Tìm x, biết:
\(6^{x+1}\)\(:12=3\)
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Đáp án là B vì 12: -3 = -4; 12: -4 = -3; 12: -6 = -2;12: -12 = -1 và đáp ứng điều kiện a< -2
\(\left(x-2\right)^3+6\left(x+1\right)^2-x^3+12=0\)
\(\Leftrightarrow x^3-6x^2+12x-8+6x^2+12x+6-x^3+12=0\)
\(\Leftrightarrow24x+10=0\)
\(\Leftrightarrow x=-\dfrac{5}{12}\)
1/ `|x|=10<=> x=\pm 10`
2/ `|x-8|=0<=>x-8=0<=>x=8`
3/ `7+|x|=12<=>|x|=5<=>x=\pm 5`
4/ `|x+1|=3`
$\Leftrightarrow\left[\begin{array}{1}x+1=3\\x+1=-3\end{array}\right.\\\Leftrightarrow\left[\begin{array}{1}x=3\\x=-4\end{array}\right.$
5/ `15-x=16-(14-42)`
`<=>15-x=16+28`
`<=>15-x=44`
`<=>x=-29`
6/ `210-(x-12)=168`
`<=>210-x+12=168`
`<=>222-x=168`
`<=>x=54`
\(\frac{3}{\frac{1}{4}}x-\frac{1}{\frac{2}{3}}=\frac{5}{12}-\frac{7}{6}x\)
\(3:\frac{1}{4}x-1:\frac{2}{3}=\frac{5}{12}-\frac{7}{6}x\)
\(1x-1.5=\frac{5}{12}-\frac{7}{6}x\)
\(1x+\frac{7}{6}x=1.5+\frac{5}{12}\)
\(\left(1+\frac{7}{6}\right)x=\frac{18}{12}+\frac{5}{12}=\frac{23}{12}\)
\(\frac{13}{6}x=\frac{23}{12}\)
\(x=\frac{23}{12}:\frac{13}{6}\)
\(x=\frac{23}{12}\cdot\frac{6}{13}=\frac{23}{26}\)
ai ko biết thì đừng milk
\(a,\dfrac{11}{12}x+\dfrac{3}{4}=-\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{11}{12}x=-\dfrac{1}{6}-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{11}{12}x=-\dfrac{11}{12}\)
\(\Leftrightarrow x=-\dfrac{11}{12}:\dfrac{11}{12}\)
\(\Leftrightarrow x=-\dfrac{11}{12}.\dfrac{12}{11}\)
\(\Leftrightarrow x=-1\)
\(b,3-\left(\dfrac{1}{6}-x\right).\dfrac{2}{3}=\dfrac{2}{3}\)
\(\Leftrightarrow3-\dfrac{2}{3}.\left(\dfrac{1}{6}-x\right)=\dfrac{2}{3}\)
\(\Leftrightarrow3-\dfrac{1}{9}+\dfrac{2}{3}x=\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{2}{3}x=\dfrac{2}{3}-3+\dfrac{1}{9}\)
\(\Leftrightarrow\dfrac{2}{3}x=-\dfrac{20}{9}\)
\(\Leftrightarrow x=-\dfrac{20}{9}:\dfrac{2}{3}\)
\(\Leftrightarrow x=-\dfrac{10}{3}\)
3x + 9 = 72 - 12x + 12
3x + 12x = 72 + 12 - 9
15x = 75
x = 75 : 15
x = 5
\(3\left(x+3\right)=12\left(6-x\right)+12\)
\(3x+9=72-12x+12\)
\(3x+12x=72+12-9\)
\(15x=75\)
\(x=5\)
=.= hok tốt !!
a) Ta có: \(x+\dfrac{1}{3}=\dfrac{2}{6}\)
\(\Leftrightarrow x+\dfrac{1}{3}=\dfrac{1}{3}\)
hay x=0
Vậy: x=0
b) Ta có: \(x-\dfrac{1}{4}=\dfrac{1}{-2}\)
\(\Leftrightarrow x-\dfrac{1}{4}=\dfrac{-1}{2}\)
\(\Leftrightarrow x=\dfrac{-1}{2}+\dfrac{1}{4}=\dfrac{-2}{4}+\dfrac{1}{4}=\dfrac{-1}{4}\)
Vậy: \(x=-\dfrac{1}{4}\)
c) Ta có: \(\dfrac{-1}{6}=\dfrac{3}{2}x\)
\(\Leftrightarrow x=\dfrac{-1}{6}:\dfrac{3}{2}=\dfrac{-1}{6}\cdot\dfrac{2}{3}\)
hay \(x=\dfrac{-1}{9}\)
Vậy: \(x=\dfrac{-1}{9}\)
-15/12. x+3/7=6/5.x-1/2
=> -15/12 . x +3/7+1/2 = 6/5 .x
=> -15/12 .x + 13/14= 6/5 .x
=> 49/20 .x = 13/14 => x= 130/343
có:
6x+1 = 12:3
6x+1 = 4
x +1 =22
=> x+1 =2
x=2-1
x=1
VAY x=1
6x+1 =3.12
6x+1=36
6x+1=62
=>x+1=2
x=2-1
x=1