13*(x-3)-26*(x-3)=0
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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/ \(x.x-8=-2.\left(-13\right)-\left(-2\right)\)
\(\Rightarrow x^2=36\) \(\Rightarrow\left|x\right|=6\) \(\Rightarrow x=\pm6\)
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 ...
-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
<=>\(\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\) <=>\(\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
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
a) x - 26 : 13 = 2017
x - 2 = 2017
x = 2017 + 2
x = 2019
b) \(\frac{2}{3}+\frac{1}{3}.\left(x+28\right)=18\)
\(\frac{1}{3}.\left(x+28\right)=18-\frac{2}{3}\)
\(\frac{1}{3}.\left(x+28\right)=\frac{54}{3}-\frac{2}{3}\)
\(\frac{1}{3}.\left(x+28\right)=\frac{52}{3}\)
\(x+28=\frac{52}{3}:\frac{1}{3}\)
\(x+28=52\)
\(x=52-28\)
\(x=24\)
c) \(\frac{7}{8}.\left(x-27\right)=\frac{9}{8}-0,125\)
\(\frac{7}{8}.\left(x-27\right)=1\)
\(x-27=1:\frac{7}{8}\)
\(x-27=\frac{8}{7}\)
\(x=\frac{8}{7}+27\)
\(x=\frac{8}{7}+\frac{189}{7}\)
\(x=\frac{197}{7}\)
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
\(\Rightarrow\)\(\orbr{\begin{cases}x+19=39\\x+19=-39\end{cases}}\)
\(\Rightarrow\)\(\orbr{\begin{cases}x=20\\x=-58\end{cases}}\)
Ta có: \(0,\left(3\right)+\frac{31}{3}+0,4\left(2\right)=\frac{3}{9}+\frac{31}{3}+\frac{42-4}{90}=\frac{1}{3}+\frac{31}{3}+\frac{19}{45}=\frac{32}{3}+\frac{19}{45}=\frac{499}{45}.\)
\(\frac{4}{9}+0,\left(13\right)=\frac{4}{9}+\frac{13}{99}=\frac{44}{99}+\frac{13}{99}=\frac{57}{99}=\frac{19}{33}\)
\(0,\left(37\right).x\Rightarrow\frac{37}{99}.x=1\)
\(\Rightarrow x=1:\frac{37}{99}=\frac{99}{37}\)
\(0,\left(26\right).x=1,2\left(31\right)\)
\(\Rightarrow\frac{26}{99}.x=\frac{1219}{990}\)
\(\Rightarrow x=\frac{1219}{990}:\frac{26}{99}=\frac{1219}{260}\)
13 * ( x - 3 ) - 26 * ( x - 3 ) = 0
-13 * ( x - 3 ) = 0
x - 3 = 0 : -13
x - 3 = 0
x = 0 + 3
x = 3
\(13\left(x-3\right)-26\left(x-3\right)=0\)
\(\left(13-26\right)\left(x-3\right)=0\)
\(-13\left(x-3\right)=0\)
\(x-3=0\)
\(x=3\)