3x + 12 =2x -10
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(2x + 10)(3x - 12) = 0
2(x + 5) . 3(x - 4) = 0
6(x + 5)(x - 4) = 0
=> (x + 5)(x - 4) = 0
=> x + 5 =0 hoặc x - 4 = 0
=> x = -5 hoặc x = 4
Vậy x E {-5; 4}
\(\left(2x+10\right)\left(3x-12\right)=0.\Leftrightarrow\left[{}\begin{matrix}2x+10=0.\\3x-12=0.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-5.\\x=4.\end{matrix}\right.\)
Vậy \(x\in\left\{-5;4\right\}.\)
a) 3x - 10 = 2x + 13
<=> 3x - 2x = 13+10
<=> x = 23
Vậy x = 23
b) x + 12 = -5 - x
<=> x + x = -5 - 12
<=> 2x = -17
<=> x = -17 : 2
<=> x = -8,5
Vậy x = -8,5
c) x + 5 = 10 - x
<=> x + x = 10 - 5
<=> 2x = 5
<=> x = 5 : 2
<=> x = 2,5
Vậy x = 2,5
d) 6x + 2 ^ 3 = 2x - 12
<=> 6x + 8 = 2x - 12
<=> 6x - 2x = -12 - 8
<=> 4x = -20
<=> x = -20 : 4
<=> x = -5
Vậy x = -5
e) 12 - x = x + 1
<=> -x - x = 1 - 12
<=> -2x = -11
<=> x = -11 : (-2)
<=> x = 5,5
Vậy x= 5,5
f) 14 - 4x = 3x + 20
<=> -4x - 3x = 20 - 14
<=> -7x = 6
<=> x = 6 : (-7)
<=> x = \(-\frac{6}{7}\)
a)(x+2).(x+3)-(x-2).(x+5)=10
( x^2 +3x+2x+6)-(x^2 +5x-2x-10)=10
x^2 +3x+2x+6-x^2 -5x+2x+10-10=0
2x+6=0
2x=-6
x=-3
a.
\(\left|5x\right|=3x+8\Leftrightarrow\left[{}\begin{matrix}-5x=3x+8\\5x=3x+8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\)
b.
\(\left|-4x\right|=-2x+11\Leftrightarrow\left[{}\begin{matrix}-4x=-2x+11\\4x=-2x+11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=\dfrac{11}{6}\end{matrix}\right.\)
c.
\(\left|3x-1\right|=4x+1\Leftrightarrow\left[{}\begin{matrix}-3x+1=4x+1\\3x-1=4x+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
d.
\(\left|3-2x\right|=3x-7\Leftrightarrow\left[{}\begin{matrix}-3+2x=3x-7\\3-2x=3x-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
e.
\(9-\left|-5x\right|+2x=0\Leftrightarrow\left[{}\begin{matrix}9-5x+2x=0\\9+5x+2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{9}{7}\end{matrix}\right.\)
f.
\(\left(x+1\right)^2+\left|x+10\right|-x^2-12=0\Leftrightarrow\left[{}\begin{matrix}x^2+2x+1-x-10-x^2-12=0\\x^2+2x+1+x+10-x^2-12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=21\\x=\dfrac{1}{3}\end{matrix}\right.\)
a) \(3x-10=2x+13\Leftrightarrow x=13+10=23\)
b) \(x+12=-5-x\Leftrightarrow2x=-17\Leftrightarrow x=-\frac{17}{2}\)
c) \(x+5=10-x\Leftrightarrow2x=5\Leftrightarrow x=\frac{5}{2}\)
d) \(6x+23=2x-12\Leftrightarrow4x=-35\Leftrightarrow x=-\frac{35}{4}\)
e) \(12-x=x+1\Leftrightarrow2x=11\Leftrightarrow x=\frac{11}{2}\)
f) \(14+4x=3x+20\Leftrightarrow x=20-14=6\)
g) \(2\left(x-1\right)+3\left(x-2\right)=x-4\Leftrightarrow2x-2+3x-6=x-4\)
\(\Leftrightarrow5x-8=x-4\Leftrightarrow4x=4\Leftrightarrow x=1\)
h ) \(3\left(4-x\right)-2\left(x-1\right)=x+20\Leftrightarrow12-3x-2x+2=x+20\)
\(\Leftrightarrow14-5x=x+20\Leftrightarrow6x=-6\Leftrightarrow x=-1\)
i ) \(4\left(2x+7\right)-3\left(3x-2\right)=24\Leftrightarrow8x+28-9x+6=24\)
\(\Leftrightarrow34-x=24\Leftrightarrow x=34-24=10\)
k) \(3\left(x-2\right)+2x=10\Leftrightarrow3x-6+2x=5x-6=10\)
\(\Leftrightarrow5x=10+6=16\Leftrightarrow x=\frac{16}{5}\)
Tự KL cho các phần
$3x+12=2x-10$
$\Rightarrow 3x-2x=-10-12$
$\Rightarrow x=-22$
x=-22