 | -2 | 4 | -4 |
| y-1 | -4 | 4 | -2 | 2 | -1 | 1 |
| x | 0 | -2 | 1/5 | -3/5 | 3/5 | -1 |
| y | -3 | 5 | -1 | 3 | 0 | 2 |
a) (x – 45).27 = 0
=> x - 45 = 0
=> x = 45
b) 23.(42- x) = 23
=> 42- x = 1
=> x = 41
c. 3x – 5=7
=> 3x = 12
=> x = 4
e. 15 – 5x=10
=> 5x = 5
=> x = 1
a) x - 7 = 5. 0 => x - 7 = 0 =>x = 7.
b) x: 3 = 47 +13 => x: 3 = 60 => x = 60.3 => x = 180.
c) x : 7 - 7 = 0 hoặc x : 12 - 12 = 0. Do đó x = 49 hoặc x = 144.
d) x : 2 = 150 - 135 => x: 2 = 15 => x = 15.2 => x = 30.
e) 100: x = 140 -120 => 100: x = 20 => x = 100:20 => x = 5.
g) x : 5 = 300 - 273 => x : 5 = 27 =>x = 27.5 => x = 135
5x( x - 7 ) - 40( x - 7 ) = 0
<=> ( x - 7 )( 5x - 40 ) = 0
<=> \(\orbr{\begin{cases}x-7=0\\5x-40=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\x=8\end{cases}}\)
x( x + 5 ) - 7( x + 5 ) = 0
<=> ( x + 5 )( x - 7 ) = 0
<=> \(\orbr{\begin{cases}x+5=0\\x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=7\end{cases}}\)
Ta có: \(5x.\left(x-7\right)-40.\left(x-7\right)=0\)
=> \(\orbr{\begin{cases}5x-40=0\\x-7=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=8\\x=7\end{cases}}\)
Ta có: \(x.\left(x+5\right)-7.\left(x+5\right)=0\)
=> \(\orbr{\begin{cases}x-7=0\\x+5=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=7\\x=-5\end{cases}}\)
Mình làm hơi tắt xíu :v ~~~