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14) (x - 2) . (x + 4) = 0
\(\Rightarrow\) x - 2 = 0 hoặc x + 4 = 0
Nếu x - 2 = 0
x = 0 + 2
x = 2
Nếu x + 4 = 0
x = 0 - 4
x = -4
Vậy x \(\in\) {2 ; -4)
15) (x - 2) . (x + 15) = 0
\(\Rightarrow\) x - 2 = 0 hoặc x + 15 = 0
Nếu x - 2 = 0
x = 0 + 2
x = 2
Nếu x + 15 = 0
x = 0 - 15
x = -15
Vậy x \(\in\) {-15 ; 2}
1) x - 2 = -6
x = -6 + 2
x = -4
2) -5 . x - ( -3 ) =13
-5 . x = 13 + ( -3 )
-5 . x = 10
x = 10 : ( -5 )
x = -2
1/(2.x-5)+17=6
=> 2x - 5 = -11
=> 2x = -6
=> x = 3
vậy_
2/10-2.(4-3x)=-4
=> 2(4 - 3x) = 14
=> 4 - 3x = 7
=> 3x = -3
=> x = -1
3/-12+3.(-x+7)=-18
=> 3(-x+7) = -6
=> -x+7 = -2
=> -x = -9
=> x = 9
4/24:(3.x-2)=-3
=> 3x - 2 = -8
=> 3x = -6
=> x = -2
5/-45:5.(-3-2.x)=3
=> 5(-3 - 2x) = -15
=> -3 - 2x = -3
=> - 2x = 0
=> x = 0
6/x.(x+7)=0
=> x = 0 hoặc x + 7 = 0
=> x = 0 hoặc x = -7
7/(x+12).(x-3)=0
=> x + 12 = 0 hoặc x - 3 = 0
=> x = -12 hoặc x = 3
8/(-x+5).(3-x)=0
=> -x + 5 = 0 hoặc 3 - x = 0
=> x = 5 hoặc x = 3
9/x.(2+x).(7-x)=0
=> x = 0 hoặc 2 + x = 0 hoặc 7 - x = 0
=> x = 0 hoặc x = -2 hoặc x = 7
10/(x-1).(x+2).(-x-3)=0
=> x - 1 = 0 hoặc x + 2 = 0 hoặc -x-3 = 0
=> x = 1 hoặc x = -2 hoặc x = -3
a)9.x + 1=73
9x=73-1
9x=72
x=72:9
x=8
b)2.x - 5 = -17 - 12
2x-5=-29
2x=-29+5
2x=-24
x=-24:2
x=-12
c)10 - x - 5 = - 5 - 7 -11
10-x-5=-12-11
10-x-5=-23
10-x=-23+5
10-x=18
x=10-18
x=-8
d)(-9) . x + 3 = (-2) . (-7) +16
-9x+3=14+16
-9x+3=30
-9x=30-3
-9x=27
x=27:(-9)
x=-3
(-12) . x - 34 =2
-12x=2+34
-12x=36
x=36:(-12)
x=-3
(-11).x + 9 =130
-11x=130-9
-11x=121
x=121:(-11)
x=11
(-5) .x + 5 = (-15) .(-4) -12
-5x+5=60-12
-5x+5=48
-5x=48-5
-5x=43
x=43:(-5)
x=-8,6
IxI -3=0
|x|=3
=>x=+3
(7 - IxI).(2.x - 4) =0
*7-|x|=0 * 2x-4=0
|x|=7 2x=4
=>x=+7 x=4:2
x=2
280-(x-140):35 =270
(x-140):35=280-270
(x-140):35=10
x-140=10.35
x-140=350
x=350+140
x=490
(1900 - 2.x ) : 35- 32 =16
1900-2x=(16+32).35
1900-2x=1680
2x=1900-1680
2x=220
x=220:2
x=110
720 :[41-(2x -5 )] =23 .5
720:[41-(2x-5)]=40
41-(2x-5)=720:40
41-(2x-5)=18
2x-5=41-18
2x-5=23
2x=23+5
2x=28
x=28:2
x=14
(x - 5).(x2 - 4 ) =0
* x-5=0 * x2-4=0
x=0+5 x2=4
x=5 x2=22
=> x=+2
a,3.x-12=125-5.5.5
3.x-12=125-125
3.x-12=0
3.x=0+12
3.x=12
x=12:3
x=4
b,x-2!+17=35
x-2=35-17
x-2=18
x=18+2
x=20
c,x.x-3.x=0
x.3-3=0
x.3=0+3
x.3=3
x=3:3
x=1
d, (2.x-4).(3.x-9)=0
khi2.x-4=0 thì 3.x-9=0
2.x-4=0 3.x-9=0
2.x=0+4 3.x=0+9
2.x=4 3.x=9
x=4:2 x=9:3
x=2 x=3
vậy x=2 hoạc x=3
h,(15-3.x).(2.x-7)=0
khi 15-3.x=0 thì 2.x-7=0
15-3.x=0 2.x-7=0
3.x=0+15 2.x=0+7
3.x=15 2.x=7
x=15:3 x=7:2
x=5 7ko chia hết cho 2 nên ko có x
vậy x =5
nhớ k cho mình nha
1/ x(x+17)=0
⇒ \(\left[{}\begin{matrix}x=0\\x+17=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-17\end{matrix}\right.\)
2/ (x+1112)(x-3)=0
⇒\(\left[{}\begin{matrix}x+1112=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1112\\x=3\end{matrix}\right.\)
3/ (-x+25)(3-x)=0
⇒\(\left[{}\begin{matrix}-x+25=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=25\\x=3\end{matrix}\right.\)
4/ x(12+x)(7-x)=0
⇒ \(\left[{}\begin{matrix}x=0\\12+x=0\\7-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-12\\x=7\end{matrix}\right.\)
5/ (x-15)(x+2)(-x-3)=0
⇒\(\left[{}\begin{matrix}x-15=0\\x+2=0\\-x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=15\\x=-2\\x=-3\end{matrix}\right.\)
\(x\left(x+17\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x+17=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-17\end{matrix}\right.\)
Vậy \(x\in\left\{0;-17\right\}\)
\(\left(x+1112\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1112=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1112\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{-1112;3\right\}\)
\(\left(-x+25\right)\left(3-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}-x+25=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=25\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{25;3\right\}\)
\(x\left(12+x\right)\left(7-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\12+x=0\\7-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-12\\x=7\end{matrix}\right.\)
Vậy \(x\in\left\{0;-12;7\right\}\)
\(\left(x-15\right)\left(x+2\right)\left(-x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-15=0\\x+2=0\\-x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=15\\x=-2\\x=-3\end{matrix}\right.\)
Vậy \(x\in\left\{15;-2;-3\right\}\)
a) x+5+2.x=17
x.1+2x=17-5
x.(2+1)=12
x.3=12
x=12:3
x=4
vay x = 4
c) (3-x).(x+5)=0
\(\Rightarrow\left[{}\begin{matrix}3-x=0\\x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3-0\\x=0-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
vay x \(\in\left\{3:-5\right\}\)
d) (2.x+2).(x-19)=0
\(\Rightarrow\left[{}\begin{matrix}2x+2=0\\x-19=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=0-2\\x=0+19\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=-2\\x=19\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2:2\\x=19\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=19\end{matrix}\right.\)
vay x \(\in\left\{-1;19\right\}\)
e) (x+2)3=(-125)
x+2=-125:3( loai )
vay x ko co gia tri nao thoa man
f) |x-3|=4
\(\Rightarrow\left[{}\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4+3\\x=-4+3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)
vay x\(\in\left\{7;-1\right\}\)
g) 2.|7-x|=16
\(\)/7-x/=16:2
/7-x/=8
\(\Rightarrow\left[{}\begin{matrix}7-x=-8\\7-x=8\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-8+7\\x=8+7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=15\end{matrix}\right.\)
vay \(x\in\left\{-1;15\right\}\)
h) 12-2.|x-10|=(-18)
2/x-10/=-18+12
2/x-10/=-6
/x-10/=-6:2
/x-10/=-3
\(\Rightarrow\left[{}\begin{matrix}x-10=3\\x-10=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3-10\\x=-3+10\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=7\end{matrix}\right.\)
vay \(x\in\left\{-7;7\right\}\)
cảm ơn bạn nha