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(2x - 7) + 17 = 6
=> 2x - 7 = 6 - 17
=> 2x - 7 = -11
=> 2x = -11 + 7
=> 2x = -4
=> x = -4 : 2
=> x = -2
+) 12 -2(3 - 3x)= -2
=> 2(3 - 3x) = 12 + 2
=> 2(3 - 3x) = 14
=> 3 - 3x = 14 : 2
=> 3 - 3x = 7
=> 3x = 3 - 7
=> 3x = -4
=> x = -4/3
\(\left(x+1\right)\left(x-3\right)=0\)
=> \(\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)
Vậy...
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
\(\Rightarrow\)\(\left[{}\begin{matrix}-x+8=0\\x-54=0\\-24-x=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}-x=8\\x=0+54\\-x=0+24\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=8\\x=54\\-x=24\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=8\\x=54\\x=-24\end{matrix}\right.\)
a. |3x-6|=0
=> 3x-6=0
=> 3x=6
=> x=6:3
=> x=2
b. |7-2x| \(\ge0\)
mà |7-2x|=-8
=> không tồn tại x
c. |5x+6|=14
=> 5x+6=14 hoặc 5x+6=-14
=> 5x=14-6 hoặc 5x=-14-6
=> 5x=8 hoặc 5x=-20
=> x=8/5 hoặc x=-4
a) => 3x - 6= 0
=> 3x = 6
=> x = 2
b) |7 - 2x| \(\ge\) 0.
Mà đề cho |7 - 2x| = -8
=> x \(\in\) rỗng
c) |5x + 6| = 14
=> 5x + 6 = 14 hoặc 5x + 6 = -14
=> 5x = 8 hoặc 5x = -20
=> x = 8/5 hoặc x = -4
a) ta có : \(\left(2x-7\right)-\left(x+135\right)=0\Leftrightarrow2x-7-x-135=0\)
\(\Leftrightarrow x-142=0\Leftrightarrow x=142\) vậy \(x=142\)
b) ta có : \(12-\left(39-3x\right)=0\Leftrightarrow12-39+3x=0\Leftrightarrow3x-27=0\)
\(\Leftrightarrow3x=27\Leftrightarrow x=\dfrac{27}{3}=9\) vậy \(x=9\)
c) ta có : \(\left(14-3x\right)+\left(6+x\right)=0\Leftrightarrow14-3x+6+x=0\)
\(\Leftrightarrow20-2x=0\Leftrightarrow2x=20\Leftrightarrow x=\dfrac{20}{2}=10\) vậy \(x=10\)
a) 3x + 12 = 2x - 4
=> 3x - 2x = -4 - 12
=> 1x = -16
=> x = -16
vậy___
b) 14 - 3x = -x + 4
=> 14-4 = -x+3x
=> 10 = 2x
=> x = 10 : 2
=> x = 5
vậy_____
\(c) ( 2x - 8 ) ( x + 6 ) = 0\)
\(\Rightarrow\orbr{\begin{cases}2x-8=0\\x+6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-6\end{cases}}\)
vậy_____