Tìm x: (-3x+21).(5.|x|-15)=0
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[ -3x +21 ] [ 5 .| x | - 15 ] =0
<=> (-3x+21)=0 hoặc (5|x|-15)=0
<=> x=7 hoặc |x|=3
<=> x={7;-3;3}
a, 2x -(-17) = 15
2x + 17 = 15
2x = 15 - 17
2x =(-2)
x = (-2) : 2
x = (-1)
a ) 2x - 1 = 7
2x = 7 + 1
2x = 8
x = 8 : 2
x = 4
b ) 3 ( x + 5 ) = 21
( x + 5 ) = 21 : 3
( x + 5 ) = 7
x = 2
c ) 20 - ( 3x - 1 ) = 15
( 3x - 1 ) = 20 - 15
( 3x - 1 ) = 5
3x = 6
x = 2
d ) ( x - 135 ) -120 = 0
( x - 135 ) = 120
x = 120 + 135
x = 255
a)2x - 1 = 7
=>2x=8
=>x=4
b) 3(x+5) = 21
=>x+5=7
=>x=2
c) 20 - (3x - 1) = 15
=>3x-1=5
=>3x=6
=>x=2
d) (x - 135) - 120 = 0
=>x-135=120
=>x=155
a) 7x+4=3x+16\(\Leftrightarrow\)4x=12\(\Leftrightarrow\)x=3
b)(x+9)(3x-15)=0\(\Leftrightarrow\)x+9=0 hoặc 3x-15=0
\(\Rightarrow\)x\(\in\){-9;5}
c) |-5x|=2x+21
Nếu x\(\le\)0 thì -5x=2x+21\(\Leftrightarrow\)x=-3 (t/m)
Nếu x>0 thì -5x=-2x-21\(\Leftrightarrow\)x=7 (t/m)
Vậy x\(\in\){-3;7}
d) 3x-5>15-x\(\Leftrightarrow\)4x>20\(\Leftrightarrow\)x>5
e) \(\dfrac{x+1}{2001}+\dfrac{x+5}{2005}< \dfrac{x+9}{2009}+\dfrac{x+13}{2013}\)
\(\Leftrightarrow\dfrac{x+1}{2001}-1+\dfrac{x+5}{2005}-1< \dfrac{x+9}{2009}-1+\dfrac{x+13}{2013}-1\)
\(\Leftrightarrow\)\(\dfrac{x-2000}{2001}+\dfrac{x-2000}{2005}-\dfrac{x-2000}{2009}-\dfrac{x-2000}{2013}< 0\)
\(\Leftrightarrow\)(x-2000)(\(\dfrac{1}{2001}+\dfrac{1}{2005}-\dfrac{1}{2009}-\dfrac{1}{2013}\))<0
Vì \(\dfrac{1}{2001}+\dfrac{1}{2005}-\dfrac{1}{2009}-\dfrac{1}{2013}>0\) nên x-2000<0
\(\Leftrightarrow\)x<2000
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x2 - 16x - 34 = 10x2 + 3x - 34
=> 10x2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0
hoặc 10x - 19 = 0 => 10x = 19 => x = 19/10
Vậy x = 0 ; x = 19/10
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x 2 - 16x - 34 = 10x 2 + 3x - 34
=> 10x 2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0 hoặc 10x - 19 = 0
=> 10x = 19
=> x = 19/10
Vậy x = 0 ; x = 19/10
a: \(\Leftrightarrow x^2+10x+25-x^2+4x=55\)
=>14x=30
hay x=15/7
b: \(\Leftrightarrow\left(x-7\right)\left(x-3\right)=0\)
hay \(x\in\left\{7;3\right\}\)
\(\left(-3x+21\right)\left(5.|x|-15\right)=0\Leftrightarrow-3x+21=0hoac5.|x|-15=0\)
\(+-3x+21=0\Rightarrow-3x=-21\Rightarrow x=7\)
\(+5.|x|-15=0\Rightarrow5.|x|=0+15=15\Rightarrow|x|=15:5=3\Rightarrow x\in\left\{-3;3\right\}\)
\(Vay:x\in\left\{7;-3;3\right\}\)