34-x =1/27
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\(\left(x+2\right)-34=27-\left(2x+1\right)\)
\(\Rightarrow x+2-34=27-2x-1\)
\(\Rightarrow x-32=26-2x\)
\(\Rightarrow x+2x=26+32\)
\(\Rightarrow3x=58\)
\(\Rightarrow x=58\div3\)
\(\Rightarrow x=19,\left(3\right)\)
X +2 -34 =27 -2X -1
X -32 = 26 -2X
X -32 -26 +2X =0
3X-58=0
a ) x − 2 5 : 2 3 = 3 4 x − 2 5 = 3 4 . 2 3 x − 2 5 = 1 2 x = 9 10
b ) x + − 3 4 − 1 2 = 2 3 . 3 5 x + − 3 4 − 1 2 = 2 5 x + − 3 4 = 9 10 x = 33 20
c ) x − 2 7 = − 9 14 . − 4 3 x − 2 7 = 6 7 +) x − 2 7 = 6 7 x = 8 7 +) x − 2 7 = − 6 7 x = − 4 7
1: Tìm x
a) Ta có: \(2\cdot3^x=3^{12}\cdot34+20\cdot27^4\)
\(\Leftrightarrow2\cdot3^x=3^{12}\cdot34+20\cdot3^{12}\)
\(\Leftrightarrow2\cdot3^x=3^{12}\cdot\left(34+20\right)\)
\(\Leftrightarrow2\cdot3^x-3^{12}\cdot54=0\)
\(\Leftrightarrow2\cdot3^x=3^{12}\cdot2\cdot27\)
\(\Leftrightarrow3^x=3^{12}\cdot3^3\)
\(\Leftrightarrow3^x=3^{15}\)
hay x=15
Vậy: x=15
b) Ta có: \(\left(2^x+1\right)^2+3\left(2^2+1\right)=2^2\cdot10\)
\(\Leftrightarrow\left(2^x+1\right)^2=40-3\cdot5=25\)
\(\Leftrightarrow\left[{}\begin{matrix}2^x+1=5\\2^x+1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2^x=4\\2^x=-6\left(loại\right)\end{matrix}\right.\Leftrightarrow x=2\)
Vậy: x=2
b)-3.!x-3!=-27
!x-1!=-27/-3
!x-1!=9
x-1=+-9
TH1 TH2
x-1=9 x-1=-9
x=9+1 x=-9+1
x=10 x=-8
a, = 34.16 - 34.44 + 44.34 - 44.16
= 16.(34 - 44 ) + 44 .( 34 - 34 )
= 16.(-10) + 44.0
= -160
\(\Leftrightarrow3^{4-x}=3^{-3}\Leftrightarrow4-x=-3\Leftrightarrow x=7\)
thank you