Tìm x
|x-1|=0
-13.|x|=-26
tìm x, biết:
3x+17=12
|x-1|=0
-13.|x|=-26
3x+17=12
=>3x=12-17
=>3x=-5
=>x=-5:3
=>x=-1,66
/x-1/=0
=>x-1=0
=>x =0+1
=>x =1
-13./x/=-26
=>/x/=-26:(-13)
=>/x/=2
=>x=-2 hoặc x=2
Tìm x
I x - 1 I = 0
-13 . I x I = - 26
|x-1|=0
\(\Rightarrow\)x-1=0
x=0+1
x=1
Vậy x=1.
-13.|x|=26
|x|=-2
Mà |x|\(\ge\)0
\(\Rightarrow x\in\varnothing\)
Vậy \(x\in\varnothing\)
Chúc bạn học tốt!
#Huyền#
a, | x + 1 | = 0
=> x + 1 = 0
=> x = - 1
Vậy x = - 1
b, - 13 . | x | = - 26
=> | x| = 2
\(\Rightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)
Vậy \(x\in\left\{2;-2\right\}\)
@@ Học toost
a) I x - 1 I=0 => x = 1
b) -13. I x I = -26
<=> IxI= -26 : (-13)=2
=> x= -2 hoặc x = 2
vậy x= -2 hoặc x=2
a,x.(x+7) = 0
b,(-x+ 5 ) . (3 - x)=0
c,|x - 1|= 3
d,-13 . |x| = -26
e,x . x - 8 = -2 . (-13) - (-2)
a/ \(x\left(x+7\right)=0\) \(\Rightarrow\left[\begin{matrix}x=0\\x+7=0\end{matrix}\right.\) \(\Rightarrow\left[\begin{matrix}x=0\\x=-7\end{matrix}\right.\)
b/ \(\left(-x+5\right)\left(3-x\right)=0\) \(\Rightarrow\left[\begin{matrix}-x+5=0\\3-x=0\end{matrix}\right.\) \(\Rightarrow\left[\begin{matrix}x=5\\x=3\end{matrix}\right.\)
c/ \(\left|x-1\right|=3\) \(\Rightarrow\left[\begin{matrix}x-1=3\\1-x=3\end{matrix}\right.\) \(\Rightarrow\left[\begin{matrix}x=4\\x=-2\end{matrix}\right.\)
d/ \(-13\left|x\right|=-26\) \(\Rightarrow\left|x\right|=2\) \(\Rightarrow x=\pm2\)
e/ \(x.x-8=-2.\left(-13\right)-\left(-2\right)\)
\(\Rightarrow x^2=36\) \(\Rightarrow\left|x\right|=6\) \(\Rightarrow x=\pm6\)
tìm x :
a. | x-1 | = 0
b. -13 . |x| = -26
c/ -16 +23 +x = -16
ai nhanh mk sẽ tick nha\
a) \(\left|x-1\right|=0\)
<=>\(x-1=0\)<=> \(x=1\)
vậy \(x=1\)
b) \(-13.\left|x\right|=-26\)
<=> \(\left|x\right|=2\)=> \(x=\pm2\)
vậy \(x=\pm2\)
c) \(-16+23+x=-16\)
<=> \(7+x=-16\)
<=>\(x=-23\)
vậy \(x=-23\)
Tìm x
a. 17-{-x-[-x-(-x)]}=16
b. 26-giá trị tuyệt đối x+19= -13
c.(3-x).(x-3)=0
d.(2.x-1)^2=25
a.17-(-x(-x(-x)))=16
17+x-x+x=16
17+x=16
x=16-17
x=-1
b.26-|x+19|=-13
|x+19|=26--13
|x+19|=26+13
|x+19|=39
\(\Rightarrow\)\(\orbr{\begin{cases}x+19=39\\x+19=-39\end{cases}}\)
\(\Rightarrow\)\(\orbr{\begin{cases}x=20\\x=-58\end{cases}}\)
Tìm x
-16 + 23 + x = - 162x – 35 = 153x + 17 = 12│x - 1│= 0-13 .│x│ = -261,-16+23+x=-16
=>23+x=-16-16
=>23+x=-32
=>x=-32-23
=>x=-55
2,2x-35=15
=>2x=15+35
=>2x=50
=>x=25
3,3x+17=12
=>3x=12-17
=>3x=-5
=>ko có x thỏa mãn
4,|x-1|=0
=>x-1=0
=>x=1
5,-13.|x|=-26
=>|x|=-26:-13
=>|x|=2
=>x=2 hoặc -2
1. -16+23+x=-16
23+x=(-16)-(-16)
23+x=(-16)+16
23+x=0
x=0-23
x=-23
2. 2x-35=15
2x=15+35
2x=50
x=50:2
x=25
3. 3x+17=12
3x=12-17
3x=-5
x=(-5):3
x=-5/3
4. /x-1/=0
/x/=0+1
/x/=1
=>x=1 hoặc x=-1
5. -13./x/=-26
/x/=(-26):(-13)
/x/=2
=>x=2 hoặc x=-2
Tìm x biết:
a) x-\(\dfrac{2}{3}\)=\(\dfrac{3}{8}\) b) x-\(\dfrac{3}{4}\)=\(\dfrac{13}{10}\):\(\dfrac{26}{5}\) c) \(\dfrac{3}{2}\)-\(\left(x+\dfrac{1}{2}\right)\)=\(\dfrac{4}{5}\) d) |x-2|-1=0
a) \(x-\dfrac{2}{3}=\dfrac{3}{8}\Rightarrow x=\dfrac{3}{8}+\dfrac{2}{3}=\dfrac{25}{24}\)
b) \(x-\dfrac{3}{4}=\dfrac{13}{10}:\dfrac{26}{5}\Rightarrow x-\dfrac{3}{4}=\dfrac{1}{4}\Rightarrow x=\dfrac{1}{4}+\dfrac{3}{4}=1\)
c) \(\dfrac{3}{2}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\Rightarrow x+\dfrac{1}{2}=\dfrac{3}{2}-\dfrac{4}{5}=\dfrac{7}{10}\)
\(\Rightarrow x=\dfrac{7}{10}-\dfrac{1}{2}=\dfrac{1}{5}\)
d) \(\left|x-2\right|-1=0\Rightarrow\left|x-2\right|=1\)
\(\Rightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
a: Ta có: \(x-\dfrac{2}{3}=\dfrac{3}{8}\)
\(\Leftrightarrow x=\dfrac{3}{8}+\dfrac{2}{3}=\dfrac{9}{24}+\dfrac{16}{24}=\dfrac{25}{24}\)
b: Ta có: \(x-\dfrac{3}{4}=\dfrac{13}{10}:\dfrac{26}{5}\)
\(\Leftrightarrow x-\dfrac{3}{4}=\dfrac{13}{10}\cdot\dfrac{5}{26}=\dfrac{1}{4}\)
hay x=1
Tìm x biết
-16+23+x=-16
2x+35=-15
-13×|x|=-26
|2x-5|=13
(x-3)×(x+2)=0
-16 + 23 + x = -16
7 + x = -16
x = -16 - 7
x = -23
2x + 35 = -15
2x = -15 - 35
2x = -50
x = -25
-13 x |x | = -26
x = -26 : (-13)
x = 2
Vậy x = 2 hoặc -2
|2x - 5| = 13
2x = 13 + 5
2x = 18
x = 9
(x - 3) x (x + 2) = 0
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
2)2x+35=-15
2x =-15-35
2x = -50
x =-25
1)-16+23+x=-16
7+x =-16
x =-16-7
x =-23
3)-13 * lxl =-26
lxl = -26:-13
lxl =2
<=>x=-2 hoac x=2
5) (x-3).(x+2) =0
<=>\(\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\) <=>\(\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
Tìm x biết:
a, (x+1)+(x+2)+...+(x+199)=0
b, (x=30)+(x-29)+(x-28)=11
c, x+(x-1)+(x-2)+...+(x-100)=0
d, 123-3.(x+4)=23
e, [(14x+26).3+55] : 5 =35
f, 720:[41-(2x-5)]= 23.5
g, (2x-4).(48-12x)2=0
h, (3-x).(4-2x).(x+10)=0
i, |x-7|+13=25
k, |x-3|-16=-4
l, 26-|x+9|=-13
Giúp mk nha!
a, ( x + 1 ) + ( x + 2 ) + ... + ( x + 199 ) = 0
x + 1 + x + 2 + ... + x + 199 = 0
( x + x + ... + x ) + ( 1 + 2 + ... + 199 ) = 0
199x + 19900 = 0
199x = 0 - 19900
199x = -19900
x = -19900 : 199
x = -100
Vậy ...
b, ( x - 30 ) + ( x - 29 ) + ( x - 28 ) = 11
x - 30 + x - 29 + x - 28 = 11
( x + x + x ) - ( 30 + 29 + 28 ) = 11
3x - 87 = 11
3x = 11 + 87
3x = 98
x = \(\frac{98}{3}\)
Vậy ...
123 - 3 ( x + 4 ) = 23
3 ( x + 4 ) = 123 - 23
3 ( x + 4 ) = 100
x + 4 = \(\frac{100}{3}\)
x = \(\frac{100}{3}-4\)
x = \(\frac{88}{3}\)
Vậy ...