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(x+3).(2x-18)=0
=>x+3=0 hoặc 2x-18 =0
+)x+3=0
x=-3
+)2x-18=0
2x =18
x =9
Vậy x \(\in\)\(_{\left\{-3;9\right\}}\)
a, 2x + 35 -x+27=0
x +62=0
x=-62
b, 2x -41 -3x + 23 =0
-x -18=0
-x=18
x=-18
c, 4x -12-3x-15= -124
x -27=-124
x= -97
d, Suy ra x+3 =0 hoặc 2x-18=0
x=-3 hoặc 2x=18 => x=9
vậy x=-3 hoặc x=9
Bài 1:
a, 2x + 35 = x - 27
2x - x = -27 - 35
x = -62
b, 2x - 41 = 3x + (-23)
2x - 3x = -23 + 41
-x = 18
x = -18
c, 4 . (x - 3) - 3 . (x - 5) = 45 . (-2) - 34
4x - 12 - 3x + 15 = -90 - 34
4x - 3x = -90 - 34 + 12 - 15
x = -127
Bài 2:
- ( -1234 + 345 - 29 ) - ( 1234 + 135 - 59 )
= 1234 - 345 + 29 - 1234 - 135 + 59
= (1234 - 1234) - (345 + 135) + (29 + 59)
= 0 - 480 + 88
= -392
Bài 1 Tìm x biết:
a)65-(29-x)=32
65 -29+x=31
x=31-65+29
x=-5
b)(x+5)-(x+23)=x-34
x+5 -x +23 = x-34
(x-x)+ (23+5)=x-34
0+28=x-34
28=x-34
28+34=x
62=x
=>x=62
c)(16-x)+(x-38)=x+44
16-x+x-38=x+44
-x+x-x=44-16+38
-x=36
=>x=-36
d)-12+3(-x+7)=-18
3(-x+7)=-18+12
3(-x+7)=-6
-x+7=-6:3
-x+7=-2
-x=-2-7
-x=-9
=>x=9
Baif 2
d)|7-x|=10
=> \(\left[{}\begin{matrix}7-x=10\\7-x=-10\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=7-10\\x=-10-7\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=-3\\x=-17\end{matrix}\right.\)
e)(x-6).(7-2x)=0
\(\Rightarrow\)\(\left[{}\begin{matrix}x-6=0\\7-2x=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0+6\\2x=7\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=6\\x=7:2\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=6\\x=3,5\end{matrix}\right.\)
f)(9-x).(2x+8)=0
\(\Rightarrow\)\(\left[{}\begin{matrix}9-x=0\\2x+8=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0+9\\2x=-8\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=9\\x=-4\end{matrix}\right.\)
g)x(-x+8).(-3x-18)=0
\(\Rightarrow\) \(\left[{}\begin{matrix}x=0\\-x+8=0\\-3x-18=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\-x=0+8\\-3x=0+18\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\-x=8\\-3x=18\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\x=-8\\x=18:\left(-3\right)\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\x=-8\\x=-6\end{matrix}\right.\)
h)(-x+8).(x-54).(-24-x)=0
\(\Rightarrow\)\(\left[{}\begin{matrix}-x+8=0\\x-54=0\\-24-x=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}-x=8\\x=0+54\\-x=0+24\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=8\\x=54\\-x=24\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=8\\x=54\\x=-24\end{matrix}\right.\)
1/\(-2020+23+x=-2020\\ \Leftrightarrow23+x=-2020+2020\\ \Leftrightarrow23+x=0\\ \Leftrightarrow x=0-23\\ \Leftrightarrow x=-23\)
Vậy...
2/\(2x-35=25\\ \Leftrightarrow2x=60\\ \Leftrightarrow x=30\)
Vậy...
3/\(3x+17=26\\ \Leftrightarrow3x=9\\ \Leftrightarrow x=3\)
Vây...
4/\(\left|\text{x}-1\right|=0\\ \Leftrightarrow x-1=0\\ \Leftrightarrow x=1\)
Vậy...
5/ \(-17.\left|x\right|=-34\\ \Leftrightarrow\left|x\right|=2\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
VẬy...
Sorry mk ko giải đc trong 15 phút đâu , vậy mk thách bạn giải đc trong 5 phút đấy . Nếu bạn giải đc mk xin bạn nhận mk 20 lậy
a)280-(x-140):35=270
(x-140):35=280-270
(x-140):35= 10
x-140 =10x35
x-140 =350
x= 350+140
x=490
b)(190-2x):35-32=16
(190-2x):35 =16+32
(190-2x):35 = 48
190-2x= 48x35
190-2x=1680
2x=190-1680
2x=-1490
x=-1490:2
x=-745
c)720:[41-(2x-5)]=\(2^3\)x5
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
a, 2x - 41 = 3x +(-23)
2x- 41 = 3x - 23
2x- 3x = -23 + 41
-x = 18
x = -18
b, 4(x-3) - 3(x +5) =45 .(-2) - 3
4x- (4.3) - 3x + ( 3. 5) = -90 - 3
4x - 12 - 3x + 15 = -93
4x -3x -12 + 15= -93
x + 3 = -93
x= -93 - 3
x= -96
\(a.2x+35=x-27\)
\(\Leftrightarrow2x-x=-27-35\)
\(\Rightarrow x=-62\). Vậy \(x=-62\)
\(b.2x-41=3x+\left(-23\right)\)
\(\Leftrightarrow2x-3x=-23+41\)
\(\Leftrightarrow-x=18\)
\(\Rightarrow x=-18\). Vậy \(x=-18\)
\(c.4\left(x-3\right)-3\left(x+5\right)=45.\left(-2\right)-34\)
\(\Leftrightarrow4x-12-3x-15=-124\)
\(\Leftrightarrow4x-3x=-124+12+15\)
\(\Rightarrow x=-97\). Vậy \(x=-97\)
\(d.\left(x+3\right)\left(2x-18\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x+3=0\\2x-18=0\end{cases}\Rightarrow\hept{\begin{cases}x=-3\\x=9\end{cases}}}\). Vậy \(x=-3;9\)