Tìm x, biết: 1/3 : (2x - 1) = -4/24
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1: =>\(5^{x-2}-9=2^4-\left(6^2-6^2\right)\)
=>\(5^{x-2}=16+9=25\)
=>x-2=2
=>x=4
2: \(\Leftrightarrow3^x+16=19^6:19^5-3=19-3=16\)
=>3^x=0
=>x=0
3: \(\Leftrightarrow2^x+2^x\cdot16=272\)
=>2^x*17=272
=>2^x=16
=>x=4
4: \(\Leftrightarrow2^{x-1}+3=24-\left(4^2-2^2+1\right)=24-\left(16-4+1\right)\)
=>\(2^{x-1}+3=24-16+4-1=8+4-1=12-1=11\)
=>2^x-1=8
=>x-1=3
=>x=4
Ta có:
3(2x-1)-5(x-3)+6(3x-4)=24
6x-3-5x+15+18x-24=24
19x-12=24
19x=36
x=36/19
a) 4(2x+7)-3(3x-2)=24
8x+28-9x-6=24
(8x-9x)+(28-6)=24
(-1)x+22=24
(-1)x=24-22=2
x=2:(-1)=-2
b) -2(x+3)+(-4)2=3(1-x)
(-2)x+(-6)+16=3-3x
(-2)x+(-6)+16+3x=3
[(-2)x+3x]+[(-6)+16]=3
x+10=3
x=3-10=-7
không ai trả lời
a,\(2\left(3x-1\right)-5\left(x-3\right)-9\left(2x-4\right)=24\)
\(< =>6x-2-5x+15-18x+36=24\)
\(< =>-29x+49=24< =>29x=25< =>x=\frac{25}{29}\)
b,\(2x^2+4\left(x^2-1\right)=2x\left(3x+1\right)\)
\(< =>2x^2+4x^2-4=6x^2+2x\)
\(< =>2x=-4< =>x=-\frac{4}{2}=-2\)
c, \(2x\left(5-3x\right)+2x\left(3x-5\right)-3\left(x-7\right)=4\)
\(< =>10x-6x^2+6x^2-10x-3x+21=4\)
\(< =>-3x=4-21=-17< =>x=\frac{17}{3}\)
d, \(5x\left(x+1\right)-4x\left(x+2\right)=1-x\)
\(< =>5x^2+5x-4x^2-8x=1-x\)
\(< =>x^2-3x+x-1=0\)
\(< =>x^2-2x-1=0\)
\(< =>\left(x-1\right)^2=2\)
\(< =>\orbr{\begin{cases}x-1=\sqrt{2}\\x-1=-\sqrt{2}\end{cases}}\)
\(< =>\orbr{\begin{cases}x=1+\sqrt{2}\\x=1-\sqrt{2}\end{cases}}\)
\(1-\left(\frac{27}{8}+a-\frac{129}{24}\right):\frac{23}{3}=0\)
\(1-\left(\frac{81+24a-129}{24}\right):\frac{23}{3}\)=0
\(1-\left(\frac{24a-48}{24}\right):\frac{23}{3}=0\)
\(1-\left(a-2\right)\frac{3}{23}=0\\ \frac{23}{23}-\frac{3a-6}{23}=0\\ \frac{23-3a+6}{23}=0\\ 29-3a=0\\ a=\frac{29}{3}\)
a/ (x-3)2 - 4 = 0
=> (x-3-2)(x-3+2)=0
=> (x-5)(x-1)=0
=> x = 5 hoặc x=1
`1/3 : (2x - 1) = (-4/24)`
`2x - 1 = 1/3 : (-4/24)`
`2x - 1 = -2`
`2x = (-2) + 1`
`2x = (-1)`
`x = 2 : (-1)`
`x = -2`
\(\dfrac{1}{3}:\left(2x-1\right)=\dfrac{-4}{24}\)
\(2x-1=\dfrac{1}{3}:\left(\dfrac{-4}{24}\right)\)
\(2x-1=-2\)
\(2x=\left(-2\right)+1\)
\(2x=-1\)
\(x=2:\left(-1\right)\)
\(x=-2\)
vậy x =...