( 1 + 3x ) 3 = 125
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4:
=>(2x+3,5)=7/12*3/14=21/168=1/8
=>2x=1/8-7/2=1/8-28/8=-27/8
=>x=-27/16
5: =>1/3:3x=-21/4
=>3x=-1/3:21/4=-1/3*4/21=-4/63
=>x=-4/189
6: =>2+7/9-3/4(x+1)=7/9
=>2-3/4(x+1)=0
=>3/4(x+1)=2
=>x+1=2:3/4=2*4/3=8/3
=>x=5/3
\(a,\frac{1}{3}x+0.25=\frac{5}{7}\)
\(\Leftrightarrow\frac{1}{3}x=\frac{13}{28}\)
\(\Leftrightarrow x=\frac{39}{28}\)
vậy...
\(b,\frac{11}{12}x+0,25=\frac{5}{6}\)
\(\Leftrightarrow\frac{11}{12}x=\frac{7}{12}\)
\(\Leftrightarrow x=\frac{7}{11}\)
vậy.....
\(c,\left(\frac{-1}{3}\right)^2+\frac{2}{3}x=\frac{1}{4}\)
\(\Leftrightarrow\frac{1}{9}+\frac{2}{3}x=\frac{1}{4}\)
\(\Leftrightarrow\frac{2}{3}x=\frac{5}{36}\)
\(\Leftrightarrow x=\frac{5}{24}\)
vậy......
\(d,\left(3x+2\right)^3=-\frac{8}{125}\)
\(\Leftrightarrow3x+2=-\frac{2}{5}\)
\(\Leftrightarrow3x=-\frac{12}{5}\)
\(\Leftrightarrow x=-\frac{4}{5}\)
vậy.......
\(\frac{1}{3x}+0,25=\frac{5}{7}\)
\(\frac{1}{3x}+\frac{1}{4}=\frac{5}{7}\)
\(\frac{1}{3x}=\frac{13}{28}\)
\(3x=\frac{28}{13}\)
\(x=\frac{28}{39}\)
\(\frac{11}{12x}+0,25=\frac{5}{6}\)
\(\frac{11}{12x}+\frac{1}{4}=\frac{5}{6}\)
\(\frac{11}{12x}=\frac{7}{12}\)
\(x=\frac{11}{12}:\frac{7}{12}\)
\(x=\frac{7}{11}\)
\(\left(-\frac{1}{3}\right)^2+\frac{2}{3x}=\frac{1}{4}\)
\(\frac{1}{9}+\frac{2}{3x}=\frac{1}{4}\)
\(\frac{2}{3x}=\frac{5}{36}\)
\(x=\frac{2}{3}:\frac{5}{36}\)
\(x=\frac{5}{24}\)
\(\left(3x+2\right)^3=\left(-\frac{8}{125}\right)\)
\(\left(3x+2\right)^3=\left(-\frac{2}{5}\right)^3\)
\(\Rightarrow3x+2=-\frac{2}{3}\)
\(3x=-\frac{8}{3}\)
\(x=-\frac{9}{8}\)
a) \(\left(2x-1\right)^5=243\)
\(\left(2x-1\right)^5=3^5\)
\(\Rightarrow2x-1=3\)
\(2x=3+1\)
\(2x=4\)
\(x=\frac{4}{2}\)
\(x=2\)
Vậy .......
b) \(\left(3x+2\right)^3=125\)
\(\left(3x+2\right)^3=5^3\)
\(\Rightarrow3x+2=5\)
\(3x=5-2\)
\(3x=3\)
\(x=\frac{3}{3}=1\)
Vậy ..........
a) \(\left(2x+1\right)^3=125\)
\(\left(2x+1\right)^3=5^3\)
\(2x+1=5\)
\(2x=4\)
\(x=2\)
b) \(3^x+25=26\times2^2+2\times3^0\)
\(3^x+25=26\times4+2\times1\)
\(3^x+25=106\)
\(3^x=106-25\)
\(3^x=81\)
\(3^x=3^4\)
\(x=4\)
(2x+1)3 = 125
a)<=> (2x+1)3 = 53
<=> 2x+1 = 5
<=> 2x = 4
<=> x = 2
3^x+25=26 . 2^2 + 2. 3^0
b)3^x+25=104 +2
3^x+25=106
3^x=106+25
3^x=81=3^4
=> x=4
ta có : \(\left(3x-1\right)^3=125\Leftrightarrow\left(3x-1\right)^3=5^3\Rightarrow3x-1=5\)
\(\Leftrightarrow3x=5+1=6\Leftrightarrow x=\dfrac{6}{3}=2\) vậy \(x=2\)
ta có : \(x^{2010}=x\Leftrightarrow x^{2010}-x=0\Leftrightarrow x\left(x^{2009}-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x^{2009}-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x^{2009}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=1\end{matrix}\right.\) vậy \(x=0;x=1\)
Ta có: \(\left(1+3x\right)^3=125\)
\(\Leftrightarrow1+3x=\sqrt[3]{125}=5\)
\(\Leftrightarrow3x=4\)
\(\Leftrightarrow x=\dfrac{4}{3}\)
Vậy: \(x=\dfrac{4}{3}\)