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\(\frac{2x+2}{2x-4}=\frac{2\left(x+1\right)}{2\left(x-2\right)}=\frac{x+1}{x-2}=\frac{\left(x-2\right)+3}{x-2}=1+\frac{3}{x-2}\)
Để \(2x+2⋮2x-4\Leftrightarrow1+\frac{3}{x-2}\in Z\)
\(\Rightarrow x-2\inƯ\left(3\right)\) = { - 3; - 1; 1; 3 }
=> x = { - 1; 1 ; 3; 5 }
2 ( x - 3 ) - 3 ( 2 - 3x ) = 4 [ ( 1 - 2x ) + 15 ]
2x - 6 - 6 + 9x = 4 [ 1 - 2 x + 15 ]
2x - 6 -6 + 9x = 4 - 8x + 60
2x + 9x + 8x = 4 + 60 + 6 + 6
19x = 76
=> x = 76 : 19
=> x = 4
Vậy x = 4
\(2\left(x-3\right)-3\left(2-3x\right)=4\left[\left(1-2x\right)+15\right]\)
\(\Rightarrow2x-6-6+9x=4\left[1-2x+15\right]\)
\(\Rightarrow2x-6-6+9x=4-8x+60\)
\(\Rightarrow2x+9x+8x=4+60+6+6\)
\(\Rightarrow19x=76\)
\(\Rightarrow x=76:19=4\)
Vậy x = 4
Ko cần đâu bn à mk mong bn đấy
a)\(\left(3x-1\right)\left(5-\frac{1}{2}x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-1=0\\5-\frac{1}{2}x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)
b)\(2\left|\frac{1}{2}x-\frac{1}{3}\right|-\frac{3}{2}=\frac{1}{4}\)
\(2\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{4}\)
\(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{8}\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\\frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{29}{12}\\x=-\frac{13}{12}\end{cases}}\)
a)\(\left(3x-1\right)\left(\frac{-1}{2}x+5\right)=0\)
\(\Leftrightarrow\)3x - 1 = 0 hay \(\frac{-1}{2}\)x + 5 = 0
\(\Leftrightarrow\)3x = 1 I\(\Leftrightarrow\)\(\frac{-1}{2}\)x = -5
\(\Leftrightarrow\) x = \(\frac{1}{3}\) I\(\Leftrightarrow\) x = 10
b) 2 I \(\frac{1}{2}x-\frac{1}{3}\)I - \(\frac{3}{2}\)=\(\frac{1}{4}\)
\(\Leftrightarrow\) 2 I\(\frac{1}{2}x-\frac{1}{3}\)I = \(\frac{7}{4}\)
\(\Leftrightarrow\) I\(\frac{1}{2}x-\frac{1}{3}\)I = \(\frac{7}{8}\)
\(\Leftrightarrow\)\(\frac{1}{2}x-\frac{1}{3}\)= \(\frac{7}{8}\) hay \(\frac{1}{2}x-\frac{1}{3}\)= \(\frac{-7}{8}\)
\(\Leftrightarrow\)\(\frac{1}{2}x\) = \(\frac{29}{24}\) I\(\Leftrightarrow\)\(\frac{1}{2}x\) = \(\frac{-13}{24}\)
\(\Leftrightarrow\) x = \(\frac{29}{12}\) I\(\Leftrightarrow\) x = \(\frac{-13}{12}\)
c) (2x +\(\frac{3}{5}\))2 - \(\frac{9}{25}\)= 0
\(\Leftrightarrow\)(2x +\(\frac{3}{5}\))2 = \(\frac{9}{25}\)
\(\Leftrightarrow\) 2x +\(\frac{3}{5}\) = \(\frac{3}{5}\) hay 2x +\(\frac{3}{5}\)= \(\frac{-3}{5}\)
\(\Leftrightarrow\) 2x = 0 I \(\Leftrightarrow\)2x = \(\frac{-6}{5}\)
\(\Leftrightarrow\) x = 0 I \(\Leftrightarrow\) x = \(\frac{-3}{5}\)
d) 3(x -\(\frac{1}{2}\)) - 5(x +\(\frac{3}{5}\)) = -x + \(\frac{1}{5}\)
\(\Leftrightarrow\)3x - \(\frac{3}{2}\)- 5x - 3 = -x + \(\frac{1}{5}\)
\(\Leftrightarrow\)-2x + x - \(\frac{9}{2}\)- \(\frac{1}{5}\)= 0
\(\Leftrightarrow\)-x = \(\frac{-47}{10}\)
\(\Leftrightarrow\) x = \(\frac{47}{10}\)
\(Tacó:3^{2x+2}=9^{x+3}\)
\(\Rightarrow3^{2x+2}=\left(3^2\right)^{x+3}\)
\(\Rightarrow3^{2x+2}=3^{2x+6}\)
\(\Rightarrow2x+2=2x+6\)
\(\Rightarrow Vôlý\)
Vậy không có giá trị nào của x thỏa mãn
(2x+3) . 5^2=5^4
2x+3 = 5^4 : 5^2
2x+3 = 5^2
2x+3 =25
2x = 25-3
2x=22
x=22:2
vậy x = 11
a) \(\left(2x+3\right)\cdot5^2=5^4\)
\(2x+3=5^2=25\)
\(2x=22\)
\(x=11\)
b) \(\left(32-2x\right)\cdot2=2^3\)
\(32-2x=2^2=4\)
\(2x=28\)
\(x=14\)