Bài 3: Tìm x biết:
a) (5+x)-(15-31)=(-20)
b) 17= (x+3)+15-l-3l = 2
c) 27-lx-5l=12-(-3)
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a) 3x-17=x+3
=>3x-x =3+17
=>2x =20
=>x =20:2
=>x =10
b) 2x-15=-47
=>2x =-47+15
=>2x =-32
=>x =-32:2
=>x =-16
c) Ix-3I-12=I-5I
=>Ix-3I-12=5
=>Ix-3I =5+12
=>Ix-3I =17
TH1 x-3=17
x =17+3
x =20
TH2 x-3=-17
x =-17+3
x =-14
Vậy x=-14;20
d) (x-3)x(2x+6)=0
Để biểu thức trên bằng 0
<=> x-3=0=>x=0+3=>x=3
x=0
2x+6=0=>2x=0-6=>2x=-6=>x=-6:2=>x=-3
Vậy x=-3;0;3
c) \(\left(34-2x\right)\left(2x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}34-2x=0\\2x-6-0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)
d) \(\left(2019-x\right)\left(3x-12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\3x=12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\x=4\end{matrix}\right.\)
e) \(57\left(9x-27\right)=0\)
\(\Rightarrow9x-27=0\)
\(\Rightarrow9\left(x-3\right)=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
f) \(25+\left(15-x\right)=30\)
\(\Rightarrow25+15-x=30\)
\(\Rightarrow40-x=30\)
\(\Rightarrow x=40-30\)
\(\Rightarrow x=10\)
g) \(43-\left(24-x\right)=20\)
\(\Rightarrow43-24+x=20\)
\(\Rightarrow19+x=20\)
\(\Rightarrow x=20-19\)
\(\Rightarrow x=1\)
h) \(2\left(x-5\right)-17=25\)
\(\Rightarrow2\left(x-5\right)=17+25\)
\(\Rightarrow x-5=21\)
\(\Rightarrow x=21+5\)
\(\Rightarrow x=26\)
i) \(3\left(x+7\right)-15=27\)
\(\Rightarrow3\left(x+7\right)=27+15\)
\(\Rightarrow x+7=14\)
\(\Rightarrow x=14-7\)
\(\Rightarrow x=7\)
j) \(15+4\left(x-2\right)=95\)
\(\Rightarrow4\left(x-2\right)=95-15\)
\(\Rightarrow4\left(x-2\right)=80\)
\(\Rightarrow x-2=20\)
\(\Rightarrow x=20+2\)
\(\Rightarrow x=22\)
k) \(20-\left(x+14\right)=5\)
\(\Rightarrow x+14=20-5\)
\(\Rightarrow x+14=15\)
\(\Rightarrow x=15-14\)
\(\Rightarrow x=1\)
l) \(14+3\left(5-x\right)=27\)
\(\Rightarrow3\left(5-x\right)=27-14\)
\(\Rightarrow3\left(5-x\right)=13\)
\(\Rightarrow5-x=\dfrac{13}{3}\)
\(\Rightarrow x=5-\dfrac{13}{3}\)
\(\Rightarrow x=\dfrac{2}{3}\)
a, |x - 5| = x - 5 ( đk : x >= 5 )
<=> x - 5 = ( x - 5 )^2
<=> x - 5 = x^2 - 10x + 25
<=> x^2 - 10x + 25 - x + 5 = 0
<=> x^2 - 11x + 30 = 0
<=> x^2 - 5x - 6x + 30 = 0
<=> ( x^2 - 5x) - ( 6x - 30) = 0
<=> x ( x- 5) - 6( x- 5 ) = 0
<=> ( x- 5).(x - 6) =0
<=> Th1 : x- 5 = 0 => x = 5
Th2 : x - 6 = 0 => x = 6
khó kinh
a)Ta có :
\(\left|x-5\right|\ge0\)
\(\Rightarrow5-x\ge0\)
Mà 5 > 0
\(\Rightarrow x\ge0\)
Nên |x - 5| = 5 - x
=> x - 5 = 5 - x
=> x + x = 5 + 5
=> 2x = 10
=> x = 5
b) Ta có :
\(\left|x+3\right|\ge0\)
\(\left|x+2\right|\ge0\)
\(\Rightarrow\left|x+3\right|+\left|x+2\right|\ge0\)
\(\Rightarrow x\ge0\)
Nên |x + 3| + |x + 2| = x
=> x + 3 + x + 2 = x
=> 2x + 5 = x
=> 2x - x = -5
=> x = -5
1.Tính
a) 7.[ 6 : 2 - 15 :(-3) - l-3l ]
= 7.[ 3 + 5 - 3]
= 7.[( 3 - 3 ) + 5]
= 7.[0 + 5]
= 7.5
= 35
b) 159.(18-59) - 59 .(18-159)
= 159 . ( - 41) - 59 . ( - 141 )
= ( - 6519 ) - ( - 8319 )
= 1800
2.Tìm x thuộc Z
a) x + 15 = 20 -4x
x+4x=20-15
5x=5
x=5:5
x=1
Vậy x=1
b) 3 - lx - 1l =0
|x-1|=3
* x-1=3 * x-1=-3
x=3+1 x=-3+1
x=4 x=-2
Vậy x=4 hoặc x=-2
c) 7(x-3) - 5 (3-x) = 11x - 5
7x-21-15+5x=11x-5
-21-15+5=11x-7x-5x
-31=-x
31=x
Vậy x=31
a, =>5+x -(-16) = -20
=> 5+x+16 = -20
=> x+21 = -20
=> x = -20 - 21 = -41
c, => 27-|x-5| = 12+3 = 15
=> |x-5| = 27 - 15 = 12
=> x-5=12 hoặc x-5=-12
=> x=17 hoặc x=-7
k mk nha
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